Commutative Algebra 13

Zariski Topology for Rings

In this article, we generalize earlier results in algebraic geometry to apply to general rings. Recall that points on an affine variety V correspond to maximal ideals \mathfrak m\subset k[V]. For general rings, we have to switch to taking prime ideals because of the following.

Lemma 1.

If f : A \to B is a ring homomorphism and \mathfrak q is a prime ideal of B, then \mathfrak p := f^{-1}(\mathfrak q) is a prime ideal of A.

Proof

The composition A \stackrel f\to B \to B/\mathfrak q has kernel f^{-1}(\mathfrak q) = \mathfrak p. Hence this induces an injective ring homomorphism A/\mathfrak p \hookrightarrow B/\mathfrak q. Since B/\mathfrak q is a domain, so is A/\mathfrak p so \mathfrak p is prime. ♦

However it is not true that if \mathfrak q is maximal in B, \mathfrak p must be maximal in A. For instance consider the inclusion i:\mathbb Z \hookrightarrow \mathbb Q. The zero ideal is maximal in \mathbb Q but i^{-1}((0)) = (0) is not maximal in \mathbb Z.

Now we define our main object of interest.

Definition.

Given a ring A, the spectrum of A is its set of prime ideals \mathrm{Spec} A. The Zariski topology on \mathrm{Spec A} is defined as follows: V\subseteq \mathrm{Spec A} is closed if and only if it is of the following form

V(S) = \{ \mathfrak p \in \mathrm{Spec A} : \mathfrak p \supset S\},

for some subset S\subseteq A.

Note that since V(S) = V(\mathfrak a) where \mathfrak a is the ideal generated by S, we lose no generality by taking S to be ideals of A. If S = {f} we will write V(f) instead of V({f}) or V((f)) to reduce the clutter.

Immediately, we prove that the above gives a bona fide topology.

Proposition 1.

For any collection of ideals (\mathfrak a_i) of A and ideals \mathfrak a, \mathfrak b of A, we have

  • V(1) = \emptyset, V(0) = \mathrm{Spec} A,
  • \cap_i V(\mathfrak a_i) = V(\sum_i \mathfrak a_i),
  • V(\mathfrak a) \cup V(\mathfrak b) = V(\mathfrak a\cap \mathfrak b) = V(\mathfrak {ab}).

Proof

The first claim is obvious. For the second, \mathfrak p contains \mathfrak a_i for each i if and only if it contains \sum_i \mathfrak a_i.

Finally, for the third claim, we have \mathfrak {ab} \subseteq \mathfrak a \cap \mathfrak b \subseteq \mathfrak a and \mathfrak a\cap \mathfrak b \subseteq \mathfrak b so we have

V(\mathfrak {ab}) \supseteq V(\mathfrak a\cap \mathfrak b) \supseteq V(\mathfrak a) \cup V(\mathfrak b).

Finally suppose \mathfrak p \not\in V(\mathfrak a)\cup V(\mathfrak b). Then \mathfrak p \not\supseteq \mathfrak a and \mathfrak p \not\supseteq \mathfrak b so there are x \in \mathfrak a - \mathfrak p and y \in \mathfrak b - \mathfrak p. Then xy \in \mathfrak{ab} - \mathfrak p since \mathfrak p is prime so \mathfrak p \not\in V(\mathfrak{ab}). Hence if  \mathfrak p \in V(\mathfrak{ab}) then \mathfrak p\in V(\mathfrak a) \cup V(\mathfrak b). ♦

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Structure of Spec A

Thanks to Zorn’s lemma, we have the following results on the structure of Spec A.

Proposition 2.

If A is a ring, every proper ideal \mathfrak a\subsetneq A is contained in a maximal ideal.

Proof

Let \Sigma be the collection of all proper ideals \mathfrak b\subsetneq A containing \mathfrak a, ordered partially by inclusion. We claim that every chain \Sigma' \subseteq \Sigma has an upper bound. Since \mathfrak a \in \Sigma we may assume without loss of generality \Sigma' \ne\emptyset.

Take \mathfrak c = \cup_{\mathfrak b \in \Sigma'} \mathfrak b. Let us show that \mathfrak c is an ideal of A.

Clearly 0 \in \mathfrak c. Let x,y\in \mathfrak c and a\in A. Then x\in \mathfrak b and y\in \mathfrak b' for some \mathfrak b, \mathfrak b' \in \Sigma'. Since \Sigma' is totally ordered, either \mathfrak b'\subseteq \mathfrak b or \mathfrak b' \subseteq \mathfrak b'. Assuming the former, this gives x, y\in \mathfrak b so ax - y\in \mathfrak b \subseteq \mathfrak c. Thus \mathfrak c is an ideal of A.

Since none of the \mathfrak b\in \Sigma' contains 1, we have 1\not\in \mathfrak c. Hence \mathfrak c \in \Sigma is an upper bound of \Sigma'.

By Zorn’s lemma \Sigma has a maximal element, which is precisely a maximal ideal of A containing \mathfrak a

We say that \mathfrak p \in \mathrm{Spec} A minimal if it is a minimal element with respect to inclusion.

Proposition 3.

Any \mathfrak p\in \mathrm{Spec} A contains a minimal prime \mathfrak q.

Proof

Let \Sigma be the collection of all prime ideals of A contained in \mathfrak p, ordered partially by reverse inclusion, i.e. \mathfrak q \le \mathfrak q' if \mathfrak q \supseteq \mathfrak q'. To apply Zorn’s lemma we need to show that every chain \Sigma'\subseteq \Sigma has a maximal element (i.e. minimal with respect to inclusion). Without loss of generality we may assume \Sigma' \ne \emptyset.

Let \mathfrak p' = \cap_{\mathfrak q \in \Sigma'} \mathfrak q, an ideal of A. We need to show that it is prime. Suppose x,y\in A - \mathfrak p' so there exist \mathfrak q_1, \mathfrak q_2\in \Sigma' such that x\not\in \mathfrak q_1 and y\not\in \mathfrak q_2. Since \Sigma' is a chain either \mathfrak q_1 \subseteq \mathfrak q_2 or \mathfrak q_2 \subseteq \mathfrak q_1; assuming the former we have x,y\not\in \mathfrak q_1. Since \mathfrak q_1 is prime we have xy\not\in \mathfrak q_1 so xy\not\in \mathfrak p'.

Now apply Zorn’s lemma to obtain a maximal element, which is a minimal prime contained in \mathfrak p. ♦

Corollary.

If A is non-trivial, \mathrm{Spec} A must contain a maximal ideal and a minimal prime ideal.

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Homomorphisms

Suppose f:A\to B is a ring homomorphism. We saw that this induces a map

f^* : \mathrm{Spec} B \to \mathrm{Spec} A,\quad \mathfrak q \mapsto f^{-1}(\mathfrak q).

Proposition 4.

The map f^* is continuous with respect to the Zariski topology; in fact

(f^*)^{-1}(V(\mathfrak a)) = V(f(\mathfrak a)).

Note that f(\mathfrak a) is not an ideal of B in general, but that is okay since we defined V on arbitrary subsets.

Proof

Let \mathfrak q \in \mathrm{Spec} B. This lies in the LHS if and only if f^*(\mathfrak q) = f^{-1}(\mathfrak q) contains \mathfrak a, if and only if \mathfrak q contains f(\mathfrak a). ♦

Clearly the identity ring homomorphism on A induces the identity map on \mathrm{Spec A}, also for ring homomorphisms f:A\to B and g:B\to C we have

(g\circ f)^* = f^* \circ g^* : \mathrm{Spec} C \to \mathrm{Spec} A.

Exercise A

Prove that for an ideal \mathfrak a\subseteq A, the canonical map \pi : A \to A/\mathfrak a induces

\pi^* : \mathrm{Spec}(A / \mathfrak a) \to \mathrm{Spec} A

which is a subspace embedding onto the closed subset V(\mathfrak a).

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Comparison with Algebraic Geometry

Proposition 5.

Let V = V(\mathfrak a) be a closed subset of \mathrm{Spec} A. Then

\bigcap \{ \mathfrak p : \mathfrak p \in V\} = r(\mathfrak a),

the radical of \mathfrak a.

Proof

(⊇) Suppose x\in A, x^n \in \mathfrak a for some n > 0. Now each \mathfrak p \in V contains \mathfrak a and hence contains x^n. Since \mathfrak p is prime we have x\in \mathfrak p.

(⊆) Suppose x\in A, \mathfrak a does not contain any power of x. We will prove the existence of \mathfrak p \supseteq \mathfrak a not containing x.

Let \Sigma be the set of all ideals \mathfrak b\supseteq \mathfrak a not containing any power of x; thus \mathfrak a \in \Sigma. Order \Sigma by inclusion. We will show that any chain \Sigma' \subseteq \Sigma has an upper bound. Indeed, let \mathfrak c = \cup_{\mathfrak b \in \Sigma'} \mathfrak b. As before, since \Sigma' is a chain of ideals ordered by inclusion, \mathfrak c is an ideal of A. It does not include any power of x so \mathfrak c\in \Sigma.

Hence by Zorn’s lemma, \Sigma has a maximal element \mathfrak p; by construction \mathfrak p contains \mathfrak a but not any power of x. It remains to show \mathfrak p is prime: suppose not so there are y, z\in A - \mathfrak p with yz \in \mathfrak p. Now \mathfrak p + (y) and \mathfrak p + (z) strictly contain \mathfrak p, so by maximality of \mathfrak p, we have \mathfrak p + (y), \mathfrak p + (z) \not\in\Sigma. Hence

\left. \begin{aligned} \exists m > 0,\ x^m \in \mathfrak p+(y) \\ \exists n>0, \ x^n \in \mathfrak p + (z)\end{aligned} \right\} \implies x^{m+n} \in (\mathfrak p + (y))(\mathfrak p + (z)) \subseteq \mathfrak p + (yz) \subseteq \mathfrak p.

a contradiction. Thus \mathfrak p is prime. ♦

Exercise B

From proposition 5, prove that we have a bijection

zariski_correspondence_rings

The correspondence reverses inclusion. For \cap V_i and V_1 \cup V_2 in Spec A, write down the corresponding operations for radical ideals of A.

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