Algebraic Geometry, Algebras over a Jacobson Ring

Integral Extension of a Jacobson Ring

Let R be a jacobson ring, where f is an integral homomorphism from R into S. We will show S is a jacobson ring.

Since f(R) is also jacobson, reduce imediately to the case of direct inclusion. Thus R is a jacobson ring inside S, with S/R an integral extension.

Let Q be any prime in S with P = Q intersect R. Write P as the intersection of maximal ideals M_i in R. These ideals lifte to corresponding maximal ideals U_i in S. Let C be the intersection over U_i. Intersect C and R and find the same set as the intersection over M_i, which is P. Thus C is an ideal lying over P.

There may be many maximal ideals U_i over each M_i. In each case, select a U_i that contains Q. We can do this by the going up theorem. Thus C contains Q.

If C equals Q we have shown Q is the intersection of maximal ideals in S, and we are done. If not we have an ideal whose contraction is P, and we can drive C up to a prime ideal in S, missing R-P, whose contraction is still P. (This is done by localizing about P and finding a maximal ideal containing C.) Now two distinct prime ideals lie over P, and one contains the other. This is impossible, hence C = Q, Q is the intersection of maximal ideals, and S is jacobson.

Finitely Generated Algebra over a Jacobson Ring

Let R be jacobson and let S be a finitely generated R algebra. Again, reduce to the case where R embeds in S.

Let Q be prime in S with contraction P in R. Mod out by Q and S/Q becomes a finitely generated R/P algebra. Both rings are now integral domains.

Since the nil radical of R/P is 0 and R/P is the homomorphic image of a jacobson ring, the jacobson radical of R/P is also 0. By the previous theorem the jacobson radical of S/Q is 0. Pull this back to S and Q is the intersection of maximal ideals, hence S is jacobson.

If R is a field R is automatically jacobson, and every finitely generated R module (integral extension), or finitely generated R algebra, is jacobson.