Graded Rings, Proj R is a Contravariant Functor

From Graded Ring Homomorphisms to Continuous Functions

Let f be a graded ring homomorphism from R into S, preserving degree. Since f is a ring homomorphism, it induces a continuous map from spec S into spec R. What happens if we restrict this map to the subspace proj S?

since f preserves degree, it maps homogeneous elements in R to homogeneous elements in S. Conversely, the preimage of a homogeneous element in f(R) has all but one of its homogeneous components mapped to 0, hence each homogeneous element in f(R) has a homogeneous pullback in R. Use this to show f(R) is a homogeneous subring of S. That is, x in f(R) implies all the components of x lie in f(R).

Restrict a homogeneous ideal of S to f(R). This is the intersection of two homogeneous sets, and it is homogeneous. It is also an ideal in f(R). Thus the contraction of a homogeneous prime ideal in S produces a homogeneous prime ideal in f(R). This pulls back to a homogeneous prime ideal in R. Therefore the map from proj S into proj R is well defined, with one caveat. The prime in S could pull back to a prime in R that contains all of R⁺, which is not in proj R. This is seen when K[x] embeds in K[x,y], and x generates P.

Let proj⁺ include all homogeneous prime ideals, to get past this little problem. Now proj⁺ S maps into proj⁺ R. Since this map carries a subspace of spec S into a subspace of spec R, it is continuous. It also respects conposition - the composition of graded ring homomorphisms yields the composition of the associated continuous functions. We have a valid contravariant functor from graded rings and homomorphisms into topological spaces and continuous maps.

Isomorphism

Beyond some degree d₀, let f become an isomorphism, mapping R_d onto S_d. In this case the continuous map becomes a homeomorphism. The spaces are equivalent.

The key is that any homogeneous prime ideal H, not containing R⁺, is determined by its homogeneous elements of degree > d₀. If H contains an element x of degree i then it contains x*R. Conversely, if H contains all the elements in x*R of degree > d₀, then H could contain x, or it could contain all the homogeneous elements in R of degree > d₀-i. If the latter holds, take the radical to find the containing prime. Every homogeneous element of degree > 0 is present. Thus H contains R⁺, which is a contradiction. Thus H contains x. Two primes in proj R that have the same homogeneous elements, beyond a certain degree, coincide.

With this in hand, the induced map from proj S into proj R is injective. The homogeneous elements in S_d and R_d agree, beyond d₀, and that determines, uniquely, the prime in S that produced the prime in R.

Is this map onto? Let Q be a prime in proj R. As mentioned earlier, Q does not contain all the homogeneous elements beyond d₀. Let W be the multiplicatively closed set of homogeneous elements in R, and in S, beyond d₀, that are not in Q. Drive the image of Q up to a prime ideal P in S missing W. The induced functor carries P onto Q, and the map is surjective.

Remember that the intersection of the primes in a closed set becomes a homogeneous ideal. Map these ideals between R and S, and the induced map is bicontinuous. As long as R and S are isomorphic beyond d₀, proj R and proj S are homeomorphic.