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Let $\mathbb{A}$ be a polynomial ring in $n$ variables over an algebraically closed field $\mathbb F$. Given a maximal ideal $\mathfrak{m}$ of $\mathbb A$, consider the quotient $\mathbb{A}/\mathfrak{m}$ as a left $\mathbb{A}$-module ($\mathbb{A}/\mathfrak{m}$ is one-dimensional over $\mathbb F$).

Suppose $M$ is a right finitely generated $\mathbb{A}$-module which is infinite dimensional (as $\mathbb F$-vector space) and such that $\dim_{\mathbb F}(M\otimes_{\mathbb{A}} \mathbb{A}/\mathfrak{m})<\infty$ for any maximal ideal $\mathfrak m$.

Question: Can we say that $\dim_{\mathbb F}(M\otimes_{\mathbb{A}} \mathbb{A}/\mathfrak{m})$ is independent of $\mathfrak m$?

The motivation of the question is to show that $M$ is projective (therefore free, by Quillen's Theorem).

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No. If $M$ is f.g. over $A$ then $M\otimes_A (A/\mathfrak m)$ will be finite dimensional over $F$ for every maximal ideal $\mathfrak m$ of $M$ (since $A/\mathfrak m = F$, and change of scalars preserves the property of being finite dimensional).

So your question amounts to asking whether every f.g. module over $A$ that is infinite dim'l over $F$ is free, and the answer is no.

E.g. Suppose that $n = 1$, and take $M = A \oplus F,$ with $x$ and $y$ acting via $0$ on the second summand.

E.g. Suppose that $n = 2$, so $A = F[x,y].$ Let $M = F[x]$, regarded as an $A$-module by having $y$ act via $0$.

E.g. Suppose that $ n =2$, and let $M = (x,y) \subset F[x,y]$.

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Thank you for the examples. My next question is very general and unprecise (sorry about that), but it is: What is the slightest hypothesis to add in $M$ in a way that $M$ becomes a projective module? –  Matt Elly Aug 14 '12 at 13:55
    
@MattElly: Dear Matt, If $n = 1$ then (since $A$ is a PID in that case) you have that f.g. and torsion-free implies projective. Otherwise, the only generally applicable hypothesis that I know is the one that you mention in your question, namely that the dimensions $M\otimes_A (A/\mathfrak m)$ should be constant. (There are other, more homological criteria, which can be applicable in certain particular circumstances: e.g. if M can be written as an $n$th syzygy, i.e. if there is an exact sequence $0 \to M \to P_{n-1} \to \cdots \to P_1 \to P_0,$ with all the $P_i$ projective, then $M$ is ... –  Matt E Aug 14 '12 at 14:37
    
... itself projective. This generalizes the torsion freeness criterion in the $n = 1$ case, but you usually have to be in very particular situations for this to be helpful.) Regards, –  Matt E Aug 14 '12 at 14:39
    
Thank you again! Do you have any reference for this last part between parentheses? –  Matt Elly Aug 15 '12 at 3:45
    
@MattElly: Dear Matt, Any text on homological algebra should discuss it; the key point is that $F[x_1,\ldots,x_n]$ has projective dimension $n$. I would try Weibel's book, or even just wikipedia. Sorry not to be more specific; I just don't have a good knowledge of accessible homological algebra texts/sources. Also, if you're interested in this criterion, you could always ask a question about it which includes a reference request! Regards, –  Matt E Aug 15 '12 at 4:29

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