# 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 . For general rings, we have to switch to taking prime ideals because of the following.

Lemma.If is a ring homomorphism and is a prime ideal of B, then is a prime ideal of A.

**Proof**

The composition has kernel . Hence this induces an injective ring homomorphism . Since is a domain, so is so is prime. ♦

However it is not true that if is maximal in *B*, must be maximal in *A*. For instance consider the inclusion . The zero ideal is maximal in but is not maximal in .

Now we define our main object of interest.

Definition.Given a ring A, the

spectrumof A is its set of prime ideals . TheZariski topologyon is defined as follows: is closed if and only if it is of the following formfor some subset .

Note that since where 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.For any collection of ideals of A and ideals of A, we have

- ,
- ,
- .

**Proof**

The first claim is obvious. For the second, contains for each *i* if and only if it contains .

Finally, for the third claim, we have and so we have

Finally suppose . Then and so there are and . Then since is prime so . Hence if then . ♦

# Structure of Spec *A*

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

Proposition.If A is a ring, every proper ideal is contained in a maximal ideal.

**Proof**

Let be the collection of all proper ideals containing , ordered partially by inclusion. We claim that every chain has an upper bound.

Take . Let us show that is an ideal of *A*.

Clearly . Let and . Then and for some . Since is totally ordered, either or . Assuming the former, this gives so . Thus is an ideal of *A*.

Since none of the contains 1, we have . Hence is an upper bound of .

By Zorn’s lemma has a maximal element, which is precisely a maximal ideal of *A* containing ♦

We say that **minimal** if it is a minimal element with respect to inclusion.

Proposition.Any contains a minimal prime .

**Proof**

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

Let , an ideal of *A*. We need to show that it is prime. Suppose so there exist such that and . Since is a chain either or ; assuming the former we have . Since is prime we have so .

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

Corollary.If A is non-trivial, must contain a maximal ideal and a minimal prime ideal.

# Homomorphisms

Suppose is a ring homomorphism. We saw that this induces a map

.

Proposition.The map is continuous with respect to the Zariski topology; in fact

Note that is not an ideal of *B* in general, but that is okay since we defined *V* on arbitrary subsets.

**Proof**

Let . This lies in the LHS if and only if contains , if and only if contains . ♦

Clearly the identity ring homomorphism on *A* induces the identity map on , also for ring homomorphisms and we have

**Exercise**

Prove that for an ideal , the canonical map induces

which is a subspace embedding onto the closed subset .

# Comparison with Algebraic Geometry

Proposition.Let be a closed subset of . Then

,

the radical of .

**Proof**

(⊇) Suppose , for some *n* > 0. Now each contains and hence contains . Since is prime we have .

(⊆) Suppose , does not contain any power of *x*. We will prove the existence of not containing *x*.

Let be the set of all ideals not containing any power of *x*; thus . Order by inclusion. We will show that any chain has an upper bound. Indeed, let . As before, since is a chain of ideals ordered by inclusion, is an ideal of *A*. It does not include any power of *x* so .

Hence by Zorn’s lemma, has a maximal element ; by construction contains but not any power of *x*. It remains to show is prime: suppose not so there are with . Now and strictly contain , so by maximality of , we have . Hence

a contradiction. Thus is prime. ♦

**Exercise**

From the above proposition, prove that we have a bijection

The correspondence reverses inclusion. For and in Spec *A*, write down the corresponding operations for radical ideals of *A*.