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Let $X$ be a locally ringed space (or a scheme) and $M,N$ two $\mathcal{O}_X$-modules such that $M \otimes N \cong \mathcal{O}_X$. Does it follow that $M$ is invertible in the usual sense, namely that $M$ is locally free of rank $1$?

It is true if $M$ is locally of finite type (which is, of course, also necessary).

Proof: Let $x \in X$. Then $M_x \otimes N_x \cong \mathcal{O}_{X,x}$. Now tensor with the residue field of $\mathcal{O}_{X,x}$ and use linear algebra to conclude that $M_x / \mathfrak{m}_x M_x$ is $1$-dimensional. Since $M_x$ is of finite type over $\mathcal{O}_{X,x}$, Nakayama shows that $M_x$ is generated by just one element. Since $M$ is of finite type in a neighborhood of $x$, it follows that the generator at $x$ is also a generator in a neighborhood of $x$. Also $N$ has one generator, and their tensor product is a generator of $M \otimes N \cong \mathcal{O}_X$, which must be free. Thus also the generators of $M$ and $N$ are free.

But I don't know what happens in the general case. Here are some intermediate questions:

  • Does it follow that $M$ is flat?
  • Is the resulting morphism $M \to Hom(N,\mathcal{O}_X)$ an isomorphism?
  • Is the claim true for $X$ a point, i.e. a local ring?
  • Is the claim true if $X$ is an affine scheme and $M,N$ are quasi-coherent? (Thus in the question, replace $\mathcal{O}_X$ by a usual ring.)
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I think the main point is that any $M'\to M$ that induces an epi after tensoring with $N$ must also induce an epi after tensoring with $N\otimes M$ and therefpre muse be an epi. And locally such a finitely generated $M'$ must exist. –  Tom Goodwillie Jul 27 '10 at 10:09

3 Answers 3

up vote 5 down vote accepted

Any $M'\to M$ that induces an epi after tensoring with $N$ must also induce an epi after tensoring with $N\otimes M$ and therefore must be an epi. And locally such a finitely generated $M'$ must exist.

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Maybe it is a dumb question, but could explain better your last sentence? Why should such a finitely generated $M'$ exist locally? –  Andrea Ferretti Jul 27 '10 at 13:06
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Because of the nature of a tensor product: on some neighborhood of a given point there exist sections $m_i$ and $n_i$ such that the sum of $m_i\otimes n_i$ is the section of $M\otimes N$ that corresponds to the generator $1$ under the given isomorphism with $\cal O_X$. Take the $m_i$ to generate $M'$ (either as a submodule of the restriction of $M$ to the neighborhood, or freely if you prefer). –  Tom Goodwillie Jul 27 '10 at 14:43
    
Thank you, it wes easier than I thought. :-) –  Andrea Ferretti Jul 27 '10 at 15:11
    
Hm, I think that the proof has a flaw. If you define $M'$ as above, then it is not clear at all why $M' \to M$ is an epi after tensoring with $N$ on some neighborhood; this is only clear at $x$. –  Martin Brandenburg Nov 21 '10 at 23:47
    
On the neighborhood where it is defined, the map $M'\otimes N\to M\otimes N$ is such that its image has in it a section that, all alone, generates that sheaf of modules. –  Tom Goodwillie Nov 22 '10 at 1:12

Yes, for quasi-coherent sheaves on a scheme it is true that if $M \otimes N \cong \mathcal{O}_X$, then $M$ is locally free of rank one. It is enough to prove the

REDUCTION Let $M,N$ be $A$- modules such that $M \otimes_A N\cong A$. Then $M$ is projective of finite type.

Proof : We are given an isomorphism $f:M \otimes_A N\cong A$ . Say $\; f( \Sigma m_i\otimes n_i)= 1$ (FINITE index set!). The composition of the isomorphisms

$g_M:M \to M\otimes (N\otimes M):m\mapsto \Sigma m\otimes (n_i\otimes m_i)$

$assoc: M\otimes (N\otimes M) \to (M\otimes N) \otimes M: m\otimes (n \otimes m') \mapsto (m\otimes n)\otimes m'$

$f_M:(M\otimes N) \otimes M \to M: (m\otimes n) \otimes m_1\mapsto f (m \otimes n).m_1 $

is the isomorphism $j:M\to M: m\mapsto \Sigma f(m\otimes n_i).m_i$

By introducing the linear forms $\nu_i:M\to A: m\mapsto f(m\otimes n_i)$ we see that we have an isomorphism

$j:M\to M: m\mapsto \Sigma \nu _i(m).m_i$ and we deduce that for all $m\in M$ we can write $m=\Sigma \nu _i(m).j^{-1}(m_i)$.

It is well known that the existence of such a dual basis $(j^{-1 }m_i, \nu_i)$ proves that $M$ is a finitely generated projective module.

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This is the same proof as Tom gave, but a bit more down-to-earth. However, the statement for modules on locally ringed spaces is more general. –  Martin Brandenburg Jul 27 '10 at 16:09

It seems to me that if $X$ is not factorial than one must be careful.

For instance, take the weighted projective space $X=\mathbb{P}(1,1,2)$, and consider the sheaf $\mathcal{O}_X(1)$, whose construction can be found for instance in Dolgachev's paper "Weighted projective varieties". Abstractely, $X$ is isomorphic to a quadric cone in $\mathbb{P}^3$, and the sheaf $\mathcal{O}_X(1)$ has only two sections, so it is not locally free (in fact, it corresponds to a single line in the cone). However, $\mathcal{O}_X(2)$ is locally free, since it corresponds to a hyperplane section (in fact, $X$ is 2-factorial); let $\mathcal{O}_X(-2):=\mathcal{O}_X(2)^*$.

Finally, take

$M:=\mathcal{O}_X(-2) \otimes \mathcal{O}_X(1), \quad N:=\mathcal{O}(1)$.

Then $M \otimes N \cong \mathcal{O}_X$, but $M$ and $N$ are not locally free.

ADDED. As pointed out by D. Arapura in the comment below, there is actually a mistake in this example, since it is $not$ true that $\mathcal{O}_X(1) \otimes \mathcal{O}_X(1) \cong \mathcal{O}_X(2)$; in fact, these sheaves only agree on a dense open set. What is true is that

$\mathcal{O}_X(2a) \otimes \mathcal{O}_X(b) \to \mathcal{O}_X(2a+b)$

is an isomorphism for all integers $a$, $b$, but of course this cannot give any counterexample!

However, it is $always$ true that

$\mathcal{O}_X(a) \to Hom (\mathcal{O}_X(b), \mathcal{O}_X(a+b))$

is an isomorphism, see the paper of DELORME "Espaces projectifs anisotropes", p. 210.

In particular,

$\mathcal{O}_X(1) \to Hom (\mathcal{O}_X(-1), \mathcal{O}_X)$

is an isomorphism, even if $\mathcal{O}_X(1) \otimes \mathcal{O}_X(-1) \neq \mathcal{O}_X$.

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I have to apologize for the way this sounds, but this seems suspect. Are you sure there isn't a double dual in there somewhere? –  Donu Arapura Jul 27 '10 at 13:36
    
I understand your doubt. Are you suggesting that $\mathcal{O}_X(1) \otimes \mathcal{O}_X(1)$ could be actually different from $\mathcal{O}_X(2)$, aren't you? Looking at the definitions, I know that these two sheaves agree in a dense open subset of $X$, but this is actually not enough to conclude. I will check it out... –  Francesco Polizzi Jul 27 '10 at 14:02
    
Right. I think that $ (\mathcal{O}_X(1) \otimes \mathcal{O}_X(1))^{**}\cong \mathcal{O}_X(2)$ should be fine, but it wouldn't be true without the double dual. –  Donu Arapura Jul 27 '10 at 14:42
    
Ad actually they were different... The behaviour of the sheaves $\mathcal{O}_X(n)$ in the weighted projective spaces is not the same as in the usual projective space, as you pointed out, so I had to be more accurate and take the double dual into account. I've edited the answer, anyway. Thank you for the remark. –  Francesco Polizzi Jul 27 '10 at 14:47

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