Avramov’s theorem

Let A —> B be a flat local homomorphism of Noetherian local rings. By the local criterion for flatness this also implies that the map on completions A* —> B* is flat. Hence, in order to prove Avramov’s theorem that I mentioned here it suffices to prove it for a flat map of Noetherian complete local rings.

Let A be a complete local Noetherian ring. Write, using the Cohen structure theorem, A = S/I where S is a regular complete local ring. The complete intersection defect of A is the nonnegative integer

cid(A) = dim_k(I/mI) – dim(S) + dim(A)

where k = S/m is the residue field of S. Note that dim_k(I/mI) is the minimal number of generators of the ideal I. Since S is regular we see that A is a complete intersection if and only if cid(A) = 0.

Next, let A —> B be a flat local homomorphism of complete local Noetherian rings. Avramov proved, among other things, that

(*) cid(B) = cid(A) + cid(B/m_AB)

in this situation. It is clear that this proves the result we mentioned in the previous post.

What does (*) mean in more elementary terms? For simplicity, let us assume that the residue fields of A and B are identified by the map A —> B. Write A = R/I. Set S = R[[x_1, …, x_n]] for some large n and choose a surjection S –> B (here we use that the residue fields are equal). Set J ⊂ S equal to the kernel of S —> B so that B = S/J. Consider the induced map

(**) I/m_RI —> J/m_SJ

The equality (*) is equivalent to  the injectivity of (**). Namely, flatness of A –> B gives dim(B) = dim(A) + dim(B/m_AB). By construction dim(S) = dim(R) + dim(k[[x_1, …, x_n]]). Finally, the map J/m_SJ —> (J + m_RS)/(m_SJ + m_RS) is surjective with kernel equal to the image of (**). (Proof omitted, but see next post.)

OK, now why is (**) injective? I claim that this is a question about the obstruction space for the deformation theory of the algebra B/m_AB over k. I will discuss this in the next post.

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