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In Bourbaki's Commutative Algebra we have the following theorem:

II.5.2 Let $A$ ba a ring and $P$ an $A$-module. TFAE:

(i) $P$ is a f.g. projective module\

(ii) $P$ is a finitely presented module and, for every maximal ideal $m$ of $A$, $P_m$ is a free $A_m$-module.\

(iii) $P$ is a f.g. module, for all $p \in$ Spec($A$), the $_p$-module $P_p$ is free and, if we denote its rank by $r_p$, the function $p \mapsto r_p$ is locally constant in the topological space Spec($A$).

There is more to this theorem but I have stated what I need for my question. Namely, unless I am missing something, for (iii) implies (i) don't we need the extra condition that $A$ is a reduced ring? I see their proof and cannot find the problem but Eisenbud suggests that there exists a counterexample in problem 20.13 in Commutative Algebra with a View Toward Algebraic Geometry. I must be missing something!

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    $\begingroup$ Could it be because the rank of $M_\mathfrak{p}$ is not necessarily the same as the dimension $dim_{R_\mathfrak{p}/\mathfrak{p}R_\mathfrak{p}}(M_\mathfrak{p} /\mathfrak{p}M_\mathfrak{p})$. Bourbaki's condition appears to be stronger. $\endgroup$ – Harry Gindi Nov 1 '10 at 8:53
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    $\begingroup$ (fishbones: you ended each item in the enumeration with a \\: hre that serves no purpose, and in LaTeX that's very wrong!) $\endgroup$ – Mariano Suárez-Álvarez Nov 1 '10 at 13:17
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Dear fishibones, an important point is that at the beginning of Chapter II, Bourbaki states that all the rings he will consider are commutative. This implies that if a free module has finite dimension, this dimension is unambiguously defined. This property is called the Invariant Basis Number property (IBN) and may fail for non-commutative rings (Counterexamples can be found in, say, Lam's Lectures on Modules and Rings, Springer GTM 189.) This unambiguously defined dimension of a finitely generated free module is what Bourbaki calls its rank.

Let me emphasize that Bourbaki only uses "rank" in the above sense i.e. for finitely generated free modules. If $M$ is such a module of rank $r$ over the ring $A$, then for any prime ideal $\frak {p}$$ \in A$ the modules $M_{\frak p}$ over $A_{\frak {p}}$ and $M_{\frak {p}} \otimes_{A_{\frak p}} \kappa (p)$ over $\kappa (p)$ also are free of rank $r$. With this definition and use of rank, Bourbaki's Théorème 1 that you mention is perfectly correct (bien sûr!)

There is no contradiction with Eisenbud's exercise: he does not assume that his module $M$ has all its localizations $M_{\frak {p}}$ free over $R_{\frak {p}}$, whereas Bourbaki does. Eisenbud only assumes that the dimension of the $\kappa ({\frak p})$- vector space $M_{\frak {p}} \otimes_{R_{\frak p}} \kappa (p)$ is locally constant, while Bourbaki does not even mention this dimension when $M_{\frak {p}}$ is not $R_{\frak {p}}$-free.

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  • $\begingroup$ I have adopted both notations $A$ and $R$ for the ground ring, faithfully following Bourbaki and Eisenbud respectively. I apologize for the resulting slightly incoherent look of my answer. – Georges Elencwajg 9 mins ago $\endgroup$ – Georges Elencwajg Nov 1 '10 at 13:01
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As H.G. points out in a comment, Bourbaki's condition deals with the localized module $P_p$ whereas Eisenbud's statement is about the quotient $P(p)=P_p/p$. For instance, to take a silly example, $A=k[x]/x^2$ and $M=k[x]/x$ is not free (so fails (iii) above) but satisfies the condition Eisenbud is considering (of course, $A$ is not reduced).

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