Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
    
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). –  manoj Feb 12 '13 at 6:17
    
$Q$ 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. –  manoj Feb 13 '13 at 5:21
    
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 $A$. –  manoj Feb 13 '13 at 5:35

1 Answer 1

up vote 3 down vote accepted
+100

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.

share|improve this answer
    
Havn't checked the details, but this looks nice ! –  user18119 Feb 20 '13 at 19:46
    
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. –  manoj Feb 21 '13 at 5:29
    
@manoj Yeah, I've forgot that the ring must be reduced! I've edited my answer. –  user26857 Feb 21 '13 at 11:06
    
So what is the answer? Not everyone has the book by Lam. –  Martin Brandenburg Feb 21 '13 at 11:29
    
@MartinBrandenburg Sometimes google can help in such situations: see here. –  user26857 Feb 21 '13 at 11:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.