Let $M$ and $N$ be $R$ modules ($R$ commutative with identity). Is it true that if for every prime ideal $P$, $M_P \cong N_P$ (as $R_P$ modules) then $M \cong N$ ? Clearly the question is true if $M$ or $N$ is zero. But what about the non-zero case !?
[Math] locally isomorphic modules
modules
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The second definition is the correct one (at least in my opinion). It is similar to the correct notion of defining torsion. For instance one does not in general want to define an abelian group A to be p-torsion iff p^nA = 0 as this rules out for instance the Prufer p-group which should certainly be a torsion group but no fixed power of p will kill all of it. In particular, it is the injective envelope of Z/pZ in the category of abelian groups and so has support = {(p)} which means its single associated prime is (p) also. This makes the Prufer p-group p-coprimary with respect to the second definition which is a "sort of extension" of the fact that a finitely generated module over a noetherian ring is coprimary iff it has at most a single associated prime.
This example with the Prufer p-group extends to indecomposable injective modules over noetherian commutative rings with unit - so I guess my justification is that it makes all such guys coprimary with respect to the relevant ideal and it lines up with the right notion of torsion.
The answer to the second question is yes I think... For instance the right notion of support is somewhat subtle for non-finitely generated modules and (although I've never thought of it this way before) being P-coprimary for some prime ideal P does come up.
Dear roger123, let $R$ be a commutative ring and $M$ an $R$-module ( which I do not suppose finitely generated). In order to minimize the risk of misunderstandings, allow me to introduce the following terminology:
Locfree The module $M$ is locally free if for every $P \in Spec (R)$ there is an element $f \in Spec(R)$ such that $f \notin P$ and that $M_f$ is a free $R_f$ - module.
Punctfree The module $M$ is punctually free if for every $P \in Spec (R)$ the $R_{P} $ - module $M_P$ is free.
Fact 1 Every locally free module is punctually free. Clear.
Fact 2 Despite Wikipedia's claim, it is false that a punctually free module is locally free.
Fact 3 However if the punctually free $R$- module $M$ is also finitely presented, then it is indeed locally free.
Fact 4 A finitely generated module is locally free if and only it is projective.
Fact 5 A projective module over a local ring is free.This was proved by Kaplansky and is remarkable in that, let me repeat it, the module $M$ is not supposed to be finitely generated.
A family of counterexamples to support Fact 2 Let R be a Von Neumann regular ring. This means that every $r\in R$ can be written $r=r^2s$ for some $s\in R$. For example, every Boolean ring is Von Neumann regular. Take a non-principal ideal $I \subset R$. Then the $R$- module $R/I$ is finitely generated (by one generator: the class of 1 !), all its localizations are free but it is not locally free because it is not projective (cf. Fact 4) .The standard way of manufacturing that kind of examples is to take for R an infinite product of fields $\prod \limits_{j \in J}K_j$ and for $I$ the set of families $(a_j)_{j\in J}$ with $a_j =0$ except for finitely many $j$ 's.
Final irony In the above section on counterexamples I claimed that the $R$ - module $R/I$ is not projective.This is because in all generality a quotient $R/I$ of a ring $R$ by an ideal $I$ can only be $R$ - projective if $I$ is principal . And I learned this fact in...Wikipedia !
Best Answer
Here is an explicit counterexample:
Let $R^3$ be euclidean 3-space and $S^2$ the 2-sphere, embedded in $R^3$ as usual. Let $A$ be the ring of all real-valued continuous functions on $S^2$. Let $T$ be the $A$-module of all $R^3$-valued continuous functions on $S^2$ (so that $T\approx A^3$ is a free $A$-module). Let $M\subset T$ consist of all those functions $f$ such that $f(x).x=0$ for all $x$ (where "dot" denotes the usual inner product in $R^3$). Let $M'\subset T$ be the submodule generated by the identity function.
Observation 1: $M\oplus M'=T$. Thus, for any prime $P\subset A$, we have $M_P\oplus M'_P\approx T_P$. But over a local ring, any direct summand of a free module is free. Therefore $M_P$ is free.
Observation 2: $M$ can't be free. If it were, it would have a basis consisting of two triples $(f_1,f_2,f_3)$ and $(g_1,g_2,g_3)$ (the entries $f_i$ and $g_i$ being real-valued functions). This basis, together with the basis consisting of the single element $(x,y,z)$ for $M'$, would form a basis for $T$. It would follow that the matrix $$\pmatrix{f_1&f_2&f_3\cr g_1&g_2&g_3\cr x&y&z\cr}$$ has unit determinant; in particular the determinant is a function on $S^2$ with no zeros.
But it is a fact from topology that if $f(x).x=0$ for all $x$, then there is some $x$ such that $f(x)=(f_1(x),f_2(x),f_3(x))=(0,0,0)$. Thus the determinant of the displayed matrix has a zero at $x$. This contradiction shows that $M$ is not free.
Now let $N$ be a free $A$-module of rank 2. Observation 1 shows that $M_P\approx N_P$ for all primes $P$; Observation 2 shows that $M$ is not isomorphic to $N$.