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Let $\mathcal{A}$ be an abelian category and $\mathcal{B}$ be a full subcategory of $\mathcal{A}$. Suppose that $\mathcal{B}$ is abelian and that the inclusion of $\mathcal{B}$ in $\mathcal{A}$ is exact. My question is: if an object $P$ of $\mathcal{B}$ is projective in $\mathcal{B}$, then is it true that $P$ is projective in $\mathcal{A}$? (And what about the injective case?)

For example, consider a noetherian ring $R$, take $\mathcal{A} = \text{Mod}R$ and $\mathcal{B} = \text{mod}R$ (subcategory of finitely generated $R$-modules). Then, if $P$ is a finitely generated $R$-module which is projective in $\text{mod}R$, is it true that $P$ is a projective object in $\text{Mod}R$? (An the injective case?)

Edit: The statement of the second paragraph is true: use the fact that a module is projective iff it is a direct summand of a free module; for the injective case we can use Baer's criterion.

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  • $\begingroup$ Do you assume $\mathcal B$ is closed under $\mathcal A$-extensions? $\endgroup$
    – Dag Oskar Madsen
    Commented Apr 29, 2020 at 21:39
  • $\begingroup$ @DagOskarMadsen Well, in the case that I am interested, yes. If that is true in this case, it will be very helpful for me. Nevertheless, what about the general case? $\endgroup$
    – user144185
    Commented Apr 29, 2020 at 21:45
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    $\begingroup$ In the general case the subcategory of finitely generated semi-simple objects is a counterexample (if $\mathcal A$ has finite length objects that are not semi-simple). $\endgroup$
    – Dag Oskar Madsen
    Commented Apr 29, 2020 at 22:04
  • $\begingroup$ Even assuming that $\mathcal B$ is closed under extensions in $\mathcal A$, the answer to the question in the first paragraph is negative. Consider a semisimple finite-dimensional Lie algebra $\mathfrak g$ over a field of characteristic $0$, take $\mathcal A$ to be the category of all $\mathfrak g$-modules and $\mathcal B$ to be the full subcategory of finite-dimensional $\mathfrak g$-modules. Then $\mathcal B$ is semisimple, so all the objects of $\mathcal B$ are projective and injective in $\mathcal B$, but they are neither projective nor injective in $\mathcal A$ (if nonzero). $\endgroup$
    – Leonid Positselski
    Commented Apr 30, 2020 at 4:07
  • $\begingroup$ The answers to the questions in the second paragraph are positive though. But this is standard material, may be better to ask on math.stackexchange.com. $\endgroup$
    – Leonid Positselski
    Commented Apr 30, 2020 at 4:39

1 Answer 1

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You need more assumptions for this to be true.

Consider the ring $$A = \begin{bmatrix} \mathbb k & \mathbb k \\ 0 & \mathbb k \end{bmatrix},$$ where $\mathbb k$ is some field, and let $\mathcal A= \operatorname{mod} A$. There is a non-split exact sequence $$0 \to \begin{bmatrix} \mathbb k \\ 0 \end{bmatrix} \to \begin{bmatrix} \mathbb k \\ \mathbb k \end{bmatrix} \to \begin{bmatrix} \mathbb k \\ \mathbb k \end{bmatrix} / \begin{bmatrix} \mathbb k \\ 0 \end{bmatrix} \to 0$$ in $\mathcal A$, so the simple module $M= \begin{bmatrix} \mathbb k \\ \mathbb k \end{bmatrix} / \begin{bmatrix} \mathbb k \\ 0 \end{bmatrix}$ is not projective in $\mathcal A$.

Let $\mathcal B$ be the full exact abelian subcategory of finite direct sums of $M$. Then $M$ is projective in $\mathcal B$.

(In this example $\mathcal B$ is closed under $\mathcal A$-extensions, so that condition was not so relevant.)

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