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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