Are finitely generated projective modules free over the total ring of fractions?

Solution 1:

For the beginning only few hints which led to the conclusion that the answer to your question is NO.

In this topic I've defined the idealization of a module. I'll repost the construction for the sake of completeness.

Let $R$ be a commutative ring and $M$ an $R$-module. On the set $A=R\times M$ one defines the following two algebraic operations:

$$(a,x)+(b,y)=(a+b,x+y)$$

$$(a,x)(b,y)=(ab,ay+bx).$$

With these two operations $A$ becomes a commutative ring with $(1,0)$ as unit element. ($A$ is called the idealization of the $R$-module $M$ or the trivial extension of $R$ by $M$).

Remarks: if every noninvertible element of $R$ kill some nonzero element of $M$, then $A$ equals its own total ring of fractions. Such an example is the following: take $(\mathfrak m_i)_{i\in I}$ a family of maximal ideals of $R$ such that every noninvertible element of $R$ belong to an $\mathfrak m_i$ and set $M=\bigoplus_{i\in I} R/\mathfrak m_i$.

Construction: take $R$ such that there exists a nonfree projective $R$-module of rank $1$. Now define $M$ as before in such a way that there exists a nonfree projective $A$-module of rank $1$.

Edit. Unfortunately I've forgot that the ring $A$ should be reduced. But an example for the reduced case can be found in Lam, Exercises in Modules and Rings, Exercise 2.35.