TITLE: Is the category of symmetric bimodules over a commutative ring closed under extensions?
QUESTION [6 upvotes]: Let $A$ be a commutative ring, and consider the category of bimodules over $A$.
An $A$-bimodule $M$ is called symmetric if $a\cdot m = m \cdot a$ for all $a \in A$, $m \in M$.
Is the category of symmetric bimodules over $A$ closed under extensions?
Namely, given an exact sequence of $A$-bimodules
$0 \to M \to N \to K \to 0$
where $M,K$ are symmetric, must $N$ also be a symmetric $A$-module?

REPLY [2 votes]: Here a general abstract argument that shows that the subcategory of symmetric bimodules is never extension closed in case $A$ is a commutative finite dimensional Frobenius algebra that is not semisimple:
Let $A$ be such an algebra and $B=A \otimes_K A$ its enveloping algebra and assume that the subcategory of symmetric bimodules is closed under extensions.
The simple module $S$ is symmetric and thus the subcategory of symmetric finite dimensional bimodules equals the module category of $B$.
It thus also contains $B$, but $Hom_B(A,B) \cong D(A) \cong A$ has dimension less than $B$ and thus $B$ is never symmetric. This is a contradiction and thus the subcategory of symmetric bimodules is never closed under extensions.