Finite Atol, Field Extensions

Trace of a Nilpotent Element

Let R be an integral domain with fraction field K. Let S be a ring and a free R module. Let x be a nilpotent element in S, and find its trace. The endomorphism x*S is a matrix over R, which has eigen values in the algebraic closure of K. Since x*S is a nilpotent endomorphism, all the eigen values are nilpotent. Yet the closure of K is an integral domain, hence all the eigen values are 0. The trace and the norm are 0.

Since x*y gives a nilpotent endomorphism, trace(xy) = 0. If S contains a nilpotent element x, φ cannot be injective. In other words, a finite atol extension of an integral domain has no nilpotent elements.

The Finite Atol Extensions of a Field K

Since S/K is a finite K vector space, it is an artinian ring. Assume S is finite atol, whence there are no nilpotent elements, and S has no nil radical. Split S into a direct product of local artinian rings. It is enough to consider each summand, a local artinian ring that is finite atol.

Let F be such a ring. The maximal ideal of F is nilpotent. Yet there are no nilpotent elements, hence F is a field, and F/K is a finite field extension.

Suppose F contains a purely inseparable element x. When computing the trace of xy, one multiplies by the degree of inseparability, which is the characteristic of K. Thus the trace is 0. With φ(x) = 0, F/K is not finite atol. This is a contradiction, hence F/K is separable.

Let b₁ through b_n be a basis for F/K. Since F/K is separable, galois theory provides n automorphisms on its normal closure. Let c₁ through c_n be the embeddings of F into the algebraic closure of K. Let M be the matrix of embeddings of basis elements. In other words, M_i,j = c_i(b_j. Since the embeddings are linearly independent, M is a nonsingular matrix.

Let W = M^T times M, in that order. Verify that W_i,j is the trace of b_ib_j.

Since det(W) = det(M)² in an integral domain, W is a nonsingular matrix. In particular, W is a nonzero matrix. Find i and j such that b_ib_j has a nonzero trace. Given a nonzero x, let y = b_ib_j/x. Now the trace of xy is nonzero, φ(x) is nonzero, and φ is injective.

Let v be a vector in K and imagine a function that maps b₁ through b_n onto v₁ through v_n. Since W is invertible, let u be a vector in K such that u*W = v. Let x be the sum over u_ib_i, and let y be the sum over c_ib_i, using arbitrary coefficients c_i. Since trace is K linear, expand the trace of xy. The result is u*W*c, which is the same as v.c, which is our original function applied to y. Therefore φ is onto, and F/K is finite atol.

In summary, S/K is finite atol iff S is the direct product of finitely many finite separable extensions of K.