A theorem for you to prove: Let B be a local ring containing a perfect field...

A theorem for you to prove: Let B be a local ring containing a perfect field k that is isomorphic to its residue field B/m, and such that B is a localization of a finitely generated k-algebra. Then the module of relative differential forms M_B/k is a free B-module of rank equal to the dimension of B if and only if B is a regular local ring.

Expert Solution

PROOF:

Let B is noetherian. Suppose that M B/k is a free B-module of rank dimB.

Then by ( Proposition A : Let B be a local ring with field of representatives k. Then the canonical morphism of k-modules ? : m/m2 ?? M B/k ?B k is an isomorphism.), rank km/m2 = dimB, so B is a regular local ring. In particular this implies that B is a normal domain (MAT,Theorem).

For the converse, suppose that B is a regular local ring of dimension r. Then rank km/m2 = r so by Proposition A, we have rank k(MB/k ?B k) = r. On the other hand, let K be the quotient field of B. Then there is an isomorphism of K-modules MB/k?B K ?= MK/k .

Since k is perfect, K is a separably generated extension field of k , So rank MK/k = tr.deg.K/k .

We claim that dimB = tr.deg.K/k.

We know that ,if k be a field, A a finitely generated k-algebra and suppose that B = Ap is a domain for some prime ideal p of A. Then B is isomorphic as a k-algebra to the localisation of an affine k-algebra at a prime ideal.

So we can find an affine kalgebra A and a prime ideal p such that B ?= Ap as k-algebras. Since k is a field of representatives of B we have Ap/pAp ?= k as k-algebras. Using the fact that A/p is an affine k-algebra with quotient field k-isomorphic to Ap/pAp, we have coht.p = dim(A/p) = tr.deg.(Ap/pAp)/k = tr.deg.k/k = 0

using the fact that the quotient field of A is k-isomorphic to K, we have dimB = dimA = tr.deg.K/k as claimed. It now follows that MB/k is a free B-module of rank dimB.

Manojponduru answered 2 years ago

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