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Let $M$ be an $R$-module ($R$ commutative with unity) and suppose that given $\mathfrak{p}\subset R$ a prime ideal we have $M_{\mathfrak{p}}\cong R_{\mathfrak{p}}^n$.

Is then true that we can find an element $f \in R$ such that $M_f \cong R^n[f^{-1}]$?

user26857
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Abellan
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  • Note: This is called generic freeness. – MooS Nov 23 '16 at 14:18
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    @MooS Wouldn't that be what you'd call if it you looked at the generic point? Generic freeness, to me, means that you're the trivial bundle on a dense open. That is related to this problem, and its solution, assuming that $M$ is coherent, say, and $\mathfrak{p}=0$ (where $R$ is a domain). For example, if $R$ is disconnected and $M$ is free on one component, but not even a vector bundle on the other, I don't think I'd call that generic freeness. – Alex Youcis Nov 23 '16 at 14:33
  • Of course it is not the exact formulation, but it is closely related and can be even considered as a generalization. One often calls generalizations by the same name, for an example look at Krull's Principal ideal theorem. – MooS Nov 23 '16 at 17:03
  • Question: Do we need $M$ to be finitely presented or just finitely generated? – Abellan Nov 23 '16 at 17:36
  • You will at least need finitely presented as the following example shows. Let $R$ be the polynomial ring in infinitely many variables $x_i,y_i$ modulo $x_iy_i$ for all $i$. Let $\mathfrak{p}$ be the prime ideal generated by all the $x_i$s and let $M=R/\mathfrak{p}$. – Mohan Nov 23 '16 at 17:44

2 Answers2

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Yes...assuming that $M$ is finitely generated (if not see Mohan's helpful comment below).

Hint: The isomorphism $R^n_{\mathfrak{p}}\to M_{\mathfrak{p}}$ sends the basis vectors $e_i$ to elements of the form $\displaystyle \sum_j f_{i,j} m_j$ where $f_i\in R_{\mathfrak{p}}$. This isomorphism really only needs to have that the $f_{i,j}$ make sense, and this really only requires inverting finitely many denominators.

Alex Youcis
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    If $M$ is not finitely generated, this is false in general. For example, take $R$ to be a domain, $\mathfrak{p}=0$, $K$ the fraction field and let $M=K$. – Mohan Nov 23 '16 at 14:28
  • @Mohan Oops, of course! I sort of assume everything in life is coherent :P Thanks--will edit. – Alex Youcis Nov 23 '16 at 14:30
  • Are you suggesting to take $f$ as the product of all denominators of $f_{i,j}$? – Abellan Nov 23 '16 at 14:37
  • @Abellan I am. $ $ – Alex Youcis Nov 23 '16 at 14:39
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    I hope that this is my last comment. Let us suppose that each $f_{i,j}=\frac{1}{g_{i,j}}$ and we define $f=\prod\limits_{(i,j)}f_{i,j}$ and $\gamma_{i,j}=\prod\limits_{(k,l)\neq (i,j)}f_{k,l}$. Then I define the following morphism,

    $R^n[g^{-1}] \to M[g^{-1}], e_i \mapsto \sum\limits_j\frac{\gamma_{i,j}}{g}m_j$. This is clearly injective. Where do I used that the module is finite generated? To show surjectivity?

     – Abellan Nov 23 '16 at 16:13
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    @Abellan Somehow, I don't see why it is clearly injective. Something one can do is the following. Suppose $M_p = R_p(m_1,\dots, m_n)$. One can assume that $m_i \in R$. Consider the exact sequence $0 \to \ker \phi \to R^n \to M \to coker \phi \to 0$, where $\phi(e_i) = m_i$. Localization is exact. Hence $(\ker \phi)_P = (coker~ \phi)_P = 0$. So there exist a,b not in $P$ such that $(\ker \phi)_a = (coker~ \phi)_b = 0$. Take $f = ab$. The finiteness condition is necessary in order to find such $a,b$ as in Mohan's comment. – Youngsu Nov 23 '16 at 19:22
  • It looked clear before your comment. Thanks a lot for your insights! – Abellan Nov 23 '16 at 20:34
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$\def\p{\mathfrak{p}}$We need $M$ finitely presented for the result to hold [ref]. Contrary to what Alex Youcis says, $M$ finite is not enough. Let's show Mohan's comment in OP's question is indeed a counterexample. Let $I$ be an infinite set and $k$ be a field. Let $\displaystyle R=\frac{k[x_i,y_i\mid i\in I]}{(x_iy_i\mid i\in I)}$. Let $\p=(x_i\mid i\in I)\subset R$ and let $M=R/\p\cong k[y_i\mid i\in I]$. Then $R_\p=k(y_i\mid i\in I)=M_\p$. We shall show that for any $f\in R\setminus\p$, the module $M[f^{-1}]$ is not $R[f^{-1}]$-free. Let $j\in I$ be an index such that $y_j$ shows up in no monomial of $f$. Then $f^n x_j\neq 0$, since the monomials in $f^n$ have no $y_j$ either. Thus $x_j$ is not zero in $R[f^{-1}]$, but $x_j\in\operatorname{Ann}(M[f^{-1}])$ (since $x_j\in\operatorname{Ann}M$). That is, $M[f^{-1}]$ has torsion over $R[f^{-1}]$. Hence it is not $R[f^{-1}]$-free.

Elías Guisado Villalgordo
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