# Flat finite type ring extensions

Encouraged by the success in studying finite flat modules, see the preceding post, let’s think a bit about flat, finite type ring extensions.

Question: For which rings R is every finite type flat ring map R —> S of finite presentation?

A Noetherian ring satisfies this property. In the paper by Raynaud and Gruson they prove that this holds if R is a domain. I recently added this result to the stacks project (with a purely algebraic proof), see Algebra, Proposition Tag 053G. If R is a local ring whose maximal ideal is nilpotent then the result is true as well. But I don’t know what happens if the maximal ideal is only assumed to be locally nilpotent, i.e., every element of the maximal ideal is nilpotent, i.e., the maximal ideal is √(0). Do you?

By the way, I still want more ideas about the question I posted here! [Edit: this question has now been answered.]

[Edit on August 23, 2010: As David Rydh points out in a comment below any ring which has finitely many associated primes satisfies the condition. This follows trivially from Raynaud-Gruson Theorem 3.4.6. Don’t know why I did not see this! Anyway, so a local ring whose maximal ideal is locally nilpotent is an example too.]

## 6 thoughts on “Flat finite type ring extensions”

1. Doesn’t this follow from the corollary of Nakayama’s lemma on p. 8 of Matsumura’s “Commutative ring theory”? Let R be a local ring whose maximal ideal m is locally nilpotent. Denote the residue field by k. Let A be R[x_1,…,x_d]. Then the ideal I = mA is also locally nilpotent, hence contained in rad(A). Now let M be an ideal in A such that the quotient B = A/M is a flat R-module. In particular, this implies that M/mM –> A/mA is injective, i.e., M/IM –> A/I is injective. But A/mA = k[x_1,…,x_d] is Noetherian. Hence M/IM is finitely generated. Lift these finitely many generators to elements in M, and let N be the finitely generated A-submodule of M generated by these lifts. Then N surjects onto M/IM, i.e., M equals N + IM. By the corollary, M equals N. Thus M is finitely generated.

2. In my notation above, “B” is what you are calling “S”.

3. Oops! I missed the hypothesis that M/N should be finitely generated. The argument above is wrong.

• Yeah… but that is exactly why your argument does work when m is nilpotent since in that case one does not need the assumption of finiteness in Nakayama’s lemma!

4. Well, Raynaud-Gruson shows that this is true if R has a finite number of associated primes (“weakly associated” primes in the terminology of Bourbaki). This is Thm 3.4.6. In fact they show the stronger result that given R->S of finite type with Ass(R) finite, the locus U of s in Spec(S) where O_{S,s} is flat, is open and that U->S is locally of finite presentation.

Note that their hypothesis is slightly different but that if R->S is of finite type, we can consider S as a finite type module over the finitely presented ring R[x_1,…,x_n].

• Johan on said:

Thanks! I edited the post to update it.