Algebraic Geometry, Nil Radical = Jacobson Radical

Nil Radical = Jacobson Radical

Let A be a finitely generated K algebra. Clearly the jacobson radical, the intersection of the maximal ideals, contains the nil radical, the intersection of the prime ideals. But in this case the nil radical contains the jacobson radical, and jac(A) = nil(A).

Let f be an element that is not in nil(A), and show it is not in jac(A) either. The powers of f never reach 0, so let B be the fraction ring of A, whose denominators are powers of f. Like the ring A, B is a finitely generated K algebra. This is less than obvious, so let's take a closer look.

If x is a unit in A, such as an element of K, and x/1 becomes 0 in B, then x times a power of f is 0, making x a zero divisor. Since unit and zero divisor are incompatible, all of K maps to K, and B is indeed a K algebra.

Carry the generators of A forward into B, and bring in the addditional generator 1/f. That's all we need to span B, hence B is a finitely generated K algebra.

Let M be a maximal ideal of B, and map B/M into C, where C is the algebraic closure of K. (Such a map was described earlier.) In other words, the field B/M becomes a finite extension L/K in C.

By applying x → x/1, A maps into B, then into B/M, then into L, then into C. This composite homomorphism has a maximal ideal D in A. Since D maps into M, a proper ideal of B, D cannot contain f. Thus D is a maximal ideal that misses f, and f is not in jac(A).

Other Rings

Why doesn't this work for every ring? Why do we need a finitely generated K algebra?

Let A be the dvr Zp, that is, Z localized about p. The jacobson radical is generated by p, and the nil radical is 0. Let f = p, an element that is not in nil(A). Let B be the fractions of A by f. This is simply Q, the rational numbers. We don't have to worry about B/M being a field, because B is already a field. So a ring homomorphism takes A into B, which is a field. Since these rings are integral domains, the homomorphism is an embedding. The preimage of 0 is guaranteed to be a prime ideal missing f, but it need not be maximal. In fact the preimage of 0 is 0, which is not a maximal ideal in A.

Well these aren't even K algebras. Perhaps a more satisfying example is A = K[x] localized about x. This is another dvr. It is a K algebra, but not finitely generated. Let B = A adjoin 1/x, thus giving K(X), or all the rational functions in x. This is a field, but not a finite extension of K. As A embeds, it is not integral over K, and we cannot upgrade it from integral domain to field. The kernel need not be a maximal ideal; and it is not.