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.)
  • $\begingroup$ 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. $\endgroup$ Commented Jul 27, 2010 at 10:09

3 Answers 3


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.

  • $\begingroup$ Maybe it is a dumb question, but could explain better your last sentence? Why should such a finitely generated $M'$ exist locally? $\endgroup$ Commented Jul 27, 2010 at 13:06
  • 1
    $\begingroup$ 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). $\endgroup$ Commented Jul 27, 2010 at 14:43
  • $\begingroup$ 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 $x$. $\endgroup$ Commented Nov 21, 2010 at 23:47
  • $\begingroup$ 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. $\endgroup$ Commented Nov 22, 2010 at 1:12
  • $\begingroup$ Alright, this makes sense. $\endgroup$ Commented Nov 22, 2010 at 12:39

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.

  • $\begingroup$ 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. $\endgroup$ Commented Jul 27, 2010 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 $\mathcal{O}_X(1) \otimes \mathcal{O}_X(-1) \neq \mathcal{O}_X$.

  • 1
    $\begingroup$ 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? $\endgroup$ Commented Jul 27, 2010 at 13:36
  • $\begingroup$ 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... $\endgroup$ Commented Jul 27, 2010 at 14:02
  • $\begingroup$ 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. $\endgroup$ Commented Jul 27, 2010 at 14:42
  • $\begingroup$ 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. $\endgroup$ Commented Jul 27, 2010 at 14:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.