Let $Q(A)$ be the total ring of fractions of a commutative reduced non-noetherian ring $A$. Let $P$ be a finitely generated projective module over $Q(A)$ which is of constant rank (i.e. locally free of constant rank for each localization). Is $P$ free? (Note that $Q(A)$ need not be zero dimensional.)

  • $\begingroup$ dimension of a reduced von Neumann regular ring $A$ is 0, and f.g. projective modules of constant rank over zero dimensional ring are free. In general, if dimension of a commutative ring $A$ is $d$ ($A$ need not be noetherian), and $P$ is a f.g. projective $A$-module of constant rank $r>d$, then $P$ is isomorphic to $Q\oplus A$. This result is due to Heitmann (1982) and the Noetherian case of this result is due to Serre (1960). $\endgroup$
    – manoj
    Feb 12, 2013 at 6:17
  • $\begingroup$ $Q$ is some projective $A$-module of rank $r-1$. Basically it says that whenever rank of $P$ is $>$ dimension of $A$, then $P$ splits a free summand, i.e. there is an $A$-linear surjection from $P$ to $A$. This result is best possible, since there are counter-examples when $r=d$. Heitmann's paper is (1984) titled "Generating non-noetherian modules efficiently", Michigan J. of Math. For counter example, we can take a Dedekind domain which is not a PID. Then non-principle ideals are rank 1 projective modules, but not free. $\endgroup$
    – manoj
    Feb 13, 2013 at 5:21
  • $\begingroup$ For higher dimensional examples, we can take $A$ to be the coordinate ring of real sphere $S^n$ when $n\not=1,3,7$ and, i.e. $A=\mathbb R[x_0,\ldots,x_n]/(x_0^2+\ldots+x_n^2-1)$, and $P$ is the projective module which is $A^{n+1}/(x_0,\ldots,x_n)A$. Then $P\oplus A$ is free, but $P$ is not free. Further, when $n$ is even, there is no surjection from $P$ to $A$. In this case, note that rank of $P=$ dimension of $A$. $\endgroup$
    – manoj
    Feb 13, 2013 at 5:35

1 Answer 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:



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.

  • $\begingroup$ Havn't checked the details, but this looks nice ! $\endgroup$
    – user18119
    Feb 20, 2013 at 19:46
  • $\begingroup$ This is a nice construction. But ring $A$ is not reduced as every element of $M$ is nilpotent in $A$. Further, $A/nil(A)=R/nil(R)$. If we do not require reduced ring, then this might work, since taking non-free projective $R$-module, say $P$ ,we can construct $Q=P\otimes_R A$. This will be projectiv $A$-module, and $Q$ can not be free, otherwise $Q/MQ=P$ will be free $R$-module. $\endgroup$
    – manoj
    Feb 21, 2013 at 5:29
  • $\begingroup$ @manoj Yeah, I've forgot that the ring must be reduced! I've edited my answer. $\endgroup$
    – user26857
    Feb 21, 2013 at 11:06
  • $\begingroup$ So what is the answer? Not everyone has the book by Lam. $\endgroup$ Feb 21, 2013 at 11:29
  • $\begingroup$ @MartinBrandenburg Sometimes google can help in such situations: see here. $\endgroup$
    – user26857
    Feb 21, 2013 at 11:38

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