Commutative Algebra 64

Segre Embedding

Throughout this article, k is a fixed algebraically closed field. We wish to construct the product in the category of quasi-projective varieties.

For our first example, let V\subset \mathbb P^3_k be the projective variety defined by the homogeneous equation T_0 T_3 - T_1 T_2 = 0. We define maps \pi_1, \pi_2 : V\to \mathbb P^1_k as follows

\pi_1 : (t_0 : t_1 : t_2 : t_3) \mapsto \begin{cases} (t_0 : t_2), \text{ if } (t_0, t_2) \ne (0, 0), \\ (t_1 : t_3), \text{ if } (t_1, t_3) \ne (0, 0),\end{cases} \\ \pi_2 : (t_0 : t_1 : t_2 : t_3) \mapsto \begin{cases} (t_0 : t_1), \text{ if } (t_0, t_1) \ne (0, 0), \\ (t_2 : t_3), \text{ if } (t_2, t_3) \ne (0, 0).\end{cases}

Note that the maps are well-defined: if (t_0, t_2), (t_1, t_3) \ne (0, 0) then since t_0 t_3 = t_1 t_2 we have (t_0 : t_2) = (t_1 : t_3).

Proposition 1.

The triplet (V, \pi_1, \pi_2) is a product in the category of quasi-projective varieties.

Proof

Let W\subseteq \mathbb P^n be a quasi-projective variety and \psi_1, \psi_2 : W \to \mathbb P^1 be morphisms. We will define the corresponding f : W \to V as follows. For each \mathbf w\in W, there is an open neighbourhood U of w such that \psi_1|_U = (F_0 : F_1) and \psi_2|_U = (G_0 : G_1) where F_0, F_1 \in k[T_0, \ldots, T_n] are homogeneous of the same degree and either F_0 or F_1 has no zero in U. Same holds for G_0, G_1 \in k[T_0, \ldots, T_n].

Now define f : U \to \mathbb P^3 by (F_0 G_0 : F_0 G_1 : F_1 G_0 : F_1 G_1). Clearly the image of f lies in V so we get a morphism f: U \to V. It is easy to see that \pi_1|_U \circ f = \psi_1|_U and \pi_2|_U \circ f = \psi_2|_U. Repeating this construction over an open cover of W, we obtain our desired f:W \to V. ♦

Using similar techniques, we can show the following.

Proposition 2.

For any m, n \ge 0, the product \mathbb P^n \times \mathbb P^m exists in the category of quasi-projective varieties and is a projective variety.

Specifically, the product is the image of the Segre embedding

\mathbb P^n \times \mathbb P^m \to \mathbb P^{mn + n + m}, \quad (a_0 : \ldots : a_n), (b_0 : \ldots : b_m) \mapsto (a_i b_j)_{0\le i \le n, 0\le j \le m},

where the projective coordinates of \mathbb P^{mn + n + m} are indxed by (i, j) with 0\le i \le n and 0\le j \le m.

We denote the image of this map by \mathbb P^{n, m}.

Exercise A

Prove that \mathbb P^{n, m} is the closed subspace of \mathbb P^{mn + n + m} defined by

T_{ij}T_{kl} - T_{il} T_{kj} over all (i, j), (k, l) \in \{0, \ldots, n\} \times \{0 ,\dots, m \}

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Products of Quasi-Projective Varieties

Proposition 3.

If W_1 \subseteq \mathbb P^n and W_2\subseteq \mathbb P^m are open (resp. closed), so is the image of W_1\times W_2 in \mathbb P^{n,m}.

In particular, the topology on \mathbb P^n \times \mathbb P^m is at least as fine as the product topology.

Proof

It suffices to prove the case where W_1\subseteq \mathbb P^n and W_2\subseteq \mathbb P^m are open. Pick any \mathbf w_1 = (a_0 : \ldots : a_n) \in W_1 and \mathbf w_2 = (b_0 : \ldots : b_m) \in W_2; without loss of generality say a_0, b_0 \ne 0.

Since W_1 is open in \mathbb P^n there exists a homogeneous F \in k[A_0, \ldots, A_n] such that \mathbf w_1 \in D(F) \subseteq W_1. Similarly, there exists a homogeneous G \in k[B_0, \ldots, B_m] such that \mathbf w_2 \in D(G) \subseteq W_2. Then

(\mathbf w_1, \mathbf w_2) \in \overbrace{D(A_0 F) \times D(B_0 G)}^{\subseteq W_1 \times W_2} \stackrel \cong \longrightarrow \overbrace{D(T_{00}F(T_{00}, \ldots, T_{n0})G(T_{00}, \ldots, T_{0m})) \cap V}^{\text{open in } W}

so the image of W_1 \times W_2 in V is open. ♦

warningAs in the product of affine varieties, the topology on \mathbb P^n \times \mathbb P^m is in general strictly finer than the product topology. This is already clear in the case mn = 1, since \mathbb P^1 has the cofinite topology.

Corollary 1.

The product of two projective (resp. quasi-projective) varieties exists and is projective (resp. quasi-projective).

Note

In the following proof, we say a subset of a topological space is locally closed if it is an intersection of an open subset and a closed subset. Thus every quasi-projective variety (resp. quasi-affine variety) is a locally closed subspace of some \mathbb P^n_k (resp. \mathbb A^n_k).

Prove the following properties as a simple exercise:

  • an intersection of two locally closed subsets is locally closed;
  • if Y is a locally closed subset of X and Z is a locally closed subset of Y then Z is a locally closed subset of X;
  • a subset Y of X is locally closed if and only if Y is open in its closure in X.

Proof

If W_1 \subseteq \mathbb P^n and W_2 \subseteq \mathbb P^m are closed (resp. locally closed), so is the image W of W_1 \times W_2 in \mathbb P^{n, m} by proposition 3. The projections \mathbb P^{n,m} \to\mathbb P^n and \mathbb P^{n,m}\to \mathbb P^m then restrict to \pi_1: W \to W_1 and \pi_2 : W\to W_2.

Let  us show that (W, \pi_1, \pi_2) is the product of W_1 and W_2 in the category of quasi-projective varieties.

If X is any quasi-projective variety and \psi_1 : X\to W_1, \psi_2 : X\to W_2 are any morphisms then \psi_1 : X \to \mathbb P^n and \psi_2 : X \to \mathbb P^m induce f : X\to \mathbb P^{n,m}; the image of f lies in W so we obtain an induced X\to W. ♦

Exercise B

1. Let W \subset \mathbb P^2 \times \mathbb P^1 be the set of points ((a_0 : a_1 : a_2), (b_0 : b_1)) satisfying a_0^2 b_0 - a_1 a_2 b_1 = 0. Find a set of homogeneous polynomials in \mathbb P^{2, 1} \subset \mathbb P^5 which define the image of W.

2. More generally prove that a subset V \subseteq \mathbb P^{n, m} is closed if and only if its corresponding subset V' \subseteq \mathbb P^n \times \mathbb P^m is the set of solutions of some bihomogeneous polynomials

F(T_0, \ldots, T_n; U_0, \ldots, U_m) = 0,

i.e. F is homogeneous as a polynomial in T_0, \ldots, T_n as well as U_0, \ldots, U_m

Dimensions

Lemma 1.

For any point \mathbf v in a quasi-projective variety V, there is an open neighbourhood U, \mathbf v \in U \subseteq V, which is affine.

Proof

Suppose V\subseteq \mathbb P^n_k is a locally closed subset. Without loss of generality, \mathbf v \in U_0 so \mathbf v is contained in W := U_0 \cap V, a locally closed subset of \mathbb A^n. Now W is open in \overline W, its closure in \mathbb A^n. By an analogue of proposition 1 here, we can pick a basis of the topological space \overline W in the form of \{D(f) : f\in k[\overline W]\}, where

D(f) = \{ \mathbf w \in \overline W : f(\mathbf w) \ne 0\}.

Thus for some f\in k[\overline W] we have \mathbf v \in D(f)\subseteq W. Now we are done since D(f) is isomorphic to the affine variety with coordinate ring k[\overline W][T]/(T\cdot f - 1). ♦

Exercise C

Prove that if V and W are irreducible quasi-projective varieties, then V\times W is also irreducible. Again, please be reminded that V\times W is not the product topology.

Proposition 4.

if V and W are quasi-projective varieties, then

\dim (V \times W) = \dim V + \dim W.

Proof

Suppose V and W are irreducible; by lemma 1 we can pick open affine subsets U_1 \subseteq V and U_2 \subseteq W. Then U_1 \times U_2 is an open affine subset of the quasi-projective variety V\times W, which is irreducible by exercise C. Now 

\dim (V \times W) = \dim (U_1 \times U_2) = \dim U_1 + \dim U_2 = \dim V + \dim W.

where the first and third equalities follow from proposition 3 here and the second is from proposition 2 here

The general case is left as an exercise (write V and W as unions of irreducible components). ♦

Finally, we consider the dimension of the cone of a projective variety.

Proposition 5.

Let V\subseteq \mathbb P^n be a non-empty closed subset. Then 

\dim (\mathrm{cone} V) = \dim V + 1.

Proof

Without loss of generality, suppose V' := U_0 \cap V \ne \emptyset; by proposition 3 here, \dim V = \dim V' and V’ is a closed subset of \mathbb A^n. Also since cone(V’) is open in cone(V) we have \dim (\mathrm{cone} V) = \dim(\mathrm{cone} V'). Now there is an isomorphism

V' \times (\mathbb A^1 - \{0\}) \longrightarrow \mathrm{cone}(V') - \{\mathbf 0\}, \quad ((1 : t_1 : \ldots : t_n), \lambda) \mapsto (\lambda, \lambda t_1, \ldots, \lambda t_n).

Hence \dim(\mathrm{cone} V') = \dim V' + \dim(\mathbb A^1 - \{0\}) = \dim V' + 1. ♦

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Commutative Algebra 63

Serre’s Criterion for Normality

Throughout this article, fix an algebraically closed field k.

In this section, A denotes a noetherian domain. We will describe Serre’s criterion, which is a necessary and sufficient condtion for A to be normal. In the following section, we will relate the results here to an interesting example from the last article.

Lemma 1.

If A is normal, then for all a \in A - \{0\}, we have

\mathfrak p \in \mathrm{Ass}_A (A/aA) \implies \mathrm{ht} \mathfrak p = 1.

Note

Since A is a domain, it has only one minimal prime: 0. Hence all associated primes of A/aA have height at least 1. Lemma 1 thus says principal ideals of a normal domain have no embedded primes.

Proof

We need to show \mathfrak p \in \mathrm{Ass}_A (A/aA) has height 1. Pick b\in A such that b+aA \in A/aA has annihilator \mathrm{Ann}_A (b+aA) = \mathfrak p, i.e. (aA : bA) = \mathfrak p. Since bA is finitely generated, we localize both sides at \mathfrak p to obtain (aA_{\mathfrak p} : bA_{\mathfrak p}) = \mathfrak m, the unique maximal ideal of B := A_{\mathfrak p}. Thus

b\mathfrak m \subseteq aB \implies \overbrace{(ba^{-1})}^{\in \mathrm{Frac} A}\mathfrak m \subseteq B

and ba^{-1} \not\in B. We claim that ba^{-1} \mathfrak m = B; if not, ba^{-1} \mathfrak m \subseteq \mathfrak m and by the adjugate matrix trick (see proof of proposition 6 here), ba^{-1} is integral over B. This contradicts the fact that B is normal. Hence \mathfrak m is an invertible ideal so B is a dvr, and \mathrm{ht} \mathfrak p = \dim B = 1. ♦

Lemma 2.

Suppose for all \mathfrak p \in \mathrm{Ass}_A (A/aA), a\in A-\{0\}, we have \mathrm{ht} \mathfrak p =1. Then

A = \bigcap_{\mathrm{ht} \mathfrak p = 1} A_{\mathfrak p},

where intersection occurs in \mathrm{Frac} A.

Proof

Let \frac a b \in \text{RHS}, where a, b \in A, b\ne 0; we need to show a \in bA. Write bA = \mathfrak q_1 \cap \ldots \cap \mathfrak q_n for its primary decomposition with associated primes \mathfrak p_i = r(\mathfrak q_i) all of height 1. For each i we have \frac a b \in A_{\mathfrak p_i} \implies a \in bA_{\mathfrak p_i}. But bA_{\mathfrak p_i} = \mathfrak q_i A_{\mathfrak p_i} since all \mathfrak p_i are minimal in V(aA). Thus

a \in \mathfrak q_i A_{\mathfrak p_i} \cap A = \mathfrak q_i

by proposition 2 here and a \in \cap_i \mathfrak q_i = bA. ♦

Exercise A

Let A = k[V] for an irreducible affine variety V, and \mathfrak p \subset A be a prime ideal with corresponding subvariety W = V(\mathfrak p) \subset V. Prove that A_{\mathfrak p} is the set of all rational functions on V which are regular at some point of W.

Note

We already know A = \cap_{\mathfrak m \text{ max.}} A_{\mathfrak m} holds for all domains; geometrically, this means if f:V \to \mathbb A^1 (for irreducible V) is regular at each point, then f can be represented by the same polynomial globally. The condition in lemma 2 is notably stronger; geometrically, it says if f is regular on an open dense subset of every codimension 1 subvariety, then it is regular everywhere.

Theorem (Serre’s Criterion).

A noetherian domain A is normal if and only if the following conditions both hold.

  1. All \mathfrak p \in \mathrm{Ass}_A (A/aA), for a\in A-\{0\}, have height 1.
  2. For each \mathfrak p of height 1, A_{\mathfrak p} is a dvr.

When that happens, A = \cap_{\mathrm{ht} \mathfrak p = 1} A_{\mathfrak p}.

Note

In the context of algebraic geometry, the first condition says “subvarieties cut out by a single equation have no embedded components” while the second says “the set of singular points has codimension at least 2” (this will be elaborated in later articles). Thus normality can fail in two different ways: hidden (embedded) components or too many singular points.

Proof

(⇒) Condition 1 follows from lemma 1; condition 2 follows from proposition 6 here.

(⇐) By lemma 2, condition 1 gives us A = \cap_{\mathrm{ht} \mathfrak p = 1} A_{\mathfrak p}, the final statement. By condition 2, each A_{\mathfrak p} is normal; hence so is A. ♦

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Hartog’s Theorem

Definition.

If V is an affine k-variety, we say V is normal if k[V] is a normal domain.

If W\subseteq V is a non-empty closed subset, corresponding to radical ideal \mathfrak a \subsetneq k[V], the codimension of W in V is \mathrm{ht} \mathfrak a.

As a consequence of the above result, we have:

Proposition 1 (Hartog’s Theorem).

Let V be an affine normal k-variety. If W\subsetneq V is a non-empty closed subset of codimension at least 2, then the inclusion V-W \subseteq V induces an isomorphism of coordinate rings

k[V-W] \cong k[V].

Note

Thus, if a rational function f on V is not regular, its “set of irregularity” has codimension 1.

Proof

Suppose f \in k[V-W], considered as a rational function on V. For each \mathfrak p \subset k[V] of height 1, f is regular on (V-W) \cap V(\mathfrak p) \ne \emptyset, a dense open subset of V(\mathfrak p). Hence by exercise A, f\in A_{\mathfrak p} and by Serre’s criterion, we have

f\in \bigcap_{\mathrm{ht} \mathfrak p = 1} A_{\mathfrak p} = A. ♦

This gives the following generalization of example 7 from the previous article.

Corollary 1.

If V, W are as in proposition 1, then V-W is not an affine variety.

Proof

The inclusion f : V-W \hookrightarrow V induces an isomorphism f^* : k[V] \to k[V-W]. If VW were affine, f would also be an isomorphism, which is absurd since f is not surjective. ♦

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Dimensionality

Finally we wish to look at the dimension of a general quasi-projective variety; first we have a general definition.

Definition.

Let X be a non-empty topological space; we consider all chains

\emptyset \ne Z_0 \subsetneq Z_1 \subsetneq \ldots \subsetneq Z_k \subseteq X

of irreducible closed subsets of X. The length of the chain is k; the Krull dimension of X (denoted \dim X) is the supremum of the lengths of all such chains.

The dimension of a quasi-projective variety is its Krull dimension.

Note that when X = \mathrm{Spec} A, the Krull dimension of X is the Krull dimension of A.

Lemma 3.

If Y is a non-empty subspace of X then \dim Y \le \dim X.

Proof

Let Z_0 \subsetneq \ldots \subsetneq Z_k be a chain of irreducible closed subsets of Y. Taking their closures in X we have

\emptyset\ne\overline Z_0 \subseteq \overline Z_1 \subseteq \ldots \subseteq \overline Z_k \subseteq X.

By proposition 2 here, each \overline Z_i is a closed irreducible subset of X. Furthermore by a general result in point-set topology (see proposition 3 here), the closure of any subset Z\subseteq Y in Y is \overline Z \cap Y. Hence \overline Z_i \cap Y = Z_i so \overline Z_i \ne \overline Z_{i+1} for any i. Thus we get a chain of irreducible closed subsets of X of length k. ♦

To proceed, we need the following correspondence.

Proposition 2.

Let U be a non-empty open subset of a topological space X. Then there is a bijective correspondence:

irred_closed_correspondence

where \overline{C'} is the closure of C’ in X.

Proof

First, \overline{C'} is an irreducible subset of X by proposition 2 here. Next note that C\cap U is a non-empty open subset of C; hence by proposition 1 here C\cap U is irreducible. It remains to show:

\overline{C \cap U} = C, \quad \overline {C'} \cap U = C'.

For the first statement, C\cap U is a non-empty open subset of C so it is dense in C. The second statement holds for any closed subset C’ of U as noted above, so we are done. ♦

In particular, we have:

Corollary 2.

Let U and X be as above; if C’ is an irreducible component of U, then \overline {C'} is an irreducible component of X.

Proof

By the correspondence in proposition 2, since C’ is a maximal irreducible closed subset of U, \overline{C'} cannot be properly contained in any irreducible closed subset of X. ♦

Let us return to results on dimensionality.

Lemma 4.

If (U_i) is an open cover for an irreducible space X, and each U_i \ne \emptyset, then \dim X = \sup \dim U_i.

Proof

(≥) holds by lemma 3. For (≤), suppose Z_0 \subsetneq \ldots \subsetneq Z_k is a chain of closed irreducible subsets of X. Pick an i such that U_i \cap Z_0 \ne\emptyset. By proposition 2 we get a chain of irreducible closed subsets of U_i of length k :

\emptyset \ne U_i \cap Z_0 \subseteq U_i \cap Z_1 \subseteq \ldots \subseteq U_i \cap Z_k.

So \dim U_i \ge k. ♦

Finally, we have:

Proposition 3.

Let V be a quasi-projective variety. If V is irreducible and W is any non-empty open subset of V, then \dim W = \dim V.

In particular, \dim \mathbb P^n_k = \dim \mathbb A^n_k = n.

Proof

If V is affine, this follows from proposition 2 here. For the general case, V is an open subset of some projective V’ so it suffices to assume V is projective, i.e. a closed subset of some \mathbb P^n.

By lemma 4, for some 0\le i \le n we have \dim V = \dim (U_i \cap V). Since U_i \cap V is affine, by what we just showed \dim (U_i \cap V) = \dim (W\cap U_i\cap V). On the other hand \dim (W\cap U_i \cap V) \le \dim W \le \dim V by lemma 3, so equality holds throughout. ♦

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Commutative Algebra 62

Irreducible Subsets of Projective Space

Throughout this article, k is an algebraically closed field.

We wish to consider irreducible closed subsets of \mathbb P^n_k. For that we need the following preliminary result.

Lemma 1.

Let A be a graded ring; a proper homogeneous ideal \mathfrak p \subsetneq A is prime if and only if:

a, b \in A \text{ homogeneous}, ab \in \mathfrak p \implies a \in \mathfrak p \text{ or } b\in\mathfrak p.

Proof

Suppose \mathfrak p is not prime so there exists a, b\in A-\mathfrak p such that ab\in \mathfrak p. Since a\not\in \mathfrak p, among all homogeneous components of a pick a_d of maximum degree such that a_d \not\in\mathfrak p; similarly pick b_e fo b so b_e \not\in\mathfrak p.

The degree-(d+e) component of ab is congruent to a_d b_e mod \mathfrak p. Since \mathfrak p is homogeneous we have a_d b_e \in \mathfrak p, and by the given condition this means a_d \in \mathfrak p or b_e \in \mathfrak p, a contradiction. ♦

Exercise A

Prove that a proper homogeneous ideal \mathfrak q of a graded ring A is primary if and only if

a, b \in A \text{ homogeneous }, ab\in\mathfrak q \implies a \in \mathfrak q \text{ or } b\in r(\mathfrak q).

Proposition 1.

Suppose the closed subset V\subseteq \mathbb P^n_k corresponds to the homogeneous radical ideal \mathfrak a \subseteq B, \mathfrak a \ne B_+, where B = k[T_0, \ldots, T_n].

Then V is irreducible if and only if \mathfrak a is prime.

Proof

(⇒) Suppose V is irreducible; let \mathfrak a = I_0(V). If f, g \in B - I_0(V) are homogeneous, then C := V_0(\mathfrak a + fB) and D := V_0(\mathfrak a + gB) are closed subsets of \mathbb P^n_k properly contained in V. Since V is irreducible the following shows fg\not\in\mathfrak a:

V \supsetneq C \cup D = V_0(\mathfrak a + fB) \cup V_0(\mathfrak a + gB) = V_0((\mathfrak a + fB)(\mathfrak a + gB)) \supseteq V_0(\mathfrak a + fgB).

(⇐) Let V = V_0(\mathfrak p) where \mathfrak p is prime. Let C, D\subseteq V be closed subsets with union V. Now write C = V_0(\mathfrak a) and D = V_0(\mathfrak b) for homogeneous radical ideals \mathfrak a and \mathfrak b. Then V = C\cup D = V_0(\mathfrak a \cap \mathfrak b). Since \mathfrak a\cap \mathfrak b is a homogeneous radical ideal, \mathfrak p = \mathfrak a \cap \mathfrak b. By exercise B here, \mathfrak a = \mathfrak p or \mathfrak b = \mathfrak p. ♦

Corollary 1.

Let V\subseteq \mathbb P^n_k be a non-empty closed subset. Then V is irreducible if and only if \mathrm{cone}(V) is irreducible.

Proof

By proposition 1, V is irreducible if and only if I_0(V) is prime. But I_0(V) = I(\mathrm{cone}(V)) (from an exercise here) so the result follows. ♦

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Quasi-projective Varieties

Recall that a projective variety is a closed subset of some \mathbb P^n_k.

Definition.

A quasi-projective variety is an open subset of a projective variety.

This merely defines it as a set; we need a geometric structure on it.

First, let F_0, \ldots, F_m \in k[T_0, \ldots, T_n] be homogeneous polynomials of the same degree. If \mathbf v \in \mathbb P^n is such that not all F_i(\mathbf v) = 0, then we can define a function on an open subset U of \mathbb P^n containing \mathbf v as follows:

U \longrightarrow \mathbb P^m,\quad \mathbf v' \mapsto ( F_0(\mathbf v') : F_1(\mathbf v') : \ldots : F_m(\mathbf v') ).

The map is well-defined: indeed if F_i(\mathbf v) \ne 0 we can find an open neighbourhood U of \mathbf v such that 0 \not\in F_i(U). Also, if we replace projective coordinates (t_0 : \ldots : t_n) with (\lambda t_0 : \ldots : \lambda t_n), then each F_i(\lambda t_0, \ldots, \lambda t_n) = \lambda^d F_i(t_0, \ldots, t_n) where d = \deg F_i so

\begin{aligned} (F_0(\lambda t_0, \ldots, \lambda t_n) : \ldots : F_m(\lambda t_0, \ldots, \lambda t_n)) &= (\lambda^d F_0(t_0, \ldots, t_n) : \ldots: \lambda^d F_m(t_0, \ldots, t_n)) \\ &= (F_0(t_0, \ldots, t_n) : \ldots : F_m(t_0, \ldots, t_n)).\end{aligned}

We write (F_0 : \ldots : F_m) : U\to \mathbb P^m for the resulting function.

Definition.

Let V \subseteq \mathbb P^n and W \subseteq \mathbb P^m be quasi-projective varieties and \phi : V\to W be a function.

We say \phi is regular at \mathbf v \in V if there is an open neighbourhood U of \mathbf v in V such that

\phi|_U = (F_0 : \ldots : F_m) for some homogeneous F_i \in k[T_0, \ldots, T_n] of the same degree.

We say \phi is regular if it is regular at every \mathbf v\in V, in which case we also say \phi : V\to W is a morphism of quasi-projective varieties.

From the above definitions, we obtain the category of all quasi-projective varieties and morphisms between them.

Example 1

First consider the case where V\subseteq \mathbb A^n and W\subseteq \mathbb A^m are closed subsets.

E.g., let V\subseteq \mathbb A^3. A regular map \phi : V\to \mathbb A^1 in the earlier sense can be expressed as a polynomial f(X, Y, Z), e.g. take f = X^3 - Y^2 + 3Z. Via embeddings \mathbb A^3 \hookrightarrow \mathbb P^3 and \mathbb A^1 \hookrightarrow \mathbb P^1 taking (x, y, z) \mapsto (1:x:y:z) and t \mapsto (1:t) respectively, f can be written in terms of homogeneous coordinates as

(T_0 : T_1 : T_2 : T_3) \mapsto (T_0^3 : T_1^3 - T_2^2 T_0 + 3 T_3 T_0^2)

since it is the homogenization of the map (\frac{T_1}{T_0}, \frac{T_2}{T_0}, \frac{T_3}{T_0}) \mapsto (\frac{T_1}{T_0})^3 - (\frac{T_2}{T_0})^2 + 3(\frac{T_3}{T_0}). This generalizes to an arbitrary regular map of closed subsets \phi : (V\subseteq \mathbb A^n) \to (W \subseteq \mathbb A^m).

Conversely we have:

Lemma 2.

Let \phi :V\to W be regular under the new definition. Then there exist polynomials f_1, \ldots, f_m \in k[X_1, \ldots, X_n] which represent \phi.

Proof

We will prove this for the case where V is irreducible.

For each of 1\le i\le m, let \pi_i : \mathbb A^m \to \mathbb A^1 be projection onto the i-th coordinate. Then \pi_i \circ \phi : V \to \mathbb A^1 is regular under the new definition, and by proposition 2 here (and its preceding discussion) \pi_i \circ \phi can be represented as a polynomial f_i(X_1, \ldots, X_n). Hence we see that

\phi(\mathbf v) = (f_1 (\mathbf v), \ldots, f_m(\mathbf v)) for polynomials f_1, \ldots, f_m \in k[X_1, \ldots, X_n]. ♦

Example 2

Take the map \phi : \mathbb P^1 \to \mathbb P^3 given by

\phi : (T_0 : T_1) \mapsto (T_0^3 : T_0^2 T_1 : T_0 T_1^2 : T_1^3)

Note that the same set of polynomials (F_0, F_1, F_2, F_3) works globally over the whole of \mathbb P^1.

Example 3

Suppose \mathrm{char} k \ne 2. Let V\subset \mathbb P^2 be the closed subset defined by T_0^2 = T_1^2 + T_2^2. We define a map \phi : V \to \mathbb P^1 as follows

  • Outside the point (1 : 1 : 0), take (T_0 : T_1 : T_2) \mapsto (T_0 - T_1 : T_2).
  • Outside the point (1 : -1 : 0), take (T_0 : T_1 : T_2) \mapsto (T_2 : T_0 + T_1).

p1_cover

The map agrees outside those two points since (T_0 - T_1 : T_2) = (T_2 : T_0 + T_1) due to the equality T_0^2 = T_1^2 + T_2^2.

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Isomorphisms

Definition.

Consider the category of all quasi-projective k-varieties, with morphisms defined as above. Two such varieties are said to be isomorphic if they are isomorphic in the category.

A quasi-projective variety is said to be

  • projective if it is isomorphic to a closed subset of some \mathbb P^n_k (this generalizes the existing definition of projective varieties);
  • affine if it is isomorphic to an affine k-variety (closed subspace of some \mathbb A^n);
  • quasi-affine if it is isomorphic to an open subset of an affine variety.

Example 4

In example 3 above, we get an isomorphism \phi : V\to \mathbb P^1 since we have the reverse map

\mathbb \psi : \mathbb P^1 \to V, \quad (U_0 : U_1) \mapsto (U_0^2 + U_1^2 : U_1^2 - U_0^2 : 2U_0 U_1).

As an exercise, prove that \phi\circ \psi = 1_{\mathbb P^1} and \psi\circ \phi = 1_V.

Definition.

The coordinate ring of a quasi-projective variety V is the set

k[V] := \{ f : V\to \mathbb A^1 : f \text{ regular } \},

taken to be a k-algebra via point-wise addition and multiplication:

f, g : V\to\mathbb A^1 \text{ regular } \implies \begin{cases} (f+g) :V \to \mathbb A^1, \ &\mathbf v \mapsto f(\mathbf v) + g(\mathbf v), \\ (fg) : V\to \mathbb A^1, \ &\mathbf v \mapsto f(\mathbf v)g(\mathbf v). \end{cases}

Note

As before, a regular map \phi:V\to W of quasi-projective varieties induces a ring homomorphism \phi^* : k[W] \to k[V]. By lemma 2, when V is affine k[V] agrees with our earlier version (we proved this in the case where V is irreducible).

Example 5

For each g\in GL_{n+1}(k), we have an automorphism 

\phi_g : \mathbb P^n_k \longrightarrow \mathbb P^n_k, \quad (t_0 : \ldots : t_n) \mapsto (\sum_{j=0}^n g_{0j} t_j : \ldots : \sum_{j=0}^n g_{nj} t_j).

Note that \phi_{gh} = \phi_g \circ \phi_h for g, h \in GL_{n+1}(k). Also \phi_g = 1 if and only if g is a scalar multiple of the identity matrix, so we get an injective homomorphism PGL_{n+1}(k) = GL_{n+1}(k)/k^* \hookrightarrow \mathrm{Aut} \mathbb P^n_k. In fact this is an isomorphism of groups.

E.g. when n = 1, we get the Möbius transformations:

\left[\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in GL_2 k \right] : (t_0 : t_1) \mapsto (at_0 + bt_1 : ct_0 + dt_1).

Example 6

We have an isomorphism between the quasi-affine variety \mathbb A^1 - \{0\} and V = \{(x,y) \in \mathbb A^2 : xy = 1\} via the maps

\mathbb A^1 - \{0\} \to V, \ x \mapsto (x, \frac 1 x), \quad V \to \mathbb A^1 - \{0\}, \ (x, y) \mapsto x.

Hence \mathbb A^1-\{0\} is an affine variety even though it is not closed in \mathbb A^1. From the isomorphism we also have:

k[\mathbb A^1 - \{0\}] \cong k[V] = k[X, Y]/(XY - 1) \implies k[\mathbb A^1 - \{0\}] = k[X, \frac 1 X].

Example 7

Let V = \mathbb A^2 -\{(0, 0)\}. We will show that V is not affine. Indeed consider the injection \phi : V \hookrightarrow \mathbb A^2 which induces

\phi^* : k[X, Y] \cong k[\mathbb A^2] \longrightarrow k[V].

The map is injective since V is dense in \mathbb A^2. Let us show that it is surjective. Suppose f \in k[V] so that f: V \to \mathbb A^1 is regular. Write

V = U \cup U', where U = (\mathbb A^1 - \{0\}) \times \mathbb A^1, \ U' = \mathbb A^1 \times (\mathbb A^1 - \{0\}).

By example 6, we have f|_U \in k[U] \cong k[X, Y, \frac 1 X] and f|_{U'} \in k[U']\cong k[X, Y, \frac 1 Y]. Since U, U', U\cap U' are all dense in V we have injections k[V] \to k[U] \to k[U\cap U'] and k[V] \to k[U'] \to k[U\cap U'] so that f \in k[X, Y, \frac 1 X] \cap k[X, Y, \frac 1 Y]. It is easy to show that this means f\in k[X, Y].

Hence \phi induces an isomorphism of the coordinate rings k[\mathbb A^2] \to k[V]. If V is affine, by proposition 1 here \phi would be an isomorphism of varieties, which is a contradiction since \phi is not surjective.

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Commutative Algebra 61

In this article, we will consider algebraic geometry in the projective space. Throughout this article, k denotes an algebraically closed field.

Projective Space

Definition.

Let n\ge 0. On the set \mathbb A^{n+1}_k - \{\mathbf 0\}, we consider the equivalence relation:

v, w \in \mathbb A^{n+1}_k - \{\mathbf 0\} \implies (v \sim w \iff \exists \lambda \in k-\{0\}, w = \lambda v).

The projective n-space \mathbb P^n_k is the set of equivalence classes under this relation. An element of \mathbb P^n_k is denoted by (t_0 : t_1 : \ldots : t_n) for any representative (t_0, \ldots, t_n) \in \mathbb A^{n+1}_k - \{\mathbf 0\}.

For example (5 : 2 : 3) = (\frac 1 6 : \frac 1 {15} : \frac 1 {10}) \in \mathbb P^2_{\mathbb C}.

The above only defines \mathbb P^n_k as a set; we need some geometric structure on it for the concept to be meaningful. Throughout this article we will fix B = k[T_0, \ldots, T_n], which is graded by degree.

Definition.

Suppose F\in B is homogeneous and \mathbf v = (t_0 : t_1 : \ldots : t_n) \in \mathbb P^n_k. We write F(\mathbf v) = 0 if  F(t_0, \ldots, t_n) = 0.

Note that the value F(\mathbf v) is in general not well-defined. Indeed if we have two representatives (t_0 : \ldots : t_n) = (\lambda t_0 : \ldots :\lambda t_n) for \mathbf v, then

F(\lambda t_0, \ldots, \lambda t_n) = \lambda^{\deg f} F(t_0, \ldots, t_n).

Despite this, F(\lambda t_0, \ldots, \lambda t_n) = 0 if and only if F(t_0, \ldots, t_n) = 0 so the definition is sensible. As in the case of the affine n-space, we will define a correspondence between ideals of B and subsets of \mathbb P^n_k.

Definition.

Let \mathfrak a\subseteq B be a graded ideal. Write

V_0(\mathfrak a) := \{ \mathbf v \in \mathbb P^n : F(\mathbf v) = 0 \text{ for all homogeneous } F \in \mathfrak a\}.

For a finite sequence of homogeneous polynomials F_1, \ldots, F_m we also write V_0(F_1, \ldots, F_m) for V_0(\mathfrak a) where \mathfrak a = (F_1, \ldots, F_m).

Also let D_0(F) := \mathbb P^n_k - V_0(F).

Exercise A

Prove the following, for any graded ideals \mathfrak a, \mathfrak b\subseteq B and collection of graded ideals (\mathfrak a_i) of B.

  • \mathfrak a \subseteq \mathfrak b \implies V(\mathfrak a)\supseteq V(\mathfrak b).
  • If (F_i) is a set of homogeneous generators of \mathfrak a, then

V_0(\mathfrak a) = \{\mathbf v \in \mathbb P^n_k : F_i(\mathbf v) = 0 \text{ for all } i\}.

  • \cap_i V_0(\mathfrak a_i) = V_0(\sum_i \mathfrak a_i).
  • V_0(\mathfrak a) \cup V_0(\mathfrak b) = V_0(\mathfrak a \cap \mathfrak b) = V_0(\mathfrak {ab}).

In the other direction, we define:

Definition.

Let V\subseteq \mathbb P^n be any subset. Then I_0(V) denotes the (graded) ideal of B generated by:

\{F \in B \text{ homogeneous } : F(\mathbf v) = 0 \text{ for all } \mathbf v \in V\}.

In summary, we defined the following maps.

proj_space_correspondence

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Zariski Topology of Projective Space

We wish to define the Zariski topology on \mathbb P^n_k; for that let us take subsets of \mathbb P^n which can be identified with the affine space \mathbb A^n. Fix 0\le i \le n; let

U_i = \{ (t_0 : t_1 : \ldots : t_n ) \in \mathbb P^n_k : t_i \ne 0\}.

Note that for (t_0 : t_1 : \ldots : t_n) \in U_i the same point can be represented by (\frac{t_0}{t_i} : \frac{t_1}{t_i} : \ldots : \frac{t_n}{t_i}) where the i-th coordinate is 1. This gives a bijection \varphi_i : U_i \to \mathbb A^n_k. E.g. for n = 2, we have:

\varphi_0 : (x : y : z) \mapsto (\frac y x, \frac z x), \quad \varphi_1 : (x : y : z) \mapsto (\frac x y, \frac z y), \quad \varphi_2 : (x : y : z) \mapsto (\frac x z, \frac y z).

Note that for any i\ne j, the intersection U_i \cap U_j maps to an open subset of \mathbb A^n_k via both \varphi_i and \varphi_j. Indeed if i < j then \varphi_j(U_i\cap U_j) \subset \mathbb A^n_k is the set of all (t_0, \ldots, t_n) satisfying t_i \ne 0 while \varphi_i(U_i\cap U_j)\subset \mathbb A^n_k is the set of all (t_0, \ldots, t_n) satisfying t_{j-1} \ne 0. Hence, the following is well-defined.

Definition.

The Zariski topology on \mathbb P^n_k is defined by specifying every U_i \subseteq \mathbb P^n as an open subset, where U_i obtains the Zariski topology of \mathbb A^n from \varphi_i: U_i \to \mathbb A^n.

projective variety is a closed subspace V\subseteq \mathbb P^n_k.

First, we have the following preliminary results.

Lemma 1.

For any homogeneous F \in B, the set V_0(F) is (Zariski) closed in \mathbb P^n_k.

Proof

It suffices to show that V_0(F) \cap U_i is closed in U_i for each 0\le i\le n. But V_0(F) \cap U_0 = V(f), where f(X_1, \ldots, X_n) = F(1, X_1, \ldots, X_n) \in k[X_1, \ldots, X_n]. The same holds for U_1, \ldots, U_n. ♦

Definition.

For any f \in k[X_1, \ldots, X_n], let

F(T_0, \ldots, T_n) := T_0^{\deg f} f(\frac {T_1} {T_0}, \ldots, \frac{T_n}{T_0})

be its homogenization.

Exercise B

1. Prove that if f,g\in k[X_1, \ldots, X_n], the homogenization of fg is the product of the homogenizations of f and g.

2. Let F\in k[T_0, \ldots, T_n] be the homogenization of a non-constant f \in k[X_1, \ldots, X_n]. Then f is irreducible if and only if F is irreducible. [ Hint: you may find lemma 2 here helpful. ]

The Zariski topology on \mathbb P^n is consistent with our earlier notions of closed subsets:

Proposition 1.

A subset V\subseteq \mathbb P^n_k is closed under the Zariski topology if and only if V = V_0(\mathfrak a) for some graded ideal \mathfrak a \subseteq B.

Proof

(⇐) Suppose \mathfrak a = (F_1, \ldots, F_m) for some homogeneous F_i \in B. Since V_0(\mathfrak a) = \cap_i V_0(F_i), by lemma 1 this is Zariski closed.

(⇒) Let U= \mathbb P^n_k - V; it suffices to show that any \mathbf v \in U is contained in D_0(F) for some homogeneous F \in B. Now \mathbf v is contained in some U_i, say U_0 without loss of generality. Hence \mathbf v \in D(f) \subseteq U \cap U_0 for some f \in k[X_1, \ldots, X_n]. If \mathbf v = (1, t_1, \ldots, t_n), then

f(t_1, \ldots, t_n) \ne 0 \implies F(\mathbf v) \ne 0 \implies \mathbf v \in D_0(F), where F = homogenization of f. ♦

Example

Let V\subseteq \mathbb P^2_k be the projective variety defined by the homogeneous equation T_0^2 + T_1^2 = T_2^2. Then

  • V \cap U_0 is cut out from \mathbb A^2_k by 1 + y^2 = z^2;
  • V \cap U_1 is cut out from \mathbb A^2_k by x^2 + 1 = z^2;
  • V \cap U_2 is cut out from \mathbb A^2_k by x^2 + y^2 = 1.

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Cone of Projective Variety

We wish to prove the bijective correspondence between graded radical ideals of B and closed subsets V\subseteq \mathbb P^n_k. For that, we can piggyback on existing results for the affine case.

Definition.

Let C\subseteq \mathbb P^n_k be any subset. The cone of C is

\mathrm{cone}(C) := \{\mathbf v = (x_0, \ldots, x_n) \in \mathbb A^{n+1}_k : \mathbf v = \mathbf 0 \text{ or } (x_0 : \ldots : x_n) \in C\}.

Note

For any non-empty collection of subsets C_i \subseteq \mathbb P^n_k we have

\bigcup_i \mathrm{cone}(C_i) = \mathrm{cone}(\cup_i C_i), \quad \bigcap_i \mathrm{cone}(C_i) = \mathrm{cone}(\cap_i C_i).

Also, we have:

Lemma 2.

If \mathfrak a \subsetneq B is a proper graded ideal then \mathrm{cone}(V_0(\mathfrak a)) = V(\mathfrak a).

In particular, by proposition 1, the cone of a closed V \subseteq \mathbb P^n_k is closed in \mathbb A^{n+1}_k.

Proof

Note: if F\in B is non-constant homogeneous, then \mathrm{cone} V_0(F) = V(F) \subseteq \mathbb A^{n+1}_k. Now pick a set of homogeneous generators F_i for \mathfrak a; each F_i is non-constant so

\mathrm{cone}(V_0(\mathfrak a)) = \mathrm{cone}(\cap_i V_0(F_i)) = \bigcap_i \mathrm{cone} V_0(F_i) = \bigcap_i V(F_i) = V(\mathfrak a).

This completes the proof. ♦

Furthermore we have:

Lemma 3.

A non-empty closed subset W\subseteq \mathbb A^{n+1} is of the form \mathrm{cone}(V) for some closed V\subseteq \mathbb P^n if and only if

\mathbf v \in W, \lambda \in k \implies \lambda \mathbf v \in W.

When that happens, we call W a closed cone in \mathbb A^{n+1}.

Proof

(⇒) is obvious; for (⇐) clearly W = cone(V) for some subset C\subseteq \mathbb P^n. Let \mathfrak a = I(W) so that V(\mathfrak a) = W. It remains to show \mathfrak a is graded, for we would get W = \mathrm{cone}(V_0(\mathfrak a)) by lemma 2.

Indeed if f\in \mathfrak a, write f = f_0 + \ldots + f_d as a sum of homogeneous components. Then for any \mathbf v \in W and \lambda \in k we have \lambda \mathbf v \in W which gives

0 = f(\lambda \mathbf v) = f_0 (\mathbf v) + \lambda \cdot f_1(\mathbf v) + \ldots + \lambda^d \cdot f_d(\mathbf v).

Thus f_0, \ldots, f_d vanish for any \mathbf v \in W, i.e. f_0, \ldots, f_d \in \mathfrak a. ♦

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Projective Nullstellensatz

Thus we have the following correspondences:

proj_nullstellensatz

The top-left column is a bijection by lemma 3; the bottom row is a bijection by Nullstellensatz. In the proof of lemma 3, we also showed that for a closed cone W\subseteq \mathbb A^{n+1}_k, the ideal I(W) is graded. Conversely, if \mathfrak a\subsetneq k[T_0, \ldots, T_n] is graded, V(\mathfrak a) is the (non-empty) solution set of a collection of graded polynomials; hence it is a closed cone too.

Hence we have a bijection between

  • closed subsets V\subseteq \mathbb P^n_k, and
  • proper homogeneous radical ideals \mathfrak a \subsetneq B = k[T_0, \ldots, T_n].

The correspondence takes V \mapsto I(\mathrm{cone}(V)) =: \mathfrak a and so

\mathrm{cone}(V) = V(\mathfrak a) = \mathrm{cone} (V_0 (\mathfrak a)) \implies V = V_0(\mathfrak a).

The last piece of the puzzle is the map I_0 which takes closed subsets of \mathbb P^n_k to homogeneous ideals of B. As an easy exercise, show that

V \subseteq \mathbb P^n_k \text{ closed non-empty } \implies I_0(V) = I(\mathrm{cone}(V)).

However I_0(\emptyset) = (1) so we modify our bijection to:

Theorem (Projective Nullstellensatz).

There is a bijection between:

  • closed subsets V \subseteq \mathbb P^n_k;
  • homogeneous radical ideals \mathfrak a \subseteq B such that \mathfrak a \ne B_+ where B_+ := (T_0, \ldots, T_n) is the irrelevant ideal of B.

proj_nullstellensatz_bij

Exercise C

Prove that if \mathfrak a \subseteq B is a homogeneous ideal then V(\mathfrak a) is empty if and only if \mathfrak a contains a power of B_+.

[ Hint: prove that the radical of 𝔞 is either (1) or B+. ]

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Commutative Algebra 60

Primary Decomposition of Ideals

Definition.

Let \mathfrak a \subsetneq A be a proper ideal. A primary decomposition of \mathfrak a is its primary decomposition as an A-submodule of A:

\mathfrak a = \mathfrak q_1 \cap \mathfrak q_2 \cap \ldots \cap \mathfrak q_n,

where each \mathfrak q_i is \mathfrak p_i-primary for some prime \mathfrak p_i \subset A, i.e. \mathrm{Ass}_A (A/\mathfrak q_i) = \{\mathfrak p_i\}.

Here is a quick way to determine if an ideal is primary.

Proposition 1.

A proper ideal \mathfrak b \subsetneq A is \mathfrak p-primary for some \mathfrak p if and only if:

x, y \in A, xy \in \mathfrak b \implies x \in \mathfrak b \text{ or } y\in r(\mathfrak b),

in which case \mathfrak p = r(\mathfrak b).

Recall that r(\mathfrak b) is the radical of the ideal \mathfrak b.

Proof

(⇒) Suppose \mathrm{Ass}_A A/\mathfrak b = \{\mathfrak p\}. By proposition 3 here, the set of zero-divisors (in A) of A/\mathfrak b is \mathfrak p. Now if xy\in \mathfrak b and x\not\in \mathfrak b, then y\in A is a zero-divisor of A/\mathfrak b so y\in \mathfrak p.

To prove that r(\mathfrak b) = \mathfrak p, suppose x^n \in \mathfrak b; then x^n \in A is a zero-divisor for A/\mathfrak b so x^n \in \mathfrak p and we have x\in \mathfrak p. Conversely, if x\in \mathfrak p, then since every minimal prime in V(\mathfrak b) is an associated prime of A/\mathfrak b (by proposition 2 here), x is contained in every minimal prime of V(\mathfrak b). By proposition 5 here, x\in r(\mathfrak b).

(⇐) Suppose \mathfrak b satisfies the given condition; we first show that r(\mathfrak b) is prime. Suppose xy \in r(\mathfrak b) and x\not\in r(\mathfrak b). For some n > 0, x^n y^n \in \mathfrak b. But x^n \not \in r(\mathfrak b) so by the given condition y^n \in \mathfrak b and y\in r(\mathfrak b).

Let \mathfrak p = r(\mathfrak b); it remains to show \mathrm{Ass}_A (A/\mathfrak b) = \{\mathfrak p\}. First, \mathfrak p is the unique minimal element of V(\mathfrak b) so \mathfrak p \in \mathrm{Ass}_A (A/\mathfrak b).

Conversely, it remains to show any zero-divisor of A/\mathfrak b as an A-module lies in \mathfrak p. But if x\in A and y\in A - \mathfrak b are such that xy \in \mathfrak b, then by the given condition x \in r(\mathfrak b) = \mathfrak p. ♦

We thus say a proper ideal \mathfrak q \subsetneq A is primary if:

x, y\in A, xy \in \mathfrak q\implies x \in \mathfrak q \text{ or } y \in r(\mathfrak q).

Note that prime ideals are primary, and by proposition 1, the radical of a primary ideal is prime.

Exercise A

1. Prove that if f:A\to B is a ring homomorphism and \mathfrak q\subset B is a primary ideal, then f^{-1}(\mathfrak q) is a primary ideal of A. Also r(f^{-1}(\mathfrak q)) = f^{-1}(r(\mathfrak q)).

2. Prove that for an ideal \mathfrak a \subseteq A, there is a bijection between primary ideals of A containing \mathfrak a and primary ideals of A/\mathfrak a.

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Primary Decomposition and Localization

Throughout this section S\subseteq A is a fixed multiplicative subset.

Proposition 2.

There is a bijection between:

  • primary ideals \mathfrak q \subset A such that \mathfrak q \cap S = \emptyset;
  • primary ideals \mathfrak q' \subset S^{-1}A

Furthermore, if \mathfrak q is in the first set and \mathfrak p = r(\mathfrak q), then

\mathfrak p\cdot S^{-1}A = r(\mathfrak q \cdot S^{-1}A).

Proof

By exercise A.2 above and proposition 3 here, it suffices to show: if \mathfrak q\subset A is primary and \mathfrak q \cap S = \emptyset then

  • \mathfrak q' := \mathfrak q (S^{-1}A) is primary, and
  • \mathfrak q' \cap A = \mathfrak q.

For the first claim, note that \mathfrak q' is a proper ideal since \mathfrak q \cap S = \emptyset. Suppose \frac x s, \frac y t \in S^{-1}A satisfy \frac{xy}{st} \in \mathfrak q'; then for some s'\in S we have s'xy \in \mathfrak q. Since \mathfrak q \cap S = \emptyset no power of s’ is contained in \mathfrak q so

xy\in \mathfrak q \implies x\in\mathfrak q \text{ or } y\in r(\mathfrak q) \implies \frac x s \in \mathfrak q' \text{ or } \frac y t \in r( \mathfrak q').

For the second claim, clearly \mathfrak q' \cap A \supseteq \mathfrak q. Conversely let a\in A satisfy \frac a 1 \in \mathfrak q'. For some s\in S we have sa \in \mathfrak q. As above s\not\in r(\mathfrak q) so a \in \mathfrak q. ♦

Finally, we will see later that unlike factorization of ideals in a Dedekind domain, primary decompositions are not unique. However we can still salvage the following.

Proposition 3.

Suppose we have minimal primary decompositions

\mathfrak a = \mathfrak q_1 \cap \ldots \cap \mathfrak q_n = \mathfrak q'_1 \cap \ldots \cap \mathfrak q'_n

with \mathfrak p_i = r(\mathfrak q_i) = r(\mathfrak q_i') for each i. If \mathfrak p_i is minimal in V(\mathfrak a) then \mathfrak q_i = \mathfrak q_i'.

Proof

We localize \mathfrak a = \mathfrak q_1 \cap \ldots \cap \mathfrak q_n at \mathfrak p := \mathfrak p_i. But since \mathfrak p is minimal in V(\mathfrak a), we have

j\ne i \implies \mathfrak p_j \not\subseteq \mathfrak p \implies \mathfrak q_j \not\subseteq \mathfrak p \implies \mathfrak q_j A_{\mathfrak p} = A_{\mathfrak p}.

Hence \mathfrak a A_{\mathfrak p} = \mathfrak q_i A_{\mathfrak p}. By proposition 2 this gives \mathfrak q_i = \mathfrak q_i A_{\mathfrak p} \cap A = \mathfrak a A_{\mathfrak p} \cap A which is uniquely determined by \mathfrak a. ♦

Exercise B

Prove the following.

1. If \mathfrak a \subseteq A is a reduced ideal, then it (i.e. A/\mathfrak a) has no embedded primes.

2. If \mathfrak m\subset A is maximal, then \mathfrak q is \mathfrak m-primary if and only if \mathfrak m^n \subseteq \mathfrak q \subseteq \mathfrak m for some n > 0.

warningA power of a prime ideal is not primary in general. E.g. let k be a field and A =k[X, Y, Z]/(Z^2 - XY) with \mathfrak p = (X, Z) \subset A, which is prime since

A/\mathfrak p \cong k[X, Y, Z]/(Z^2 - XY, X, Z) \cong k[Y].

Then \mathfrak p^2 is not primary because XY = Z^2 \in \mathfrak p^2 but X\not\in \mathfrak p^2 and Y\not\in r(\mathfrak p^2) = \mathfrak p.

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Worked Examples

Throughout this section, k denotes a field.

Example 1

In an earlier example, we saw that for A = k[X, Y] and \mathfrak a = (X^2, XY), the A-module A/\mathfrak a has two associated primes: (X) and (XY). The following are primary decompositions, both of which are clearly minimal:

(X^2, XY) = (X) \cap (X^2, XY, Y^2) = (X) \cap (X^2, Y).

To check that these ideals are primary:

  • (X) is prime and hence primary;
  • (X^2, XY, Y^2) = (X, Y)^2 is a power of the maximal ideal (X, Y); by exercise B.2 it is primary;
  • (X, Y)^2 \subseteq (X^2, Y) \subseteq (X, Y) so (X^2, Y) is primary by exercise B.2.

Note that (X^2, XY, Y^2) \subsetneq (X^2, Y). Geometrically, the k-scheme with coordinate ring A/\mathfrak a looks like the following.

embedded_diagram_1

Example 2

Let A = k[X, Y, Z] with \mathfrak a = (X - YZ, XY). Then

k[X, Y, Z]/(X - YZ) \cong k[Y, Z], \ X \mapsto YZ \implies k[X, Y, Z]/\mathfrak a \cong k[Y, Z]/(Y^2 Z).

We have (Y^2 Z) = (Y^2) \cap (Z), an intersection of primary ideals with r((Y^2)) = (Y) and r((Z)) = (Z). This translates to

\mathfrak a = (X - YZ, Y^2) \cap (X - YZ, Z) = (X - YZ, Y^2) \cap (X, Z)

with r((X - YZ, Y^2)) = (X, Y) and (X, Z) is already prime.

Example 3

Let A = k[X, Y] with \mathfrak a = (X) \cap (X, Y)^2 \cap (X, Y-1)^2. Then each of (X), (X,Y)^2 and (X, Y-1)^2 is primary (the first ideal is prime; the remaining two ideals are powers of a maximal ideal). Hence this gives a primary decomposition of \mathfrak a so its associated primes are (X), (X, Y) and (X, Y-1), with the latter two embedded.

Geometrically, the k-scheme with coordinate ring A/\mathfrak a looks like:

embedded_diagram_2

Computing an explicit set of generators for \mathfrak a is not trivial, but it can be done with Buchberger’s algorithm.

Exercise C (from Atiyah & MacDonald, Exercise 4.5)

Let A = k[X, Y, Z] and \mathfrak p_1 = (X, Y), \mathfrak p_2 = (X, Z), \mathfrak m = (X, Y, Z) be ideals of A. Set \mathfrak a := \mathfrak p_1 \mathfrak p_2. Prove that

\mathfrak a = \mathfrak p_1 \cap \mathfrak p_2 \cap \mathfrak m^2

is a minimal primary decomposition of \mathfrak a.

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Commutative Algebra 59

Prime Composition Series

Throughout this article, A is a noetherian ring and all A-modules are finitely generated.

Recall (proposition 1 here) that if M is a noetherian and artinian module, we can find a sequence of submodules whose consecutive factors are simple modules. Correspondingly we have:

Proposition 1.

For finitely generated M, there exists a sequence of submodules

0 = M_0 \subsetneq M_1 \subsetneq \ldots \subsetneq M_n = M

such that each M_{i} / M_{i-1} \cong A/\mathfrak p_i as A-modules for prime ideals \mathfrak p_i\subseteq A.

Proof

Assume M\ne 0. Since \mathrm{Ass}_A M \ne \emptyset by proposition 3 here, there is an embedding of A-modules A/\mathfrak p \hookrightarrow M. If equality holds, we are done. Otherwise, let M_1 be the image of the map and repeat with M/M_1 to obtain a submodule M_2/M_1 \cong A/\mathfrak p'. Repeating this process, this must eventually terminate since we cannot have an infinite ascending chain of submodules M_0 \subsetneq M_1 \subsetneq M_2 \subsetneq \ldots. ♦

Corollary 1.

Let 0 = M_0 \subset M_1 \subset \dots \subset M_n = M be as in proposition 1 with M_i / M_{i-1} \cong A/\mathfrak p_i for some prime \mathfrak p_i \subset A. Then

\mathrm{Ass}_A M \subseteq \{\mathfrak p_1, \ldots, \mathfrak p_n\}.

In particular if M is finitely generated then \mathrm{Ass}_A M is finite.

Proof

Repeatedly applying proposition 4 here, we have

\mathrm{Ass}_A M \subseteq \mathrm{Ass}_A (M_1/M_0) \cup \ldots \cup \mathrm{Ass}_A (M_n / M_{n-1}).

Since each \mathrm{Ass}_A M_i/M_{i-1} = \mathrm{Ass}_A (A/\mathfrak p_i) = \{\mathfrak p_i\} for each i, we are done. ♦

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Associated Primes and Support

If \mathfrak p\in \mathrm{Ass}_A M, then there is a map A/\mathfrak p \hookrightarrow M; upon localizing at \mathfrak p we get an A-linear k(\mathfrak p)\hookrightarrow M_{\mathfrak p} and so M_{\mathfrak p} \ne 0. Hence we have shown:

Lemma 1.

For any module M, \mathrm{Ass}_A M \subseteq \mathrm{Supp}_A M.

For a partial reverse inclusion, we have:

Proposition 2.

If \mathfrak p is a minimal element of \mathrm{Supp}_A M then \mathfrak p \in \mathrm{Ass}_A M.

Proof

By proposition 5 here, we have

\mathrm{Ass}_{A_{\mathfrak p}} M_{\mathfrak p} = \{\mathfrak q A_{\mathfrak p} : \mathfrak q \in \mathrm{Ass}_A M \text{ contained in } \mathfrak p\},

which is non-empty since M_{\mathfrak p} \ne 0. By lemma 1 this set lies in \mathrm{Supp}_{A_{\mathfrak p}} M_{\mathfrak p}; but since \mathfrak p is minimal in \mathrm{Supp}_A M, by exercise B.2 here \mathrm{Supp}_{A_{\mathfrak p}} M_{\mathfrak p} has exactly one element: \mathfrak p A_{\mathfrak p} so

\mathrm{Ass}_{A_{\mathfrak p}} M_{\mathfrak p} = \{\mathfrak p A_{\mathfrak p}\}

so \mathfrak p \in \mathrm{Ass}_A M by proposition 5 here again. ♦

Definition.

If \mathfrak p \in \mathrm{Ass}_A M is not a minimal prime of \mathrm{Supp}_A M, we call it an embedded prime of M.

Example

Let k be a field, A = k[X, Y] and M = k[X, Y]/(X^2, XY), considered as an A-module. Note that \mathrm{Ann}_A M = (X^2, XY) with radical (X) so we get (by proposition 1 here)

\mathrm{Supp}_A M = V(\mathrm{Ann}_A M) = V((X)).

On the other hand, we claim that \mathrm{Ass}_A M = \{(X), (X, Y)\}. Indeed we have:

\begin{aligned} f = Y \in M &\implies \mathrm{Ann}_A f = (X), \\ g = X \in M &\implies \mathrm{Ann}_A g = (X, Y),\end{aligned}

so (X), (X, Y) \in \mathrm{Ass}_A M and (X, Y) is an embedded prime of M.

Conversely, we pick the chain of submodules 0 \subset M_1 \subset M with M_1 generated by g = X\in M. Then M_1 = (X\cdot A)/(X^2\cdot A + XY\cdot A); as shown above, M_1 \cong A/(X,Y). Also M/M_1 \cong A/(X) so by corollary 1

\mathrm{Ass}_A M \subseteq \{(X), (X,Y)\}.

Exercise A

Find an A-module M and prime ideals \mathfrak p_1 \subsetneq \mathfrak p_2 \subsetneq \mathfrak p_3 with \mathfrak p_1, \mathfrak p_3 \in \mathrm{Ass}_A M but \mathfrak p_2 \not\in \mathrm{Ass}_A M.

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Existence of Primary Decomposition

In this section we fix an ambient non-zero A-module M (finitely generated of course) and consider its submodules. For each prime \mathfrak p \subset A, take the set \Sigma(\mathfrak p) of all submodules N\subseteq M such that \mathfrak p \not\in \mathrm{Ass}_A N. Note that

0\in \Sigma(\mathfrak p) \implies \Sigma(\mathfrak p) \ne \emptyset.

Now for each prime \mathfrak p, fix a maximal element E(\mathfrak p) \in \Sigma(\mathfrak p).

Lemma 2.

We have \cap_{\mathfrak p} E(\mathfrak p) = 0.

Note

If \mathfrak p\not\in \mathrm{Ass}_A M then E(\mathfrak p) = M so the above intersection only needs to be taken over \mathfrak p\in \mathrm{Ass}_A M, i.e. a finite number of terms.

Proof

Let N := \cap_{\mathfrak p} E(\mathfrak p). If N\ne 0 it has an associated prime \mathfrak p. Since N\subseteq E(\mathfrak p) we have \mathfrak p \in \mathrm{Ass}_A E(\mathfrak p), a contradiction. ♦

Lemma 3.

Let \mathfrak p \in \mathrm{Ass}_A M. Then \mathrm{Ass}_A (M/E(\mathfrak p)) = \{\mathfrak p \}.

Proof

By proposition 4 here, we have \mathrm{Ass}_A M \subseteq \mathrm{Ass}_A E(\mathfrak p) \cup \mathrm{Ass}_A (M/ E(\mathfrak p)). Since \mathfrak p \not\in \mathrm{Ass}_A E(\mathfrak p) we have \mathfrak p \in \mathrm{Ass}_A M/E(\mathfrak p). On the other hand if \mathfrak q \in \mathrm{Ass}_A (M/E(\mathfrak p)) - \{\mathfrak p\}, then we have an injection A/\mathfrak q \hookrightarrow M/E(\mathfrak p) whose image is of the form Q/E(\mathfrak p). But now

\mathrm{Ass}_A Q \subseteq \mathrm{Ass}_A E(\mathfrak p) \cup \mathrm{Ass}_A Q/E(\mathfrak p) = \mathrm{Ass}_A E(\mathfrak p) \cup \{\mathfrak q\}

which does not contain \mathfrak p, contradicting maximality of E(\mathfrak p). Hence no such \mathfrak q exists. ♦

Definition.

For an associated prime \mathfrak p of M, a \mathfrak pprimary submodule of M is an N\subseteq M such that \mathrm{Ass}_A M/N = \{\mathfrak p\}.

A primary decomposition of M is an expression

0 = M_1 \cap M_2 \cap \ldots \cap M_n

where each M_i is a \mathfrak p_i-primary submodule of M.

The decomposition is irredundant if for any 1\le i\le n, \cap_{j\ne i} M_j \ne 0. It is minimal if it is irredundant and all \mathfrak p_i are distinct.

By lemmas 2 and 3, a minimal primary decomposition exists for every non-zero module.

warningIt is not true that in every primary decomposition 0 =\cap_{\mathfrak p} M(\mathfrak p), M(\mathfrak p) must be a maximal element of \Sigma(\mathfrak p). We will see an example in the next article.

Exercise B

Prove that if N_1, N_2 \subseteq M are \mathfrak p-primary submodules, so is N_1\cap N_2 \subseteq M.

Hence given any primary decomposition, we can get a minimal one by first removing the redundant terms then taking the intersection of all M_i with the same corresponding \mathfrak p_i.

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Properties of Primary Decomposition

Here, we will discuss properties of a general primary decomposition. Throughout this section we fix:

0 = M_1 \cap M_2 \cap \ldots \cap M_n where M_i is \mathfrak p_i-primary in M.

Proposition 3.

Every \mathfrak p \in \mathrm{Ass}_A M must occur among the \mathfrak p_i.

Proof

Since \cap_i M_i = 0 the canonical map M\hookrightarrow M/M_1 \oplus \ldots \oplus M/M_n is injective. Hence

\mathrm{Ass}_A M \subseteq \cup_{i=1}^n \mathrm{Ass}_A (M/M_i) = \{\mathfrak p_1, \ldots, \mathfrak p_n\}.

Proposition 4.

If the primary decomposition is irredundant, then every \mathfrak p_i is an associated prime of M.

Proof

If \mathfrak p_i \not\in \mathrm{Ass}_A M then among the injections M \hookrightarrow \oplus_j M/M_j and M/M_i \hookrightarrow \oplus_j M/M_j, we have M \cap (M/M_i) = 0 since \mathrm{Ass}_A M and \mathrm{Ass}_A (M/M_i) = \{\mathfrak p_i\} are disjoint. Thus

M\longrightarrow \oplus_{j\ne i} M/M_j is injective

and \cap_{j\ne i} M_j = 0, contradicting the condition of irredundancy. ♦

Note

Thus proposition 4 gives us a way to compute \mathrm{Ass}_A M: find a primary decomposition and remove terms until it becomes irredundant. However, note that proposition 4 does not require \mathfrak p_1, \ldots, \mathfrak p_n to be distinct.

Corollary 2.

If the primary decomposition is minimal, then

\mathrm{Ass}_A M = \{\mathfrak p_1, \ldots, \mathfrak p_n\}

has exactly n elements.

Proof

Apply propositions 3 and 4. ♦

Finally, we define the primary decomposition of submodules.

Defintion.

Let N\subseteq M be a submodule. A primary decomposition of N in M is an expression

N = M_1 \cap M_2 \cap \ldots \cap M_n

such that M_i\subseteq M is a \mathfrak p_i-primary submodule of M.

Note

A primary decomposition of N in M corresponds bijectively to a primary decomposition of M/N. We say the primary decomposition of N in M is irredundant or minimal if the corresponding primary decomposition of M/N is so.

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Commutative Algebra 58

We have already seen two forms of unique factorization.

  • In a UFD, every non-zero element is a unique product of irreducible (also prime) elements.
  • In a Dedekind domain, every non-zero ideal is a unique product of maximal ideals.

Here, we will introduce yet another type of factorization, called primary decomposition. The main idea is that in a noetherian ring, every ideal (even the zero ideal) is an intersection of primary ideals. E.g. in \mathbb Z, every non-zero ideal is an intersection of p^m \mathbb Z \subset \mathbb Z, ideals generated by prime powers.

Annihilators

Let A be a fixed ring and M be an A-module.

Definition.

The annihilator of m\in M in A is

\mathrm{Ann}_A m := \{ a \in A : am = 0\},

an ideal of A. Similarly, the annihilator of M in A is

\mathrm{Ann}_A M := \{a \in A : \forall m\in M, am = 0\},

also an ideal of A. If \mathrm{Ann}_A M = 0, we say M is a faithful A-module.

Note that if \mathfrak a = \mathrm{Ann}_A M, then M can be regarded as a faithful A/\mathfrak a-module.

Exercise A

1. Given A-submodules N, P \subseteq M, let

(N : P) = \{ a \in A: aP\subseteq N\}.

State and prove the analogue of proposition 2 here for submodules of M. Observe that we can write the annihilators as

\mathrm{Ann}_A m = (0 : Am), \quad \mathrm{Ann}_A M = (0 : M).

Conversely, express (N: P) in terms of annihilators.

2. Prove that (S^{-1}N : S^{-1}P) = S^{-1}(N : P) if P is a finitely generated submodule of M.

In particular if M is finitely generated then

\mathrm{Ann}_{S^{-1}A} S^{-1}M = S^{-1} (\mathrm{Ann}_A M).

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Support of a Module

Definition.

The support of an A-module M is:

\mathrm{Supp}_A M := \{ \mathfrak p \in \mathrm{Spec} A : M_{\mathfrak p} \ne 0\}.

Geometrically, these are points in Spec A at which the module does not vanish.

Proposition 1.

If M is a finitely generated A-module, then

\mathrm{Supp}_A M = V(\mathrm{Ann}_A M).

In particular, the support of M is a closed subspace of \mathrm{Spec} A.

Proof

Let m\in M. For \mathfrak p \in \mathrm{Spec} A we have \frac m 1 \in M_{\mathfrak p} is zero if and only if there exists a \in A - \mathfrak p such that am = 0, equivalently if there exists a \in (\mathrm{Ann}_A m) - \mathfrak p. Hence \frac m 1 \ne 0 if and only if \mathrm{Ann}_A m \subseteq \mathfrak p.

In the general case, let m_1, \ldots, m_n generate M as an A-module. Then

\begin{aligned} M_{\mathfrak p} \ne 0 &\iff \text{for some } i,\ \tfrac {m_i} 1 \in M_{\mathfrak p} \text{ is not zero } \\ &\iff \text{for some } i,\ \mathfrak p \supseteq \mathrm{Ann}_A m_i \\ &\iff \mathfrak p \supseteq \cap_i \mathrm{Ann}_A m_i = \mathrm{Ann}_A M.\end{aligned}

For the last equivalence, recall that \mathfrak p contains \mathfrak a \cap \mathfrak b if and only if it contains either \mathfrak a or \mathfrak b. ♦

Proposition 2.

If 0 \to N \to M \to P \to 0 is a short exact sequence of A-modules, then \mathrm{Supp}_A M = (\mathrm{Supp}_A N) \cup (\mathrm{Supp}_A P).

If N, P are finitely generated A-modules, then

\mathrm{Supp}_A (N\otimes_A P) = (\mathrm{Supp}_A N) \cap (\mathrm{Supp}_A P).

Note

Philosophically, if we imagine the first case as NP  and the second case as N × P, then the result says N_{\mathfrak p} + P_{\mathfrak p} = 0 if and only if both terms are zero whereas N_{\mathfrak p} \times P_{\mathfrak p} = 0 if and only if at least one term is zero.

Proof

Let \mathfrak p \subset A be prime.

For the first claim, we have a short exact sequence 0 \to N_{\mathfrak p} \to M_{\mathfrak p} \to P_{\mathfrak p} \to 0 and it follows that M_{\mathfrak p} = 0 if and only if N_{\mathfrak p} = P_{\mathfrak p} = 0.

For the second claim, we have

(N \otimes_A P)_{\mathfrak p} = N_{\mathfrak p} \otimes_{A_{\mathfrak p}} P_{\mathfrak p}

If N_{\mathfrak p} = 0 or P_{\mathfrak p} = 0, clearly the RHS is zero. Conversely if both are non-zero, since they are finitely generated A_{\mathfrak p}-modules, Nakayama’s lemma gives k(\mathfrak p) \otimes_A N = N_{\mathfrak p}/\mathfrak p N_{\mathfrak p} \ne 0 and k(\mathfrak p) \otimes_A P \ne 0. But these are vector spaces over k(\mathfrak p) so we have

k(\mathfrak p) \otimes_A N \otimes_A P = [k(\mathfrak p) \otimes_A N] \otimes_{k(\mathfrak p)} [k(\mathfrak p) \otimes_A P] \ne 0 \implies (N\otimes_A P)_{\mathfrak p} \ne 0.

Exercise B

1. Let f:M\to N be a homomorphism of finitely generated A-modules; prove that \mathrm{Supp} f = \{\mathfrak p \in \mathrm{Spec} A : f_{\mathfrak p} \ne 0\} is a closed subset of Spec A.

2. Prove: if S\subseteq A is multiplicative then

\begin{aligned} \mathrm{Supp}_{S^{-1}A} S^{-1}M &= (\mathrm{Supp}_A M) \cap (\mathrm{Spec} S^{-1}A) \\ &= \{ \mathfrak p (S^{-1}A) : \mathfrak p \in \mathrm{Supp}_A M, \mathfrak p \cap S = \emptyset\}. \end{aligned}

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Associated Primes

Definition.

Let \mathfrak p \subset A be a prime ideal. We say \mathfrak p is associated to the A-module M if \mathfrak p = \mathrm{Ann}_A m for some m\in M.

Let \mathrm{Ass}_A M be the set of prime ideals of A associated to M.

Note

We have: \mathfrak p \in \mathrm{Ass}_A M if and only if there is an injective A-linear map A/\mathfrak p \hookrightarrow M. Observe that if the annihilator of m \in M is \mathfrak p, then so is that of any non-zero multiple of m; this follows immediately from the definition of prime ideals.

Next suppose M is an A/\mathfrak a-module; we can compute both \mathrm{Ass}_A M and \mathrm{Ass}_{A/\mathfrak a} M. There is a bijection

\mathrm{Ass}_A M \cong \mathrm{Ass}_{A/\mathfrak a} M, \quad \mathfrak p \mapsto \mathfrak p/\mathfrak a \subset A/\mathfrak a.

In particular, we can compute \mathrm{Ass} (A/\mathfrak a) by considering A/\mathfrak a as an A-module or a module over itself. The above shows that there is effectively no difference.

Proposition 3.

Suppose A is noetherian. The union of all associated primes of M is its set of zero-divisors

\{a \in A : am = 0 \text{ for some } m \in M, m\ne 0\}.

In particular, any non-zero A-module M has an associated prime.

Proof

Fix an a\in A and m' \in M - \{0\} such that am' = 0; we need to show a is contained in an associated prime of M.

So let \Sigma be the set of ideals of A containing a of the form \mathrm{Ann}_A m for m \in M-\{0\}. Since A is noetherian and \mathrm{Ann}_A m' \in \Sigma, there is a maximal \mathfrak p \in \Sigma. We will show \mathfrak p is prime; first write \mathfrak p = \mathrm{Ann}_A m_0 for some m_0 \in M-\{0\}.

Pick b,c\in A such that bc\in \mathfrak p and b \not\in \mathfrak p. Since bm_0 \ne 0, we have \mathfrak a := \mathrm{Ann}_A (bm_0) \in \Sigma. And since \mathfrak a \supseteq \mathfrak p, by maximality of \mathfrak p we have \mathfrak a = \mathfrak p so since bc m_0 = 0 we have c\in \mathfrak a = \mathfrak p. ♦

Henceforth, A denotes a noetherian ring. However, we do not assume all modules are finitely generated at first.

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Properties of Associated Primes

Proposition 4.

Let 0\to N \to M \to P \to 0 be a short exact sequence of A-modules. Then

\mathrm{Ass}_A N \subseteq \mathrm{Ass}_A M \subseteq \mathrm{Ass}_A N \cup \mathrm{Ass}_A P.

Proof

We may assume N\subseteq M is a submodule and P = M/N. The first containment in the claim is obvious. For the second, suppose we have an A-linear map f : A/\mathfrak p \hookrightarrow M.

If (\mathrm{im } f )\cap N \ne 0, then any non-zero m\in (\mathrm{im }f) \cap N has annihilator \mathfrak p so \mathfrak p \in \mathrm{Ass}_A N. If (\mathrm{im } f)\cap N = 0, then composing A/\mathfrak p \stackrel f\to M \to M/N is still injective, so \mathfrak p \in \mathrm{Ass}_A (M/N). ♦

Corollary 1.

We have \mathrm{Ass}_A (M \oplus N) = \mathrm{Ass}_A M \cup \mathrm{Ass}_A N.

Proof

Follows immediately from proposition 4. ♦

Next, we show that taking the set of associated primes commutes with localization.

Proposition 5.

Let S\subseteq A be a multiplicative subset. Then

\begin{aligned} \mathrm{Ass}_{S^{-1}A} (S^{-1}M) &= \mathrm{Ass}_A M \cap \mathrm{Spec} S^{-1}A \\ &= \{\mathfrak p S^{-1}A : \mathfrak p \in \mathrm{Ass}_A M, \ \mathfrak p \cap S = \emptyset\}.\end{aligned}

Proof

(⊇) If A/\mathfrak p \hookrightarrow M and \mathfrak p \cap S = \emptyset, then localization at S gives an A-linear map S^{-1}A/(\mathfrak p S^{-1}A) \hookrightarrow S^{-1}M.

(⊆) Suppose we have an S^{-1}A-linear map (S^{-1}A)/\mathfrak q \hookrightarrow S^{-1}M where \mathfrak q \subset S^{-1}A is prime. By theorem 1 here, we can write \mathfrak q = \mathfrak p S^{-1}A for some prime \mathfrak p \subset A such that \mathfrak p \cap S = \emptyset. Now in the injection (S^{-1}A)/\mathfrak p(S^{-1}A) \hookrightarrow S^{-1}M we let \frac m s be the image of 1. It remains to show: there exists t\in S such that \mathrm{Ann}_A (tm) = \mathfrak p.

For each a\in \mathfrak p, we have \frac{am}s = 0 in S^{-1}M so that s'am = 0 in M for some s'\in S, i.e. a \in \mathrm{Ann}_A (s'm). Pick a generating set a_1, \ldots, a_n of \mathfrak p; then for each i there exists s_i' \in S with a_i \in \mathrm{Ann}_A (s_i' m). Then t := s_1' \ldots s_n' satisfies \mathfrak p \subseteq \mathrm{Ann}_A (tm).

Conversely if a\in \mathrm{Ann}_A (tm) then atm = 0 so \frac{at}1 \in S^{-1}A lies in the annihilator of \frac m s, i.e. in \mathfrak q. Then at \in \mathfrak q \cap A = \mathfrak p; since \mathfrak p \cap S = \emptyset we have a\in \mathfrak p. ♦

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Commutative Algebra 57

Continuing from the previous article, A denotes a noetherian ring and all A-modules are finitely generated. As before all completions are taken to be \mathfrak a-stable for a fixed ideal \mathfrak a \subseteq A.

Noetherianness

We wish to prove that the \mathfrak a-adic completion of a noetherian ring is noetherian. First we express:

Lemma 1.

If \mathfrak a = (a_1, \ldots, a_n), then

\hat A \cong A[[X_1, \ldots, X_n]] / (X_1 - a_1, \ldots, X_n - a_n).

Proof

Let B = A[X_1, \ldots, X_n], still noetherian, with ideal \mathfrak a' = (X_1 - a_1, \ldots, X_n - a_n). We have a ring isomorphism f : B/\mathfrak a' \stackrel \cong\to A taking X_i \mapsto a_i. Let \mathfrak b = (X_1, \ldots, X_n) be an ideal of B; we will take the \mathfrak b-adic completion on both sides of f, treated as B-modules.

  • On the LHS, \hat B = A[[X_1, \ldots, X_n]] and \hat{\mathfrak a}' is generated by X_1 - a_1, \ldots, X_n - a_n by proposition 5 here.
  • On the RHS, we get the f(\mathfrak{b})-adic completion on A; but f(\mathfrak b) = (a_1, \ldots, a_n), so we get the \mathfrak a-adic completion.

This completes our proof. ♦

Now all it remains is to prove this.

Proposition 1.

If A is a noetherian ring, so is A[[X]].

Proof

For a formal power series f\in A[[X]] where f = b_n X^n + b_{n+1} X^{n+1} + \ldots with b_n \ne 0, we say the lowest coefficient of f is b_n and its lowest term is b_n X^n. We also write \deg f = n.

Now suppose \mathfrak b\subseteq A[[X]] is a non-zero ideal.

Step 1: find a finite set of generators of 𝔟.

Now for each n = 0, 1, …, let \mathfrak a_n \subseteq A be the set of all b\in A for which b=0 or b X^n occurs as a lowest term of some f \in \mathfrak b. We get an ascending chain of ideals \mathfrak a_0 \subseteq \mathfrak a_1 \subseteq \ldots. Since A is noetherian, for some n we have \mathfrak a_n = \mathfrak a_{n+1} = \ldots.

For each of 0\le i \le n, pick a finite generating set S_i of \mathfrak a_i comprising of non-zero elements; for each b\in S_i pick f \in \mathfrak b whose lowest term is bX^i. This gives a finite subset T_i \subseteq \mathfrak b of degree-i power series whose lowest coefficients generate \mathfrak a_i. Note that if \mathfrak a_i = 0 then T_i = \emptyset.

Let T := \cup_{i=0}^n T_i.

Step 2: prove that T generates 𝔟.

Now suppose f\in \mathfrak b has lowest term bX^m so b\in \mathfrak a_m. By our choice of T we can find g_1, \ldots, g_k \in T such that

f - (a_1 X^{d_1}) g_1 -\ldots - (a_k X^{d_k})g_k = b' X^{m+1} + (\text{higher terms}),

for some a_i \in A, d_i = m - \deg g_i \ge 0. Since T is finite, in fact we can assume T = \{g_1, \ldots, g_k\}, setting a_i = 0 for unneeded g_i. Repeating the process with the RHS polynomial, we obtain

f - (a_1 X^{d_1} + b_1 X^{d_1 + 1}) g_1 - \ldots - (a_k X^{d_k} + b_k X^{d_k + 1}) g_k = b'' X^{m+2} + (\text{higher terms}).

Repeating this inductively, we obtain formal power series h_1, \ldots, h_k \in A[[X]] such that f - h_1 g_1 - \ldots - h_k g_k = 0. ♦

Hence, by proposition 1 and lemma 1 we have:

Corollary 1.

The \mathfrak a-adic completion of A is noetherian.

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Completion of Local Rings

Next suppose (A,\mathfrak m) is a noetherian local ring.

Definition.

The completion of A is its \mathfrak m-adic completion.

Since \hat A/\hat {\mathfrak m} \cong A/\mathfrak m =: k is a field, \hat{\mathfrak m} is a maximal ideal of \hat A. Also we have:

Lemma 2.

(\hat A, \hat {\mathfrak m}) is a local ring.

Proof

By lemma 2 here, \hat{\mathfrak m} is contained in the Jacobson radical of \hat A, so it is contained in all maximal ideals of \hat A. But \hat{\mathfrak m} is already maximal. ♦

Hence, we see that the noetherian local ring (\hat A, \hat {\mathfrak m}) inherits many of the properties of (A, \mathfrak m). E.g. they have the same Hilbert polynomial

P(r) = \dim_{A/\mathfrak m} \mathfrak m^r/\mathfrak m^{r++1}

since \hat{\mathfrak m}^n / \hat{\mathfrak m}^{n+1} \cong \mathfrak m^n /\mathfrak m^{n+1} as k-vector spaces.

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Hensel’s Lemma

Here is another key aspect of complete local rings, which distinguishes them from normal local rings.

Proposition (Hensel’s Lemma).

Suppose (A, \mathfrak m) is a complete local ring. Let f(X) \in A[X] be a polynomial. If there exists \alpha\in A such that f(\alpha) \equiv 0 \pmod {\mathfrak m} and f'(\alpha) \not\equiv 0 \pmod {\mathfrak m}, then there is a unique a\in A such that

a\equiv \alpha \pmod {\mathfrak m}, \quad f(a) = 0.

Note

Hence most of the time, if we can find a root \alpha for f(X) \in A[X] in the residue field A/\mathfrak m, then \alpha lifts to a root a\in A.

There are more refined versions of Hensel’s lemma to consider the case where f'(\alpha) \equiv 0 \pmod {\mathfrak m}. One can even generalize it to multivariate polynomials. For our purpose, we will only consider the simplest case.

Proof

Fix y\in A such that f'(\alpha)y \equiv 1 \pmod {\mathfrak m}.

Set a_1 = \alpha. It suffices to show: we can find a_2, a_3, \ldots \in A such that

i\ge 1 \implies a_{i+1} \equiv a_i \pmod {\mathfrak m^i}, \ f(a_i) \equiv 0 \pmod {\mathfrak m^i},

so that \lim_{n\to\infty} a_n \in A gives us the desired element. We construct this sequence recursively; suppose we have a_1, \ldots, a_n (n\ge 1). Write

f(X) = c + d(X - a_n) + (X - a_n)^2 g(X),\quad g(X) \in A[X],\ c = f(a_n), \ d = f'(a_n).

Since a_n \equiv \alpha \pmod {\mathfrak m} we have f'(a_n) \equiv f'(\alpha) \pmod {\mathfrak m} so f'(a_n)y \equiv 1 \pmod {\mathfrak m}. Hence setting a_{n+1} := a_n + x with x = -f(a_n)y \in \mathfrak m^n gives

\begin{aligned} f(a_{n+1}) &\equiv \overbrace{f(a_n)}^c + \overbrace{f'(a_n)}^d x \pmod {\mathfrak m^{2n}} \\ &= f(a_n) - f'(a_n) f(a_n) y \\ &\equiv 0 \pmod {\mathfrak m^{n+1}}.\end{aligned},

which gives us the desired sequence. ♦

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Applications of Hensel’s Lemma

Now we can justify some of our earlier claims.

Examples

1. In the complete local ring \mathbb C[[X, Y]] with maximal ideal \mathfrak m = (X, Y), consider the equation f(T) = T^2 - (1+X). Modulo \mathfrak m, we obtain f(T) \equiv T^2 - 1 which has two roots: +1 and -1. Since f'(1) = 2 \ne 0, by Hensel’s lemma there is a unique g \in \mathbb C[[X, Y]] with constant term 1 such that g^2 = 1+X.

2. Similarly, consider the ring \mathbb Z_p of p-adic integers with p > 2. Let f(X) = X^2 - a for a \in \mathbb Z_p outside p\mathbb Z_p. If a mod p has a square root b, then there is a Hensel lift of b to a square root of a in \mathbb Z_p.

3. Next, we will prove an earlier claim that the canonical map

\mathbb C[[Y]] \longrightarrow \mathbb C[[X, Y]]/(Y^2 - X^3 + X)

is an isomorphism. Consider the polynomial f(X) = X^3 - X - Y^2 as a polynomial in X with coefficients in \mathbb C[[Y]]. Modulo \mathfrak m, we have f(X) \equiv X^3 - X which has roots -1, 0, +1. Hence

X^3 - X - Y^2 = (X - \alpha_1) (X - \alpha_2)(X - \alpha_3) where \alpha_i \in \mathbb C[[Y]]

with \alpha_1, \alpha_2, \alpha_3 \equiv -1, 0, +1 \pmod Y respectively. But X-\alpha_1 and X-\alpha_3 are invertible in \mathbb C[[X, Y]] so

\mathbb C[[X, Y]]/(Y^2 - X^3 +X) \cong \mathbb C[[X, Y]]/(X - \alpha_2) \cong \mathbb C[[Y]].

4. In \mathbb Z_p, consider the equation f(X) = X^{p-1} - 1. Modulo p, this has exactly p – 1 roots; in fact any a \in \mathbb F_p - \{0\} is a root of f. Now f'(X) = (p-1)X^{p-2} so f'(a) \ne 0 in \mathbb F_p for all a \in \mathbb F_p - \{0\}.

Hence by Hensel’s lemma, for each 1 \le a \le p-1, there is a unique lift of a to an \omega_a \in \mathbb Z_p which is a (p – 1)-th root of unity. We call \omega_a the Teichmuller lift of a\in \mathbb F_p - \{0\}.

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Analysis in Completed Rings

Warning: the purpose of this section is to give the reader a flavour of the subject matter. It is not meant to be comprehensive. In particular, there are no proofs here.

Hensel’s lemma gives us an effective criterion to determine if a polynomial over a complete local ring have roots. Although its proof gives us a method to effectively compute these roots to arbitrary precision, there are other techniques we can borrow from real analysis.

Example 1: Binomial Expansion

In \mathbb C[[X]], we can compute the square root of (1+X) with the binomial expansion:

\begin{aligned}(1+X)^{\frac 1 2} &= 1+ \tfrac 1 2 X + \tfrac{\frac 1 2 (\frac 1 2 - 1)}{2!} X^2 + \tfrac{\frac 1 2 (\frac 1 2 -1)(\frac 1 2 - 2)}{3!} X^3 + \ldots\\ &= 1 + \tfrac 1 2 X - \tfrac 1 8 X^2 + \tfrac 1 {16} X^3 + \ldots \in \mathbb C[[X]].\end{aligned}

By the same token, we can compute square roots in \mathbb Z_p by taking binomial expansion of (1+\alpha)^n, as long as the convergence is “fast enough”. For example, to compute \sqrt 2 \in \mathbb Z_7, binomial expansion gives

2\sqrt 2 = \sqrt 8 = (1 + 7)^{\frac 1 2} = 1 + \frac 1 2 (7) + \frac{\frac 1 2 (\frac 1 2 - 1)}{2!} (7^2) + \frac{\frac 1 2 (\frac 1 2 -1)(\frac 1 2 - 2)}{3!} (7^3) + \ldots \in \mathbb Z_7.

Taking the first four terms we have 2\sqrt 2 \equiv 470 \pmod {7^4} so that \sqrt 2 \equiv 235 \pmod {7^4}. Indeed, we can easily check that m = 235 is a solution to m^2 \equiv 2 \pmod {7^4}.

Example 2: Fixed-Point Method

While solving equations of the form x = f(x) in analysis, it is sometimes effective to start with a good estimate x_0 then iteratively compute x_{n+1} = f(x_n). We can do this in complete local rings too.

For example, let us solve X^3 - X = Y^2 as a polynomial in X with coefficients in \mathbb C[[Y]]. We saw above there is a unique root x \equiv 0 \pmod Y. Start with x_0 = 0 then iteratively compute x_{n+1} = x_n^3 - Y^2. This gives

\begin{aligned} x_0 &= 0, \\ x_1 &= -Y^2,\\ x_2 &= -Y^6 - Y^2, \\ x_3 &= -Y^{18} - 3Y^{14} - 3Y^{10} - Y^6 - Y^2,\end{aligned}

where x_3 is accurate up to Y^{13}.

Example 3: Newton Method

To solve an equation of the form f(x) = 0, one starts with a good estimate x_0 then iterate x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.

Exercise A

Use the Newton root-finding method to obtain \sqrt 2 \in \mathbb Z_7 to high precision (in Python).

Completion in Geometry

As another application of completion, consider A = \mathbb C[X, Y]/(Y^2 - X^3 - X^2) with \mathfrak m = (X, Y). Taking the \mathfrak m-adic completion, we obtain

\hat A = \mathbb C[[X, Y]]/(Y^2 - X^3 - X^2)

by proposition 5 here. Note that since 1+X has a square root in \mathbb C[[X, Y]], the ring \hat A is no longer an integral domain, but a “union of two lines” since Y^2 - X^3 - X^2 = (Y - \alpha X)(Y + \alpha X) where \alpha = \sqrt{1+X} \in \mathbb C[[X, Y]] is a unit.

This reflects the geometrical fact that when we magnify at the origin, we obtain a union of two lines.

completion_geometry

[ Image edited from GeoGebra plot. ]

Exercise B

Let (A,\mathfrak m) be a local ring. Prove that if the \mathfrak m-adic completion of A is an integral domain, then so is A.

[ Hint: use a one-line proof. ]

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Commutative Algebra 56

Throughout this article, A denotes a noetherian ring and \mathfrak a \subseteq A is a fixed ideal. All A-modules are finitely generated.

Consequences of Artin-Rees Lemma

Suppose we have an exact sequence of finitely generated A-modules

0 \longrightarrow N \longrightarrow M \longrightarrow P \longrightarrow 0.

Let M be given the 𝔞-adic filtration; the induced filtration on P is 𝔞-adic so its completion is the 𝔞-adic completion. By the Artin-Rees lemma, the induced filtration on N is 𝔞-stable and by proposition 2 here its completion is also the 𝔞-adic completion. Hence we have shown:

Proposition 1.

The following functor is exact:

completion_functor

From general properties of exact functors, this has the following properties.

1. If N \subseteq M is a submodule, then \hat N can be identified as a submodule of \hat M.

2. If N_1, N_2 \subseteq M are submodules, then

(N_1 \cap N_2)\hat{} \cong \hat N_1 \cap \hat N_2, \quad (N_1 + N_2)\hat{} \cong \hat N_1 + \hat N_2.

3. If f:N\to M is a map of A-modules, then \hat f : \hat N \to \hat M satisfies

\mathrm{ker } \hat f = (\mathrm{ker } f)\hat{}, \quad \mathrm{im } \hat f = (\mathrm{im } f)\hat{}.

In particular, for a fixed m\in M, take the A-linear map f : A\to M, 1 \mapsto m. Taking the \mathfrak a-adic completion gives \hat f : \hat A \to \hat M, 1 \mapsto i(m) as well, where i : M\to \hat M is the canonical map. Hence

\hat A \cdot i(m) = \mathrm{im } \hat f = (\mathrm{im } f)\hat{} = (Am)\hat{}.

From property 2, we obtain, for m_1, \ldots, m_n \in M,

\hat A \cdot i(m_1) + \ldots + \hat A \cdot i(m_n) = (Am_1 + \ldots + Am_n)\hat{}.

Thus we have shown:

Proposition 2.

Identifying M with its image in \hat M,

\hat A \cdot M = \hat M.

In particular, if M is finitely generated, so is \hat M.

We also have:

Corollary 1.

For any ideal \mathfrak b\subseteq A and A-module M

(\mathfrak b M)\hat{} = \hat{\mathfrak b} \hat M.

Proof

By proposition 2, \hat {\mathfrak b}\hat M = (\hat A \mathfrak b)(\hat A M) = \hat A(\mathfrak b M) = (\mathfrak b M)\hat{}. ♦

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Krull’s Intersection Theorem

Another interesting consequence of the Artin-Rees lemma is as follows.

Krull’s Intersection Theorem. 

Suppose (A,\mathfrak m) is local and noetherian. If M is a finitely generated A-module, then \cap_n \mathfrak m^n M = 0.

In particular, the canonical map M \to \hat M is injective where \hat M is the \mathfrak m-adic completion of M.

Proof

Let N = \cap_n \mathfrak m^n M, a submodule of M. By the Artin-Rees lemma, the \mathfrak m-adic filtration on M induces a \mathfrak m-stable filtration on N so for some n,

\mathfrak m (N \cap \mathfrak m^n M) = N \cap \mathfrak m^{n+1} M \implies \mathfrak m N = N \implies N = 0

by Nakayama’s lemma. ♦

In particular, A \to \hat A is an injective ring homorphism when we take the \mathfrak m-adic completion of a local ring (A, \mathfrak m).

warningWe give an example where Krull’s intersection theorem fails when A is not noetherian. Take the set of all infinitely differentiable functions f : I\to \mathbb R, where I is an open interval containing 0; let A be the set of equivalence classes under the relation: f : I \to \mathbb R and g : I' \to \mathbb R are equivalent if f|_J = g|_J for some J\subseteq I\cap I' containing 0.

Now A is a ring with addition and product given by pointwise addition and product. Its unique maximal ideal is \mathfrak m = \{f \in A : f(0) = 0\}. Then \cap_n \mathfrak m^n \ne 0 since it contains \exp(-\frac 1 {x^2}).

Exercise A

1. Find a noetherian ring A and a proper ideal \mathfrak a \subsetneq A such that \cap_n \mathfrak a^n \ne 0.

2. Prove that if A is a noetherian integral domain, then any proper ideal \mathfrak a\subsetneq A satisfies \cap_n \mathfrak a^n = 0. [ Hint: follow the proof of Krull’s Intersection Theorem; use the “adjugate matrix” trick. ]

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Tensoring with Â

Proposition 3.

For any finitely generated M, we have a natural isomorphism \hat A \otimes_A M \cong \hat M.

Note

In short, for finitely generated module M, taking its completion is the same as taking the induced Â-module of M.

Proof

Since \hat M is an \hat A-module with a canonical A-linear M\to \hat M, by universal property of induced modules we have a map \hat A \otimes_A M \to \hat M which is natural in M. And since M is a noetherian module, it is finitely presented so we can find an exact sequence of the form A^m \to A^n \to M \to 0. This gives a commutative diagram of maps:

tensor_completion_diagram

where the top row is exact because tensor product is right-exact and the bottom row is exact from proposition 1. Since the first two vertical maps are isomorphisms, so is the third one. ♦

Hence the functor \hat A \otimes_A - is exact when restricted to the category of finitely generated A-modules. To see that \hat A is A-flat, we apply:

Lemma 1.

Let A be any ring (possibly non-noetherian) and M be an A-module.

M is A-flat if and only if for any injective map of finitely generated A-modules N_1 \to N_2, the resulting N_1 \otimes_A M \to N_2 \otimes_A M is also injective.

Proof

(⇒) Obvious. (⇐) Let P\subseteq Q be a submodule of any module. We need to show that P\otimes_A M\to Q\otimes_A M is injective. Let \Sigma be the set of all pairs (N_1, N_2) where N_2 \subseteq Q and N_1 \subseteq P\cap N_2 are finitely generated A-submodules, ordered by inclusion (in both terms). Clearly \Sigma is a directed set; since N_1 runs through all finitely generated submodules of P, we have direct limits

\varinjlim_{(N_1, N_2) \in \Sigma} N_1 \cong P, \quad \varinjlim_{(N_1, N_2)\in \Sigma} N_2 \cong Q.

By the given condition, N_1\otimes_A M \to N_2 \otimes_A M is injective for each (N_1, N_2) \in \Sigma. By proposition 3 here, taking the direct limit gives an injective

\varinjlim_{(N_1, N_2)} (N_1 \otimes_A M) \to \varinjlim_{(N_1, N_2)} (N_2 \otimes_A M).

By exercise B.4 here, the LHS is isomorphic to (\varinjlim_{(N_1, N_2)} N_1) \otimes_A M \cong P\otimes_A M. Likewise the RHS is isomorphic to Q\otimes_A M so P\otimes_A M \to Q\otimes_A M is injective. ♦

Corollary 2.

\hat A is a flat A-algebra.

Exercise B

Prove that in lemma 1, we can weaken the flatness condition to:

  1. For each ideal \mathfrak a\subseteq A, \mathfrak a\otimes_A M \to A\otimes_A M \cong M is injective.
  2. For each finitely generated ideal \mathfrak a\subseteq A, \mathfrak a\otimes_A M \to A\otimes_A M \cong M is injective.

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Completion and Quotients

Recall that for any submodule N\subseteq M we have (M/N)\hat{} \cong \hat M / \hat N. In particular if \mathfrak b\subseteq A is an ideal then

\hat A / \hat {\mathfrak b} \cong (A/\mathfrak b)\hat{} as \hat A-modules.

But (A/\mathfrak b)\hat{} also has a ring structure! Indeed by definition it is the completion obtained from the \mathfrak a-adic filtration as an A-module

A/\mathfrak b = A_0' \supseteq A_1' \supseteq \ldots, where A'_n := (\mathfrak a^n + \mathfrak b)/\mathfrak b

which is also a filtration of A/\mathfrak b as a ring since A_i' A_j' \subseteq A_{i+j}'. The construction which gives us (A/\mathfrak b)\hat{} = \varprojlim [(A/\mathfrak b)/A_n'] as A-modules also gives us the inverse limit as rings. One easily verifies that \hat A \to (A/\mathfrak b)\hat{} is a ring homomorphism so:

Proposition 4.

We have an isomorphism of rings

\hat A / \hat {\mathfrak b} \cong (A/\mathfrak b)\hat{},

where (A/\mathfrak b)\hat{} is its (\mathfrak a + \mathfrak b)/\mathfrak b-adic completion as a ring.

Furthermore, by proposition 2, if \mathfrak b is generated (as an ideal) by a_1, \ldots, a_n, then \hat{\mathfrak b} is generated by the images of a_i in \hat A. Thus we have shown:

Proposition 5.

Suppose \mathfrak b \subseteq A is an ideal generated by a_1, \ldots, a_n. Then the completion of A/\mathfrak b is the quotient of \hat A by the ideal generated by (the images of) a_1, \ldots, a_n.

Example

Take the example A = \mathbb C[X, Y]/(Y^2 - X^3 + X) with \mathfrak m = (X, Y) from an earlier example; we wish to compute the \mathfrak m-adic completion  of A. By the proposition,  is the quotient of \mathbb C[X, Y]^\wedge (the (X, Y)-adic completion) by (Y^2 - X^3 + X). But we clearly have \mathbb C[X, Y]^\wedge \cong \mathbb C[[X, Y]] so

\hat A \cong \mathbb C[[X, Y]]/(Y^2 - X^3 + X)

as we had claimed. In the next article, we will show that this ring is isomorphic to \mathbb C[[Y]].

Completion of Completion

As a special case, we have

\hat A / (\hat {\mathfrak a})^n = \hat A / (\mathfrak a^n)\hat{} \cong A / \mathfrak a^n,

where the equality is from corollary 1 and the isomorphism from lemma 1 here.

Hence, the \hat a-adic completion of \hat A is isomorphic to \hat A. We also have the following.

Lemma 2.

For each x \in \hat{\mathfrak a}, 1-x is invertible in \hat A.

In particular, (by proposition 4 here) \hat{\mathfrak a} is contained in the Jacobson radical of \hat A.

Proof

Since x^n \in \hat{\mathfrak a}^n, we can take the infinite sum

y = 1 + x + x^2 + \ldots \in \hat A.

Then (1-x)y \in \cap_n \hat{\mathfrak a}^n so (1-x)y = 0 in \hat A. ♦

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Commutative Algebra 55

Exactness of Completion

Throughout this article, A denotes a filtered ring.

Proposition 1.

Let 0 \to N \to M \to P \to 0 be a short exact sequence of A-modules. Suppose M is filtered, inducing filtrations on N and P. Then

0 \longrightarrow \hat N \longrightarrow\hat M \longrightarrow \hat P \longrightarrow 0

is also exact as \hat A-modules.

Proof

Without loss of generality, assume N is a submodule of M and PM/N. Each term in the filtration gives a short exact sequence

0 \longrightarrow \overbrace{N/(M_i \cap N)}^{N/N_i} \longrightarrow M/M_i \longrightarrow \overbrace{M/(M_i + N)}^{P/P_i} \longrightarrow 0

since N/(M_i \cap N) \cong (M_i + N)/M_i by the second isomorphism theorem. By proposition 1 here, taking (inverse) limit is left-exact so we obtain an exact sequence

0\longrightarrow \hat N \longrightarrow \hat M \longrightarrow \hat P.

To show that \hat M \to \hat P is surjective, we pick an element of \hat P. Since P/P_k \cong M/(M_k + N), the element is represented by a sequence (m_k) in M such that m_{k+1} - m_k \in M_k + N. We need to show there is a sequence (x_k) in M such that

k\ge 0 \implies x_{k+1} - x_k \in M_k, x_k - m_k \in M_k + N.

When k = 0, just pick any x_0. Suppose we have x_0, \ldots, x_k; we need x_{k+1} \in M such that x_{k+1} - x_k \in M_k and x_{k+1} - m_{k+1} \in M_{k+1} + N. But observe that m_{k+1} - x_k = (m_{k+1} - m_k) + (m_k - x_k) \in M_k + N. If we write m_{k+1} - x_k = m + n for m\in M_k, n\in N, then x_{k+1} := m_{k+1} - n works. ♦

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Completion of Completion

Lemma 1.

We have \hat M /\hat M_n \cong M/M_n, where M_n has the filtration induced from M.

Proof

Let P = M/M_n. From proposition 1 we get an exact sequence

0 \to \hat M_n \to \hat M \to \hat P \to 0.

But we also have P/P_m = M/(M_m + M_n) which is M/M_n for all m\ge n. Thus \hat P = M/M_n and we are done. ♦

Hence if we let \hat M take the filtration given by

\hat M = \hat M_0 \supseteq \hat M_1 \supseteq \ldots

then by lemma 1, the completion of \hat M with respect to this filtration is still \hat M.

If m_1, m_2, \ldots \in \hat M is a Cauchy sequence, from the previous article we have its limit

(\lim_{n\to \infty} m_n) \in \hat{\hat M} = \hat M

Since the map from \hat M to its completion is injective, we have \cap_n \hat M_n = 0 so as shown in exercise A.3 here, we can define an (ultra)metric on \hat M such that the resulting topology has a basis comprising of the set of all cosets \{m + \hat M_n\}. From the above, every Cauchy sequence converges in \hat M. Thus:

Summary.

\hat M is a complete metric space.

Furthermore, the image of M \to \hat M is dense; indeed any basic open subset of \hat M is of the form m + \hat M_n for m\in \hat M and n\ge 0. Since \hat M / \hat M_n\cong M/M_n, we see that m can be represented by an element of M. Thus any non-empty open subset of \hat M contains an element of M.

Thus \hat M is the completion of M even in the topological sense.

Note

For visualization, one can show that \mathbb Z_2 is homeomorphic to the Cantor set:

cantor_2-adic

E.g. the point above corresponds to a 2-adic integer ending at (\ldots 0010)_2.

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The 𝔞-adic Filtration

Now instead of arbitrary filtrations on M, we will focus our attention to the 𝔞-adic filtrations on A and M for a fixed ideal \mathfrak a:

M_n = \mathfrak a^n M \implies \hat M = \varprojlim M/\mathfrak a^n M.

Clearly if M is given the 𝔞-adic filtration, so is any quotient, because \mathfrak a^n(M/N) = (\mathfrak a^n M + N)/N, so the induced filtration on M/N is also 𝔞-adic. On the other hand, the induced filtration on a submodule N is \mathfrak a^n M \cap N\ne \mathfrak a^n N.

But the situation is salvageable when A is noetherian. Instead of the 𝔞-adic filtration, let us loosen our definition a little.

Definition.

A filtration (M_n) of M is said to be 𝔞-stable if for some n, we have M_{n+k} = \mathfrak a^k M_n for all k\ge 0.

In other words, an 𝔞-stable filtration is “eventually 𝔞-adic”. When we take the completion, we get the same thing.

Proposition 2.

Suppose M is an A-module with an 𝔞-stable filtration. Its completion is canonically isomorphic to the 𝔞-adic completion of M.

Proof

Since (M_n) is a filtration for M we have A_i M_j \subseteq M_{i+j}, i.e. \mathfrak a^i M_j \subseteq M_{i+j}. Now fix an n such that M_{n+k} = \mathfrak a^k M_n for all k\ge 0. We get

k\ge 0 \implies M_k \supseteq \mathfrak a^k M \supseteq \mathfrak a^k M_n = M_{n+k} \supseteq \mathfrak a^{n+k}M

and hence maps M/\mathfrak a^{n+k}\to M/M_{n+k} \to M/\mathfrak a^k M \to M/M_k. Taking the inverse limit:

\varprojlim_k M/\mathfrak a^{n+k}M \to \varprojlim_k M/M_{n+k} \to \varprojlim M/\mathfrak a^k M \to \varprojlim M/M_k.

By explicitly writing out elements of inverse limits, we see that the above give isomorphisms \varprojlim_k M/M_{n+k} \cong \varprojlim M/M_k and \varprojlim_k M/\mathfrak a^{n+k} \cong \varprojlim M/\mathfrak a^k; thus

\hat M \cong \varprojlim M/\mathfrak a^k M. ♦

Exercise A

1. Fill in the last step of the proof.

2. Show that in any category, the inverse limit of the diagram

limit_diagram_N

remains the same when we drop finitely many terms on the right.

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Artin-Rees Lemma

The main result we wish to prove is the following.

Artin-Rees Lemma.

Let A be a noetherian ring with the \mathfrak a-adic filtration, and N a submodule of a finitely generated A-module M. If M has an \mathfrak a-stable filtration, the induced filtration on N is also \mathfrak a-stable.

Proof

Step 1: define blowup algebra and module.

Definition.

Given any filtered module M over a filtered ring A, the blowup algebra and blowup module are defined by

B(A) := A_0 \oplus A_1 \oplus \ldots, \quad B(M) := M_0 \oplus M_1 \oplus \ldots.

We define a product operation A_i \times A_j \to A_{i+j} from multiplication in A. Hence, B(A) has a canonical structure of a graded ring.

Similarly, since M is a filtered module, we obtain a product operation A_i \times M_j \to M_{i+j} which gives B(M) a structure of a graded B(A)-module. When A and M are given the 𝔞-adic filtration, we write B_{\mathfrak a}(A) and B_{\mathfrak a}(M) for their blowup algebra and module.

Step 2: if A is a noetherian ring, so is B𝔞(A).

Since A is noetherian, we can write \mathfrak a = x_1 A + \ldots + x_k A for some x_1, \ldots, x_k \in \mathfrak a. It follows that \mathfrak a^n is a sum of x_1^{d_1}\ldots x_k^{d_k} A where \sum_{i=1}^k d_i = n. Hence the map

A[X_1, \ldots, X_k] \longrightarrow B_{\mathfrak a}(A), \quad X_i \mapsto (x_i \in A_1)

is a surjective ring homomorphism so B_{\mathfrak a}(A) is also noetherian.

Now we suppose A is noetherian and is given the 𝔞-adic filtration. Let M be a finitely generated filtered A-module.

Step 3: B(M) is finitely generated if and only if the filtration on M is 𝔞-stable.

(⇐) For some n we have B(M) = M_0 \oplus M_1 \oplus \ldots \oplus M_n \oplus \mathfrak a M_n \oplus \mathfrak a^2 M_n \oplus \ldots. Since M is a noetherian A-module, each M_i (0\le i \le n) is finitely generated as an A-module by, say m_{i1}, \ldots, m_{iN}. Now we take the set of m_{ij}, as homogeneous elements of B(M) of degree i.

 generators_of_BM

In the above, each homogeneous element of M_0, \ldots, M_n is an A-linear combination of these generators. Furthermore, M_{n+k} = \mathfrak a^k M_n = A_k M_n so m_{n1}, \ldots, m_{nN} \in B(M)_n generate (over B_{\mathfrak a}(A)) the homogeneous elements in B(M) of degree n and higher.

(⇒) Suppose B(M) is finitely generated over B_{\mathfrak a}(A) by homogeneous elements x_1, \ldots, x_k; let d_i = \deg x_i and N = \max d_i. We claim that M_{n+1} = \mathfrak a M_n for all n\ge N. Since M is filtered, we have \mathfrak a M_n \subseteq M_{n+1}

Conversely take y\in M_{n+1}, regard it as an element of B(M)_{n+1} and write y = a_1 x_1 + \ldots +a_k x_k with a_i \in B_{\mathfrak a}(A). Since y and x_i are homogeneous, we may assume a_i is homogeneous of degree e_i := n+1 - d_i > 0. So a_i \in B_{\mathfrak a}(A)_{e_i} = \mathfrak a^{e_i}. Write

a_i = b_{i1} c_{i1} + b_{i2} c_{i2} + \ldots + b_{ik} c_{ik}, \quad b_{ij} \in \mathfrak a, c_{ij} \in \mathfrak a^{e_i-1} \subseteq A.

Now y is a sum of b_{ij}c_{ij} x_i, with c_{ij} x_i \in M_n so y \in \mathfrak a M_n.

Step 4: prove the Artin-Rees lemma.

By step 2, B_{\mathfrak a}(A) is a noetherian ring; since M has an \mathfrak a-stable filtration, by step 3 B(M) is a noetherian B_{\mathfrak a}(A)-module. And since B(N) \subseteq B(M) is a B_{\mathfrak a}(A)-submodule it is also noetherian. By step 3 again, this says the induced filtration on N is \mathfrak a-stable. ♦

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