# Are f.g. projective modules free over total quotient ring of a reduced non-noetherian commutative ring

Are f.g. projective modules (of constant rank) free over the ring $A$ which is the total quotient ring of a reduced non-Noetherian commutative ring. Note that dimension of $A$ need not be $0$.

• If you had a counterexample for a given ring $A$, you could describe the module and demonstrate its projectiveness using only finitely many elements of $A$. This would lead to a counterexample for a finitely generated, therefore Noetherian, subring. – Tom Goodwillie Feb 15 '13 at 12:54
• @Martin: Why does Tom's comment answer the question? – Fred Rohrer Feb 15 '13 at 13:07
• Sure, but that still doesn't work. That is, if $A$ is non-Noetherian and its own total quotient ring, it does not follow that a finitely generated subring is also its own total quotient ring. – Neil Epstein Feb 15 '13 at 14:40
• Fred's comment is appropriate, as Neil notes. More specifically, an inclusion between rings doesn't necessarily extend to a ring homomorphism between total quotient rings, since a nonzero element of a subring that isn't a zero-divisor there may be a zero-divisor in a bigger ring. For example, the inclusion $k[x] \subset k[x,y]/(xy)$ doesn't extend to the total quotient rings as a ring homomorphism (because the composite map $k[x] \hookrightarrow k[x,y]/(xy) \twoheadrightarrow k[y]$ kills $x$ and thus doesn't extend to a map of rings $k(x) \rightarrow k(y)$). – user30379 Feb 15 '13 at 14:46
• Fred: one can descend the module to a f.g. module $M_n$ over some $k[x_1,..,x_n]$. As $k[x_1,..,x_n,...]=k[x_1,...,x_n][x_{n+1},...]$ is faithfully flat over $k[x_1,..,x_n]$, $M_n$ is flat hence free. So the answer is yes (any number of indeterminates). – Qing Liu Feb 17 '13 at 17:55

• It is an excercise in Lam's book (see YACP's reference) that any ring $S$ can be embedded in a ring $R$ with $R=Q(R)$, $nil(S)=nil(R)$ and $S$ is a retract of $R$. This answers my question in NEGATIVE. I would like to know if the dimension of $R$ in Lam's exercise can be finite. – manoj Feb 21 '13 at 12:31
• I guess dim $R$ is finite. Since $R$ is union of $S=S_0\subset S_1\subset S_2\ldots$ and dimension of $S_i$ is max (1,dim $S_{i-1}$). – manoj Mar 1 '13 at 5:35
I can give an example of ring $A$ which is reduced and every non-unit is a zero-divisor. It may be helpful to find the answer over $A$. Let $R=k[x_1,x_2,...]$ be polynomial ring (in infinitely many variables) over a field; let $m =(x_1,x_2,...)$ be a maximal ideal. Fix an integer $n>0$ and define $I$ to be the ideal generated by products $x_l x_j$ with $l$, $j$ distinct and $l> n$. Define $A= R_m /I$. Then $A$ is reduced, total quotient field of $A$ is $A$, and dimension of $A$ is $n$.
The dimension of a reduced commutative Noetherian ring having no non-zerodivisor is $0$. Hence Tom's comment will not work in general, as commented by others.
• Well, I guess one could always look at $B=R/I$ and then let $C$ be the total ring of quotients of $B$. Does $C=A$? I kinda think so, but if not, then there's no reason to think $C$ is local, and it could theoretically be a useful example in this context. However, in general it can be hard to find non-free f.g. projective modules (e.g. Serre's conjecture=Quillen-Suslin theorem). – Neil Epstein Feb 16 '13 at 18:11
• Thanks Laurent. I did not realize that $A$ is local. Further, $A=C$, since any element of $B-(x_1,...)$ has non-zero constant and is not a zerodivisor in $B$, hence a unit in $C$. – manoj Feb 19 '13 at 9:58