# Finitely generated flat modules that are not projective

Over left noetherian rings and over semiperfect rings, every finitely generated flat module is projective. What are some examples of finitely generated flat modules that are not projective?

Compare to our question f.g. flat not free where all the answers are f.g. projective not free.

• Take a non-semisimple von Neumann regular ring $R$ and choose a right ideal $I$ that is not a direct summand of $R_R$. Then $R/I$ is flat and cyclic, but not projective. // Here's Exercise 17 from p. 161 Lam's Lectures on Modules and Rings giving an explicit example due to Vasconcelos of a f.g. flat module that isn't projective. – Martin May 24 '13 at 0:43
• I encourage Martin and anyone else to include answers. For instance, Martin points out a nice phrasing from the exercise where a principal ideal is flat but not projective, which is also very strange. – Jack Schmidt May 24 '13 at 3:53
• I'm sorry but I don't understand what you encourage me and others to do. I was merely trying to give you some pointers where I would start looking if I had more time to really think about it: 1) von Neumann regular rings so that flatness is automatic and 2) Lam's books which are in general an excellent source for all sorts of examples of this type. // In case you were asking for a solution of this exercise, here's Lam's own solution from the solution manual Exercises on Modules and Rings. – Martin May 24 '13 at 8:42
• @Martin: Sorry, I meant "Thanks!" and "Other people are more likely to benefit if you post your fine answer as an answer rather than a comment." I'm also waiting on someone else to post an example they deleted from another question. – Jack Schmidt May 24 '13 at 11:03

Over a von Neumann regular ring, every right module (and every left module) is flat. Let $V$ be an countable dimensional $F$ vector space, and let $R$ be the ring of endomorphisms of that vector space. It's known that $R$ is a von Neumann regular ring with exactly three ideals.

The nontrivial ideal $I$ consists of the endomorphisms with finite dimensional image. Then $R/I$ is flat but it cannot be projective. If it were projective, then $I$ would be a summand of $R$... but it is not, because it's an essential ideal.

A second example over any non-Artinian VNR ring: you can take $R/E$ for any maximal essential right ideal $E$ to get a nonprojective, simple flat module. The reasons are very much the same, since a proper essential right ideal can't be a summand of the ring.

You can even make a commutative version: take an infinite direct product of fields $\prod F_i$ (this is von Neumann regular). The ideal $I=\oplus F_i$ is an essential ideal, and $R/I$ is flat, nonprojective. (This one also has the added benefit of supplying examples of ideals which are projective but not free. Any summand of the ring will do, since the ring has IBN. The argument at the other post can be carried out again.)

Incidentally, Puninski and Rothmaler have written a nifty paper investigating which rings have all f.g. flat modules projective.

Following Jack's gentle nudging, I'm posting the part of my comments not covered by rschwieb's nice post more prominently as an answer:

In general, this type of question can be answered quickly by consulting Lam's Lectures on Modules and Rings, Springer Graduate Texts, Vol. 189, whose numerous exercises are solved and put into further context by Lam himself in the accompanying text Exercises in Modules and Rings. The section §4E, Finitely generated flat modules of the Lectures contains some sufficient conditions on when f.g. flat modules are projective.

Here's exercise 4.17 with solution, giving Vasconcelos's example of a principal ideal $(a)$ in a commutative ring such that $R/(a)$ is flat but not projective:

This appears as Example 3.2 of Wolmer V. Vasconcelos, On finitely generated flat modules, Trans. Amer. Math. Soc. 138 (1969), 505-512.

• Thanks! I agree that Lam's texts are invaluable, and I appreciate the reference to the original paper. – Jack Schmidt May 24 '13 at 12:05