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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Commutative Algebra 54

Filtered Rings

Definition.

Let A be a ring. A filtration on A is a sequence of additive subgroups

A = A_0 \supseteq A_1 \supseteq A_2 \supseteq \ldots

such that A_i A_j \subseteq A_{i+j} for any i, j\ge 0. A filtered ring is a ring with a designated filtration.

Note

Since A\cdot A_i = A_0 \cdot A_i \subseteq A_i, in fact each A_i is an ideal of A.

Examples

1. If A = \oplus_{i=0}^\infty A_i is a grading, we can form a filtration by taking n \mapsto A_n \oplus A_{n+1} \oplus \ldots (where A_n, A_{n+1}, \ldots refers to the grading).

2. Let \mathfrak a\subseteq A be an ideal. The \mathfrak aadic filtration is given by A_i = \mathfrak a^i. E.g. we can take A = \mathbb Z and \mathfrak a = 2\mathbb Z, then A_i is the set of integers divisible by 2^i. More generally, in a dvr with uniformizer \pi, we can take A_i = (\pi^i).

Definition.

Suppose A is a filtered ring. A filtration on an A-module M is a sequence of additive subgroups

M = M_0 \supseteq M_1 \supseteq M_2 \supseteq \ldots

such that A_i M_j \subseteq M_{i+j} for any i, j\ge 0. A filtered module is a module with a designated filtration.

Note

Since A\cdot M_i = A_0 M_i \subseteq M_i, each M_i is an A-submodule of M.

Also, we need a fixed filtration on the base ring A before we can talk about filtrations on A-modules.

Example

Again, for an ideal \mathfrak a \subseteq A, we obtain the \mathfrak aadic filtration of M, where M_i = \mathfrak a^i M.

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Induced Filtrations

For the rest of this article, A denotes a filtered ring.

Definition.

Let M, N be filtered A-modules. A linear map f:M\to N is said to be filtered if f(M_i) \subseteq N_i for each i.

We also have:

Definition.

Let f:M\to N be a linear map of A-modules.

  • If (M_i) is a filtration of M, the induced filtration on N via f is given by N_i = f(M_i).
  • If (N_i) is a filtration of N, the induced filtration on M via f is given by M_i = f^{-1}(N_i).

Note

Let us show that the induced filtrations are legitimate. In the first case,

A_i N_j = A_i f(M_j) = f(A_i M_j) \subseteq f(M_{i+j}) = N_{i+j}.

And in the second,

f(A_i M_j) = A_i\cdot f(M_j) \subseteq A_i N_j \subseteq N_{i+j} \implies A_i M_j \subseteq M_{i+j}.

In particular, if M is a filtered A-module and N\subseteq M is a submodule, the induced filtrations on N and M/N are given by:

N_i = M_i \cap N, \quad (M/N)_i = (M_i + N)/N.

So far everything is natural, but beneath all this a danger lurks.

warningIf f:M\to N is a filtered A-linear map, then M/\mathrm{ker} f and \mathrm{im }f have induced filtrations via quotient module of M and submodule of N. Although M/\mathrm{ker } f \cong \mathrm{im }f as A-modules, it is not an isomorphism in the category of filtered A-modules!

Let us write everything out explicitly. The filtration on the LHS and RHS are given respectively by

(M / \mathrm{ker } f)_i = (M_i + \mathrm{ker } f)/\mathrm{ker } f, \quad (\mathrm{im } f)_i = (\mathrm{im} f) \cap N_i

which gives

(M_i + \mathrm{ker } f)/\mathrm{ker } f \cong M_i / (M_i \cap \mathrm{ker } f) = M_i / \mathrm{ker} (f|_{M_i}) \cong \mathrm{im} (f|_{M_i})

which is not isomorphic to (\mathrm{im} f) \cap N_i in general.

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Completion

Definition.

Let M be a filtered A-module, the completion of M is the following (inverse) limit of A-modules:

\hat M := \varprojlim M/M_n.

completion_diagram

By the description of inverse limits in the category of A-modules, \hat M comprises of the set of all (\ldots, x_n, \ldots, x_2, x_1) \in \prod_{i=1}^\infty (M/M_i) such that for each i, x_{i+1} maps to x_i under the canonical map M/M_{i+1}\to M/M_i.

The canonical maps M\to M/M_n induce an A-linear map M \to \hat M by the universal property of inverse limits. This can be described as follows: for each m\in M, let m_i be its image in M/M_i; then the map takes (m \in M) \mapsto (\ldots, m_n, \ldots, m_2, m_1) \in \hat M. From this we see that:

Lemma 1.

The map i:M\to \hat M is injective if and only if \cap_n M_n = 0, in which case we say the filtration is Hausdorff.

Next, by setting MA, we also have \hat A = \varprojlim A/A_n. Although we defined \hat A as an A-module, one sees by the explicit construction that \hat A has a ring structure. To be specific,

(a_n)_{n=1}^\infty, (b_n)_{n=1}^\infty \in \hat A \implies (a_n) \times (b_n) := (a_n b_n)_{n=1}^\infty.

Exercise A

1. Prove that \hat M has a canonical structure as an \hat A-module.

2. Prove that \hat A is the inverse limit of A/A_n in the category of rings. In particular, the canonical map i : A\to \hat A is a ring homomorphism.

3. Prove that if the filtration on M is Hausdorff, we can define a metric on M via:

d(x, y) := 2^{-|x-y|}, where |z| := \sup\{ n : z\in M_n\},

such that the collection of all cosets \{ m + M_n : m\in M, n \ge 0\} forms a basis for the resulting topology. In fact, d is an ultrametric.

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Limits in Completion

The completion \hat M enables us to take limits of infinite sequences and sums of infinite series in modules.

Definition.

Let m_1, m_2, \ldots \in M be a sequence in a filtered module M. We say the sequence is Cauchy if: for any i, there exists n such that

x_n \equiv x_{n+1} \equiv x_{n+2} \equiv \ldots \pmod {M_i}.

The astute reader would note that if the filtration is Hausdorff, then (m_n) is Cauchy here if and only if it is Cauchy with respect to the metric of exercise A.3.

Definition.

The limit of a Cauchy sequence m_1, m_2, \ldots \in M is defined as follows. For each i, pick n such that x_n \equiv x_{n+1} \equiv \ldots modulo M_i, with image y_i \in M/M_i. Now define:

\lim_{n\to\infty} m_n := (\ldots, y_n, \ldots, y_2, y_1) \in \hat M.

From the definition, it is clear that if (m_n), (m_n') (resp. (a_n)) are Cauchy sequences in M (resp. in A), then (m_n + m_n') and (a_n m_n) are also Cauchy and we have

\begin{aligned}\lim_{n\to\infty} (m_n + m_n') &= (\lim_{n\to\infty} m_n) + (\lim_{n\to\infty} m_n') \in \hat M, \\ \lim_{n\to\infty} (a_n m_n) &= (\lim_{n\to\infty} m_n) (\lim_{n\to\infty} m_n') \in \hat M.\end{aligned}

Example: 𝔪-adic Completion

Pick a ring A with maximal ideal \mathfrak m; we will take the \mathfrak m-adic filtration A_n = \mathfrak m^n. Given a sequence x_i \in A_i for i=0, 1, \ldots, let

y_i := x_0 + \ldots + x_{i-1} \pmod {\mathfrak m^i}, an element of A/A_i.

Then (y_i) is a Cauchy sequence in A and we write:

\sum_{i=0}^\infty x_i := \lim_{n\to\infty} y_n \in \hat A.

Arithmetic Example.

Let A = \mathbb Z and A_n = p^n \mathbb Z for a prime p. The completion \hat A is called the ring of p-adic integers and denoted by \mathbb Z_p. Let us take p = 2 and the element:

y = 1 + 2^2 + 2^4 + 2^6 + \ldots \in \mathbb Z_2.

Then 3y = 3 + 3(2^2) + 3(2^4) + \ldots is congruent to -1 modulo any 2^n. Thus y = -\frac 1 3. In general, an element of \mathbb Z_p can be regarded as having an infinite base-p expansion. Thus the above y \in \mathbb Z_2 would have base-2 expansion (\ldots 1010101)_2. One easily checks that three times this value is (\ldots 1111)_2, which is -1.

Geometric Example

Let A be a ring, B = A[X, Y] and B_n = \mathfrak m^n where \mathfrak m = (X, Y). Again, we can take the infinite sum \alpha = 1 + XY + (XY)^2 + \ldots and check that \alpha(1-XY) = 1. Note that \hat B \cong A[[X,Y]], the ring of formal power series with coefficients in A.

Definition.

Let A be any ring. A formal power series in X with coefficients in A is an expression

f(X) = a_0 + a_1 X + a_2 X^2 + \ldots, where a_i \in A.

Unlike polynomials, we allow infinitely many a_i to be non-zero. Addition and multiplication of formal power series are defined as follows. For f(X) = \sum_{i=0}^\infty a_n X^n and g(X) = \sum_{j=0}^\infty b_n X^m,

\begin{aligned} f(X) + g(X) &=(a_0 + b_0) + (a_1 + b_1)X + (a_2 + b_2)X^2 + \ldots \\ f(X)\times g(X) &= (a_0 b_0) + (a_0 b_1 + a_1 b_0)X + (a_0 b_2 + a_1 b_1 + a_2 b_0)X^2 + \ldots \end{aligned}

This gives a ring structure on the set A[[X]] of formal power series. To define rings of formal power series in multiple variables, we set recursively

A[[X_1, \ldots, X_n]] := (A[[X_1, \ldots, X_{n-1}]])[[X_n]]].

As another example, let us take the \mathfrak m-adic completion for A = \mathbb C[X, Y]/(Y^2 - X^3 + X) and \mathfrak m = (X, Y). We will prove later that

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

and that the map \mathbb C[[Y]] \to \hat A is an isomorphism. Geometrically, this means when we project E: Y^2 = X^3 - X to the Y-axis, the map is locally invertible at the origin.

completion_diagram_v1

[ Image edited from GeoGebra plot. ]

warningFunctorially, the ring A[[X, Y]] behaves quite differently from A[X, Y], because as an A-module, it is a countably infinite direct product of copies of A, unlike A[X, Y] which is a direct sum. If we follow the earlier guideline, it is (for example) generally false that B \otimes_A A[[X, Y]] \cong B[[X, Y]] for any A-algebra B.

Exercise B

1. Prove that the canonical map \mathbb C[[X]] \to \mathbb C[[X, Y]]/(Y^2 - X^3 + X) is not an isomorphism.

2. Let A be a filtered ring. Prove that if f(X) \in A[[X]] is a_0 + a_1 X + a_2 X^2 + \ldots, then f defines a map

A_1 \to \hat A, \quad f(\alpha) := \sum_{n=0}^\infty a_n \alpha^n \in \hat A.

Prove that if we fix \alpha \in A_1, we get a ring homomorphism A[[X]] \to \hat A, f\mapsto f(\alpha).

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Commutative Algebra 53

Graded Rings

Definition.

A grading on a ring A is a collection of additive subgroups A_0, A_1, \ldots \subseteq A such that

A = A_0 \oplus A_1 \oplus A_2 \oplus \ldots

as abelian groups, and A_i A_j \subseteq A_{i+j} for any i, j\ge 0, i.e..

a\in A_i, b\in A_j \implies ab \in A_{i+j}.

graded ring is a ring A with a specified grading.

Note

The notation A = \oplus_i A_i means every a\in A can be uniquely written as a finite sum a_0 + a_1 + \ldots + a_n for some a_i \in A_i (uniqueness holds up to appending or removal of 0\in A_m). Then a_d is called the degree-d component of a.

An a\in A is said to be homogeneous of degree d if a \in A_d; we write \deg a = d. Note that d is unique if a\ne 0.

Example

The standard example of a graded ring is the ring of polynomials B = A[X_1, \ldots, X_n] over some ring A, where B_d has A-basis given by the set of monomials X_1^{m_1} \ldots X_n^{m_n} satisfying m_i \ge 0 and \sum_i m_i = d.

Lemma 1.

A_0 is a subring of A and each A_i is a module over A_0.

Proof

For the first statement, since A_0 A_0 \subseteq A_0 it suffices to show 1 \in A_0; we may assume 1\ne 0. Write 1 = a_0 + \ldots + a_n with a_i \in A_i and a_n \ne 0. Suppose n > 0. For any b_m \in A_m we have

b_m = b_m\cdot 1 = b_m(a_0 + \ldots + a_n) = b_m a_0 + \ldots + b_m a_n.

Since the LHS is homogeneous of degree m, we have b_m a_n = 0. Thus a_n A_m = 0 for any m so we have a_n A = 0. This gives a_n = 0, a contradiction.

The second statement is clear. ♦

By definition if a, b\in A are homogeneous of degrees m and n, then ab is homogeneous of degree m+n. We also have:

Lemma 2.

Suppose A is an integral domain with grading. If a,b\in A are non-zero elements such that ab is homogeneous, then so are a and b.

Proof

Write a = \sum_m a_m and b = \sum_n b_n as sums of their components. Let m (resp. m’) be the minimum (resp. maximum) degree for which a_m \ne 0 (resp a_{m'} \ne 0). Similarly, let n (resp. n’) be the minimum (resp. maximum) degree for which b_n \ne 0 (resp b_{n'} \ne 0). By definition a_m b_n, a_{m'}b_{n'} \ne 0. Since ab is homogeneous we have m+n = m'+n' and thus m = m', n = n'. So a and b are homogeneous. ♦

Exercise A

Find a graded ring A, with a,b \in A-\{0\} such that ab is homogeneous but a and b are not. [ Hint: come back to this exercise after finishing the whole article. ]

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Graded Modules

For the rest of this article, we fix a graded ring A.

Definition.

A grading on an A-module M is a collection of additive subgroups M_0, M_1, \ldots \subseteq M such that

M = M_0 \oplus M_1 \oplus M_2 \oplus \ldots

as additive groups, and A_i M_j \subseteq M_{i+j} for any i,j \ge 0.

Note

We need a fixed grading on the base ring A before we can talk about grading on A-modules.

For example, every graded ring is a graded module over itself (with the same grading).

  • As before if we write m\in M as a sum m_0 + m_1 + \ldots + m_n with m_i \in M_i, then m_d is called the degree-d component of m.
  • Also m\in M is homogeneous of degree d if m \in M_d; again write \deg m = d.

The following result is quite important for grading of submodules.

Proposition 1.

Let M be a graded A-module and N\subseteq M be a submodule. The following are equivalent.

  1. There is a generating set for N comprising of homogeneous elements.
  2. If n\in N, all homogeneous components of n lie in N.
  3. We have N = (N \cap M_0) \oplus (N \cap M_1) \oplus \ldots.

Proof

(1⇒3) It suffices to show N = \sum_i (N \cap M_i). Pick a generating set S for N comprising of non-zero homogeneous elements. For each n\in N, write n = \sum_{i=1}^k a_i m_i with a_i \in A and m_i \in S with \deg m_i = d_i. Fix d\ge 0 and let n’ be the degree-d component of n. Then n’ is the sum of the degree-d components of a_i m_i. Hence

n' = \sum_{i=1}^k a_i' m_i, where a_i' is the degree-(d-d_i) component of a_i

so n'\in N \cap M_d.

(3⇒2) Let n\in N; we can write n = n_0 + n_1 + \ldots + n_d where n_i \in N \cap M_i. Since n_i \in M_i, it is homogeneous of degree i; thus n_i is the degree-i component of n and it lies in N.

(2⇒1) Pick any generating set S of N, then take the homogeneous components of all m\in S to obtain a homogeneous generating set.

Definition.

Let M be a graded A-module. A submodule N\subseteq M is said to be graded if it satisfies the conditions of proposition 1.

An ideal \mathfrak a\subseteq A is said to be graded if it is graded as a submodule.

Proposition 1 immediately gives the following.

Corollary 1.

Given a graded module M, with collection of graded submodules (N_i), graded submodule N, and graded ideal \mathfrak a\subseteq A,

\sum_i N_i, \quad \cap_i N_i, \quad \mathfrak a N

are all graded submodules of M.

Proof

Since each N_i is generated by homogeneous elements, so is \sum N_i; thus \sum N_i is graded. Suppose n\in \cap_i N_i, if n’ is a homogeneous component of n, then for each i, n \in N_i \implies n' \in N_i and hence n' \in \cap N_i. Thus \cap N_i is graded. Finally, if S (resp. T) is a generating set of \mathfrak a (resp. N) comprising of homogeneous elements, then \{an : a\in S, n\in T\} is a generating set of \mathfrak aN comprising of homogeneous elements. ♦

Exercise B

Decide if each statement is true.

  • If \mathfrak a\subseteq A is a graded ideal of A, then r(\mathfrak a) is graded.
  • If N, P \subseteq M are graded submodules, then (N : P) = \{a \in A: aP \subseteq N\} is a graded ideal.

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Quotient

Proposition 2.

Let N be a graded submodule of a graded module M. Then M/N has a canonical grading given by

(M/N)_i := M_i / (N\cap M_i) \hookrightarrow M/N.

Proof

Note that we have A_i (M/N)_j \subseteq (M/N)_{i+j}. It remains to show M/N is a direct sum of (M/N)_i.

Let m\in M. Write m = \sum_i m_i with m_i \in M_i. Then m+N \in M/N is the sum of images of m_i + (N \cap M_i) \in M/(N\cap M_i) in M/N. So M/N = \sum_i (M/N)_i.

Suppose m + N can be written as \sum_i [m_i + (N\cap M_i)] and \sum_i [m_i' + (N\cap M_i)]. Then

m+N = (\sum_i m_i) + N = (\sum_i m_i') + N \implies \sum_i (m_i - m_i') \in N.

By condition 2 of proposition 1, each m_i - m_i' \in N so m_i + (N\cap M_i) = m_i' + (N\cap M_i). ♦

Corollary 2.

If \mathfrak a\subseteq A is a graded ideal, then A/\mathfrak a is a graded ring under the above grading.

Proof

Since A_i A_j \subseteq A_{i+j}, multiplication gives

A_i / (\mathfrak a \cap A_i) \times A_j / (\mathfrak a \cap A_j) \longrightarrow A_{i+j} / (\mathfrak a \cap A_{i+j}).

Example

1. Suppose A is the coordinate ring of a variety V\subseteq \mathbb A^n, so that A \cong k[X_1, \ldots, X_n]/I(V). If I(V) is homogeneous, then A is a graded ring. In particular, the group of units in A is k^*, the multiplicative group k - \{0\}. This does not work for non-homogeneous I(V), e.g. the unit group of k[X, Y]/(XY - 1) contains \{ c X^n : n \in \mathbb Z, c \in k^*\} (does equality hold?).

2. We will do exercise C here, i.e. show that A = \mathbb C[X, Y, Z]/(Z^2 - X^2 - Y^2) is not a UFD. Indeed we have the equality Z \cdot Z = (X+iY)(X - iY) in A; we claim that Z, X+iY, X-iY \in A are all irreducible. By lemma 2, since Z is homogeneous of degree 1, we can only factor it as a product of a degree-0 and a degree-1 element. But all non-zero degree-0 elements of A are units (see example 1). Similarly X+iY, X-iY are irreducible and Z, X+iY, X-iY are clearly not associates.

3. By the same reasoning, \mathbb C[W, X, Y, Z]/(Z^2 - W^2 - X^2 - Y^2) is not a UFD.

Note: however \mathbb C[X_1, \ldots, X_n]/(X_1^2 + \ldots + X_n^2) is a UFD for all n\ge 5, as we will see later.

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Graded Maps

Definition.

Let M and N be graded A-modules, and f:M\to N be A-linear. We say f is graded if f(M_i) \subseteq N_i for each i.

Immediately we have:

Lemma 3.

If f:M\to N is a graded map of graded A-modules, then \mathrm{ker} f is a graded submodule of M and \mathrm{im} f is a graded submodule of N.

Proof

If m \in M and m = m_0 + \ldots + m_d with m_i \in M_i, then f(m_i) \in N_i and hence f(m) = f(m_0) + \ldots + f(m_d) is the unique decomposition of f(m) into its homogeneous components. The rest is an easy exercise. ♦

Proposition 3 (First Isomorphism Theorem).

For a graded map f:M\to N of graded modules, we have an isomorphism

g : M/\mathrm{ker} f \longrightarrow \mathrm{im } f, \quad m + \mathrm{ker } f \mapsto f(m).

in the category of graded A-modules.

Proof

Let us show that g is graded. The grading on the LHS is given by i\mapsto M_i / (\mathrm{ker} f \cap M_i). Since f is graded, g takes the LHS into f(M_i) = N_i.

Since g is bijective, it remains to show that g^{-1} is also graded. For n\in N_i, let m = g^{-1}(n). Write m = m_0 + \ldots + m_d as a sum of homogeneous components; since g(m_i) is the degree-i homogeneous component of g(m) we have g(m) = g(m_i) so m = m_i. ♦

Exercise C

State and prove the remaining two isomorphism theorems.

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Commutative Algebra 52

Direct Limits of Rings

Let ((A_i), (\beta_{ij})) be a directed system of rings. Regard them as a directed system of abelian groups (i.e. ℤ-modules) and take the direct limit A.

Proposition 1.

The abelian group A has a natural structure of a commutative ring.

Note

General philosophy of the direct limit: “if something happens at index i and another thing happens at index j, then by picking k greater than i and j, both things may be assumed to happen at the same index”. The cautious reader is advised to fill in the gaps in the following proof.

Sketch of Proof

For each i\in J, let \epsilon_i : A_i \to A be the canonical map.

Given a,b \in A, by proposition 2 here there exists j\in J and a', b' \in A_j such that a = \epsilon_j(a'), b = \epsilon_j(b'). Now define multiplication in A by a\times b := \epsilon_j(a'b'). This does not depend on our choice of j and a',b' \in A_j. Indeed, if we have another index i\in J and a_1, b_1 \in A_i such that a = \epsilon_i(a_1), b = \epsilon_i(b_1), by proposition 2 here again pick k\in J greater than i and j such that

\beta_{ik}(a_1) = \beta_{jk}(a'), \ \beta_{ik}(b_1) = \beta_{jk}(b') \implies \beta_{ik}(a_1 b_1) = \beta_{jk}(a'b').

This gives us the desired equality

\epsilon_i(a_1 b_1) = \epsilon_k(\beta_{ik}(a_1 b_1)) = \epsilon_k (\beta_{jk}(a'b')) = \epsilon_j(a'b').

Clearly product in A is commutative. To show associativity, given a,b,c\in A, pick j\in J and a',b',c' \in A_j such that a = \epsilon_j(a'), b = \epsilon_j(b') and c = \epsilon_j(c'). Then a(bc) = \epsilon_j(a'(b'c')) = \epsilon_j((a'b')c') = (ab)c.

To define 1\in A, we pick any index i\in J and set 1_A := \epsilon_i(1_{A_i}). ♦

Exercise A

1. Prove that 1_A in the above proof is well-defined, and 1_A \times a = a for all a\in A.

2. Prove that the resulting ring A with the canonical \epsilon_i : A_i \to A gives the direct limit of A_i in the category of rings.

3. Prove that if the direct limit of rings A_i is zero, then A_i = 0 for some i. [ Hint: a ring is zero if and only if 1 = 0. ]

4. Suppose (B_i)_{i\in I} is an arbitrary collection of A-algebras. For each finite subset L = \{i_1, \ldots, i_n\} \subseteq I, define B_L := B_{i_1} \otimes_A \ldots \otimes_A B_{i_n}. Define a directed system of B_L over the directed set of all finite subsets of I, ordered by inclusion L\subseteq L'. The tensor product of B_i over A is defined to be the direct limit of this system. Prove that this gives the coproduct of (B_i)_{i\in I} in the category A-algebras.

Note

Since direct limits are denoted by \varinjlim, we will write \varprojlim for the earlier limits and call them inverse limits. For most cases of interest, inverse limits will be taken over J such that J^{op} is directed.

warningEven over directed sets, taking the inverse limit is not exact. A useful criterion for determining exactness is given by the Mittag-Lefler condition, which we will not cover (for now).

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Taking Stock

We have seen many constructions which commute and some which do not. In the following examples, (M_i) is an arbitrary collection of modules; M and N are modules, B is an A-algebra and S\subseteq A is a multiplicative subset.

Case 1 (proposition 1 here): S^{-1} (\oplus_i M_i) = \oplus_i S^{-1} M_i but S^{-1}(\prod_i M_i) \ne \prod_i S^{-1}M_i in general.

Case 2 (exercise A here): \mathfrak a (\oplus_i M_i) = \oplus_i \mathfrak a M_i but \prod_i \mathfrak a M_i \ne \mathfrak a (\prod_i M_i).

Case 3 (corollary 1 here): a direct sum of projective modules is projective. A direct product of projective modules is not projective in general, but a counter-example is not too easy to construct.

Case 4 (exercise B here): for any collection (N_i) of submodules of M, we have S^{-1}(\sum_i N_i) = \sum_i S^{-1}N_i but S^{-1}(\cap_i N_i) \ne \cap_i S^{-1} N_i in general.

Case 5 (exercise A here): if M is a flat A-module, then M^B is a flat B-module.

Case 6 (proposition 1 here): we have (\oplus_i M_i) \otimes_A N \cong \oplus_i (M_i \otimes_A N).

Case 7: more generally, we have (\mathrm{colim}_{i\in J} M_i) \otimes_A N \cong \mathrm{colim}_{i\in J} (M_i \otimes_A N) if (M_i)_{i\in J} is a diagram of A-modules of type J.

Case 8: hence we have an isomorphism (\mathrm{colim}_{i\in J} M_i)^B \cong \mathrm{colim}_{i\in J}(M_i)^B of B-modules; in particular S^{-1}(\mathrm{colim}_{i\in J} M_i) \cong \mathrm{colim}_{i\in J} (S^{-1} M_i).

Case 9 (proposition 3 here): if f: M_1 \to M_2 is surjective, then f\otimes_A 1_N : M_1 \otimes_A N \to M_2 \otimes_A N is surjective; however, if f is injective f\otimes_A 1_N is not injective in general.

Exercise B

Prove that there is always a canonical map between (\prod_i M_i) \otimes_A N and \prod_i (M_i \otimes_A N). Find an example where the map is not an isomorphism.

If M is a projective A-module, must M^B = B\otimes_A M be a projective B-module?

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Duality Principle

Remembering all the above relations may seem like a pain: in general if we have n constructions, we have about O(n^2) relations to learn. It turns out most of these constructions can be classified as either “left-adjoint-like” or “right-adjoint-like”, which saves us a whole lot of effort in remembering them.

In the following table, constructions on the same side tend to commute or have consistent properties. Constructions on different sides may commute under specific additional conditions (e.g. finiteness, noetherianness).

Left-adjoint-like Right-adjoint-like
Sum of submodules Intersection of submodules
Coproducts Products
Right-exact functors Left-exact functors
Pushouts Pullbacks, fibre products
Direct sum of modules Direct product of modules
Tensor products of modules Hom modules
HomA(-, M) HomA(M, -)
Colimits / Direct limits Limits / Inverse limits
Injective maps Surjective maps
Quotient modules Submodules
Induced modules M\mapsto M^B. (Coinduced modules)
Projective / free modules (Injective modules)
Localization
Multiplying ideal by module: \mathfrak a M

warningDo consider this table as a very rough guide. For example, if 0 \to N \to M \to P \to 0 is a short exact sequence of A-modules, we do not get a right-exact sequence \mathfrak a N \to \mathfrak a M \to \mathfrak a P \to 0. [ Take 0\to 2\mathbb Z \to \mathbb Z \to \mathbb Z / 2\mathbb Z \to 0 and \mathfrak a = 2\mathbb Z. ]

Also note that the terms in brackets have not been defined yet.

Further Examples

1. We have \mathrm{Hom}_A(\oplus_i N_i, M) \cong \prod_i \mathrm{Hom}_A(N_i, M).

2. The functor \mathrm{Hom}_A(-, M) takes a right-exact sequence to a left-exact sequence.

3. Recall that the colimit of a diagram of A-modules was constructed by taking a quotient of the direct sum of these modules. Dually, its limit can be constructed by taking a submodule of the direct product.

4. The tensor product was constructed by taking a quotient of a free (hence projective) module.

Exercise C (Coinduced Modules)

1. Let B be an A-algebra. Prove that for an A-module M, M_B := \mathrm{Hom}_A(B, M), the set of all A-linear maps B\to M, has a natural structure of a B-module.

2. Prove that we get a functor

F :A\text{-}\mathbf{Mod} \longrightarrow B\text{-}\mathbf{Mod}, \quad M \mapsto M_B

such that there is a natural bijection

\mathrm{Hom}_A(N, M) \cong \mathrm{Hom}_B(N, M_B).

for any B-module N. Thus F = \mathrm{Hom}_A(B, -) is right-adjoint to the forgetful functor B\text{-}\mathbf{Mod} \to A\text{-}\mathbf{Mod}.

We call M_B the coinduced B-module from M.

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Commutative Algebra 51

Limits Are Left-Exact

By example 6 and proposition 2 in the previous article, one is inclined to conclude that taking the colimit in \mathcal C = A\text{-}\mathbf{Mod} is a right-exact functor, but there is a rather huge issue here: the functors are between \mathcal C and \mathcal D = \mathcal C^J, the category of diagrams in \mathcal C while we only defined exactness of functors between categories of modules. The proper way to do this is to introduce the framework of abelian categories and extend our concept of additive functors and exact functors there. However, doing this will take us too far afield so we will prove it directly (which is, admittedly, a bit of a cop out).

Proposition 1.

Let J be an index category, and D', D, D'' : J\to A\text{-}\mathbf{Mod} be diagrams of type J. For concreteness, write these diagrams as

((N_i), (\beta_e^N : N_i \to N_j)), \ ((M_i), (\beta_e^M : M_i \to M_j)),\ ((P_i), (\beta_e^P : P_i \to P_j))

where i\in J and e:i\to j. Let D'\to D \to D'' be morphisms, written as a collection of N_i\stackrel {\phi_i} \to M_i \stackrel {\psi_i} \to P_i over i\in J. Then

\left( \begin{aligned}0\to N_i\stackrel {\phi_i} \to M_i \stackrel {\psi_i} \to P_i\\ \text{exact for each } i\in J\end{aligned}\right) \implies 0 \to \lim N_i \to \lim M_i \to \lim P_i \text{ exact.}

\left( \begin{aligned}N_i\stackrel {\phi_i} \to M_i \stackrel {\psi_i} \to P_i \to 0\\ \text{exact for each } i\in J\end{aligned}\right) \implies \mathrm{colim} N_i \to \mathrm{colim} M_i \to \mathrm{colim} P_i \to 0\text{ exact.}

Note

In summary, taking the limit is left-exact while taking the colimit is right-exact.

Proof

We prove the second claim, leaving the first as an exercise. By proposition 1 here, \mathrm{colim} M_i is concretely described as follows. Take the quotient of \oplus_{i\in J} M_i by all m_i - \beta_e^M(m_i), where e:i\to j is an arrow in J, m_i \in M_i and \beta_e^M(m_i)\in M_j are identified with their images in \oplus_i M_i.

With this description, clearly \mathrm{colim} M_i \to \mathrm{colim} P_i is surjective. Also, composing \mathrm{colim} N_i \to \mathrm{colim} M_i \to \mathrm{colim} P_i is the zero map so \mathrm{im}(\mathrm{colim} \phi_i) \subseteq \mathrm{ker}(\mathrm{colim} \psi_i). Now write \phi : \oplus_i N_i \to \oplus_i M_i for \oplus \phi_i and \psi : \oplus_i M_i \to \oplus_i P_i for \oplus \psi_i.

Conversely, let x\in \oplus_i M_i represent an element in the kernel of \mathrm{colim} \psi_i. Thus \psi(x) \in\oplus_i P_i is a finite sum of p_i - \beta_e^P(p_i). Since \psi_i is surjective, we can write such a term as

\psi_i(m_i') - \beta_e^P(\psi_i(m_i')) = \psi_i(m_i') - \psi_j(\beta_e^M(m_i')) = \psi(m_i' - \beta_e^M(m_i'))

for some m_i'\in M_i. Since \psi(x) is a finite sum of \psi(m_i' - \beta_e^M(m_i')), we can replace x by another representative such that \psi(x) = 0. Then x = \phi(y) for some y\in \oplus N_i. ♦

warningNeither the limit nor the colimit functor is exact in general. For the colimit case, consider the following commutative diagram of A-modules

pushout_counterexample

where all maps A\to A are identities. The rows are short exact sequences and the squares all commute, but taking the colimit of the columns gives

0 \longrightarrow A^2 \longrightarrow A \longrightarrow 0 \longrightarrow 0

which is not exact.

Exercise A

Find an example for the case of limits.

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Direct Limits

We will describe a special case where taking the colimit is exact.

Given a poset (S, \le), we recall the category \mathcal C(S) whose objects are elements of S, and between any x,y\in S, |\mathrm{hom}(x, y)| \le 1 with equality if and only if x \le y. Composition is the obvious one.

Definition.

A poset (S, \le) is called a directed set if for any a,b\in S, there is a c\in S such that a\le c and b\le c.

In other words, a poset is directed if every finite set has an upper bound.

Definition.

If J is an index category obtained from \mathcal C(S) for some directed set S, then a diagram in \mathcal C of type J is called a directed system. The colimit of (A_i)_{i\in J} is called the direct limit and denoted by

\varinjlim_{i\in J} A_i.

In other words, direct limit = colimit over directed set. We will abuse notation a little and regard J as the directed set itself.

To avoid set-theoretic difficulties, the directed set J is always assumed to be non-empty.

Example

In exercise C.3 here, for a multiplicative S\subseteq A and A-module M, we have an isomorphism of A-modules

\mathrm{colim}_{f\in S} M_f \cong M_S

where f\le g if g is a multiple of f. Since S is multiplicative, any {fg} has an upper bound fg. Hence M_S is the direct limit of M_f over f\in S:

\varinjlim_{f\in S} M_f \cong M_S.

Similarly, we have the following direct limit in the category of rings:

\varinjlim_{f\in S} A_f \cong A_S.

Next we will discuss the general direct limit in the categories A-Mod and Ring.

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Direct Limit of Modules

Let A be a fixed ring; the following holds for direct limits in the category of A-modules.

Proposition 2.

Suppose ((M_i)_{i\in J}, (\beta_{ij})_{i\le j}) is a directed system of A-modules over a directed set J. Let

M = \varinjlim_{i\in J} M_i, with canonical \epsilon_i : M_i \to M for each i\in J.

Then for each m\in M, there exists an i\in J and m_i \in M_i such that \epsilon_i(m_i) = m.

Also if m_i\in M_i satisfies \epsilon_i(m_i) = 0, then there exists j\ge i such that \beta_{ij}(m_i) = 0 \in M_j.

Note

The philosophy is that “whatever happens in the direct limit happens in M_j for some sufficiently large index j“.

Proof

By proposition 1 here, the colimit M is described concretely by taking the quotient of P = \oplus_{i\in J} M_i (with canonical \nu_i : M_i \to P) by relations of the form

\nu_k(m_k) - \nu_l\beta_{kl}(m_k),\ m_k \in M_i,\ k\le l\ (k, l\in J).

Hence any m\in M can be written as \epsilon_{i_1}(m_1) + \ldots + \epsilon_{i_N}(m_N) for m_1 \in M_{i_1}, \ldots, m_N \in M_{i_N}. But J is a directed set, so we can pick index j\in J such that j\ge i_1, \ldots, i_N; then

m = \epsilon_j(m_j), where m_j = \beta_{i_1 k}(m_1) + \ldots + \beta_{i_N k}(m_N),

proving the first claim.

For the second claim, if \epsilon_i(m_i) = 0 then \nu_i(m_i) \in \oplus_i M_i is a finite sum of the above relations. Pick an index j\in J larger than i and all indices kl in the sum; then \beta_{ij}(m_i) is the sum of the images of these relations in M_k. But each such relation has image \beta_{kj}(m_k) - \beta_{lj}\beta_{kl}(m_k) = 0 in M_k, so \beta_{ij}(m_i) = 0 as desired. ♦

Corollary 1.

If ((M_i), (\beta_{ij})) is a directed system of A-modules such that \beta_{ij} are all injective, then

\epsilon_j : M_j \longrightarrow \varinjlim_i M_i

is also injective for each j\in J.

Finally we have:

Proposition 3.

Let ((N_i), (\beta_{ij})) and ((M_i), (\gamma_{ij})) be directed systems of A-modules and \phi_i : N_i \to M_i be a morphism of the directed systems, i.e. for any i\le j, we have \gamma_{ij} \circ \phi_i = \phi_j \circ \beta_{ij} : N_i \to M_j.

If each \phi_i is injective, so is \phi : \varinjlim M_i \to \varinjlim N_i.

Since taking the colimit is right-exact by proposition 1, we see that taking the direct limit is exact. 

Proof

Write \epsilon_i^M : M_i \to \varinjlim M_i and \epsilon_i^N : N_i \to \varinjlim N_i for the canonical maps.

Suppose \phi(m) = 0 for m \in \varinjlim M_i. By proposition 2, we have m = \epsilon_i^M(m_i) for some m_i \in M_i; then

0 = \phi(m) = \phi(\epsilon_i^M(m_i)) = \epsilon_i^N(\phi_i(m_i))

so by proposition 2 again, there exists j\ge i such that \gamma_{ij}(\phi_i(m_i)) = 0, so \phi_j(\beta_{ij}(m_i)) = 0. Since \phi_j is injective we have \beta_{ij}(m_i) = 0 so m =  \epsilon_i^M(m_i) = \epsilon_j^M\beta_{ij}(m_i) = 0. ♦

Exercise B

Describe the direct limit of sets (S_i) over J. State and prove an analogue of proposition 2.

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