Spec R, A Contravariant Functor

From Ring Homomorphisms to Continuous Functions

If you are familiar with category theory, then you know commutative rings and ring homomorphisms form a concrete category that we will call C_R. Topological spaces and continuous functions form another category C_T. Spec R implements a contravariant functor from C_R into C_T.

We already know how to carry rings to topological spaces; let's see how a ring homomorphism can induce a continuous function between those spaces. The adjective contravariant means the functions run in the other direction. If a ring homomorphism h maps R into S, the induced function g carries spec S into spec R.

Every ideal in S has a preimage in R, which is called the contraction of that ideal. When rings are commutative, prime ideals pull back to prime ideals. this establishes the function g from spec S to spec R.

Note that every prime in the image of g contains the kernel of h. In other words, the image of spec S lies in the closed set v_kernel.

Let v_E be closed in spec R. If E does not contain the kernel of h then it does not intersect the image of g at all. Its preimage under g is empty, which is closed. This isn't very interesting, so assume E contains the kernel of h. As usual, E is an ideal, and it determines a well defined image set F in S. A prime ideal Q in S pulls back to a prime P in R containing E iff Q contains F. In other words, the preimage of v_E under g is v_F. This is a closed set, hence g is continuous.

Verify that the composition of ring homomorphisms upstairs induces the composition of continuous functions downstairs. This completes the characterization of spec R as a functor.

The Closure of the Image of a Closed Set

Normally g is not bicontinuous, but we can still examine the closure of v_F under g. Let E be the preimage of F under h. The primes containing F pull back to primes containing E, hence the image of v_F is a subspace of v_E. Let K be rad(F), whence v_K = v_F. Let J be the preimage of K under h. Again, the primes containing K pull back to primes containing J. The eintersection of the primes containing K is precisely K. Pull these primes back to R and suppose their intersection includes more than J. The additional element maps to something in S that is outside of K. This is a contradiction, hence J is a radical ideal. The closure of the pullback primes containing J is the smallest closed set that contains these primes, which is v_J. Thus the closure of the image of v_F is v_J.

Suppose there is a prime ideal P in R, containing E, that does not contain J. Let x lie in J-P, so that h(x) lies in K-h(P). Since K = rad(F), h(x)ⁿ is in F for some n. This means xⁿ is in E, and hence in P, thus x is in P after all. Since v_E = v_J, the closure of the image of v_F is v_E, where E is the preimage of the ideal F.

Apply the above to a single point in spec S. If g(Q) = P, then g applied to the closure of Q has closure equal to the closure of P.

Let e be an element of R, with image f in S. Every prime Q in S missing f pulls back to a prime P in R missing e. Conversely if Q contains f then P contains e. Combine these results and the preimage, under g, of the base open set o_e, is o_f. Similarly, the preimage of v_e is v_f.

A Dense Image

Is g(spec S) dense in spec R? It is, iff the kernel of h lies in nil(R). If the kernel is not in the nil radical then some prime ideal does not contain the kernel and cannot be the image of anything under g. Therefore o_kernel is a nontrivial open set disjoint from the image of g. Conversely, assume nil(R) contains the kernel. let o_E be an open set in R, and ratchet E up to a radical ideal, whence E contains the kernel. To be nontrivial there are prime ideals in R that do not contain E. Suppose all prime ideals in S contain h(E), whence every element in h(E) is nilpotent. This means every element of E, raised to some power, winds up in the kernel, which is in the nil radical, hence everything in E is nilpotent. This contradicts the fact that some prime ideals do not contain E. Therefore there must be some prime ideal in S not containing h(E), that maps to a prime ideal in R not containing E, and the image of g intersects every open set in spec R.

Spec S embeds in Spec R

If h is surjective then g embeds spec S into spec R. Primes in S correspond 1-1 with primes of R containing the kernel. Thus the image is the closed subspace v_kernel. Since ideals in R and S correspond, closed sets correspond, and g embeds spec S into spec R.