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Let $A$ be a commutative Noetherian ring, and let $F$ be a flat $A$-module. We can assume $A$ is local, so $F$ is projective.

Question 1. When does $F\otimes_A-$ preserve injective objects?

Question 2. When does $F\otimes_A-$ have a left adjoint? And when is this adjoint exact?

The two questions are linked as follows. I learnt from this thread that, given an adjoint pair $F:\mathcal C\leftrightarrows\mathcal D:G$ where $\mathcal C$ and $\mathcal D$ are abelian categories and $\mathcal D$ has enough injectives, then $F$ is exact if and only if $G$ preserve injectives.

I would like to know when the functor $G=F\otimes_A-$ preserve injectives. So the question reduces to find out when $G$ has an exact left adjoint.

Any thoughts or comments are very welcome!

Community
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Brenin
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2 Answers2

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For question 2, $F\otimes_A -$ has a left adjoint iff $F$ is finitely generated, and the left adjoint is always exact. For if $(f_i)\in F^I$ is an element of an infinite product of copies of $F$, then it is easy to see $(f_i)$ is in the image of the canonical map $F\otimes_A A^I\to F^I$ iff $\{f_i\}$ is contained in a finitely generated submodule of $F$. If $F\otimes_A -$ has a left adjoint, then it this canonical map must be an isomorphism, and it follows that $F$ must be finitely generated. Conversely, if $F$ is finitely generated (and hence projective), $F^{\vee}\otimes_A -$ is adjoint to $F\otimes_A -$ on both sides.

(By a similar argument using instead the injectivity of the map $F\otimes_A A^I\to F^I$, you can show that $F$ must be finitely presented, even if you don't assume $A$ is Noetherian. So for non-Noetherian $A$, you still get that $F\otimes_A -$ has a left adjoint iff $F$ is finitely presented and flat, or equivalently finitely generated and projective.)

Note that in this case you can also easily see directly that $F\otimes_A -$ preserves injectives, since $F$ is a direct summand of a finitely generated free module and tensoring with a finitely generated free module obviously preserves injectives. In general, however, $F\otimes_A -$ might preserve injectives without having a left adjoint. For instance, it is a well-known theorem that a ring is Noetherian iff any (possibly infinite) direct sum of injective modules is injective. So since you're assuming $A$ is Noetherian, tensoring with any free module preserves injectives, and hence so does tensoring with any projective module.

Eric Wofsey
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  • Thank you very much for your answer! In the beginning of the third paragraph, what assumptions (if any) allow you to say that $F$ is a direct summand of a finitely generated free module? – Brenin Nov 30 '15 at 14:20
  • A finitely generated projective module is a direct summand of a finitely generated free module (namely, the free module corresponding to any finite set of generators). We know $F$ is projective because any finitely presented (=finitely generated if $A$ is Noetherian) flat module is projective. – Eric Wofsey Dec 01 '15 at 02:05
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Let me remove most of your assumptions and work with an arbitrary module $M$ over an arbitrary commutative ring $R$. We'd like to know when the functor $M \otimes_R (-)$ has a left adjoint.

The answer is iff $M$ is finitely presented projective, in which case the left adjoint is $M^{\ast} \otimes_R (-)$ where $M^{\ast} = \text{Hom}_R(M, R)$. You can extract a proof from this blog post. Since $M^{\ast}$ is also finitely presented projective, this functor is always exact.

Qiaochu Yuan
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  • I was thinking of using Eilenberg-Watts, but that would give you that $M\otimes_R -$ has an adjoint iff $M$ has a dual as a bimodule. Is it easy to see that this is the same as a dual as a module? – Eric Wofsey Nov 28 '15 at 21:51
  • Oh, I guess you can just explicitly compute that if you have an $(R,R)$-bimodule where the two actions of $R$ are the same, then the two actions of $R$ on the dual bimodule are also the same. – Eric Wofsey Nov 28 '15 at 22:00
  • @Eric: right, you should be using a "relative" version of Eilenberg-Watts where everything (the algebras and the bimodules) is over a base commutative ring $k$, and then you can set $k = R$. – Qiaochu Yuan Nov 28 '15 at 22:01
  • But if you use the relative version, you only get a characterization for when $M\otimes_R -$ has an $R$-linear left adjoint. So you have to check that if it has any left adjoint, the left adjoint is automatically $R$-linear. – Eric Wofsey Nov 28 '15 at 22:03
  • @QiaochuYuan: Thanks very much for your answer, and congratulations for the very nice blog :) – Brenin Dec 01 '15 at 22:59