Is the category of symmetric bimodules over a commutative ring closed under extensions?

Question

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?

Answer 1

Easier counterexample than the one Mare gave:

Let $A=k[X]$ and $N:=k^2$ where $X$ acts as $id_N$ on the right and as $\begin{pmatrix}1&1\\&1\end{pmatrix}$ on the left. Then $M:=\langle e_1\rangle$ is a symmetric sub-bimodule ($X$ acts as $id$ on both sides) and the quotient $K:=M/N=\langle e_2+M\rangle$ is a symmetric bimodule (again, $X$ acts as $id$ on both sides), but $N$ is not symmetric.

Answer 2

Here a counterexample found by computer:

Let $A=k[x]/(x^2)$ for a field $k$. Let $B:=A^{op} \otimes_k A$ be the enveloping algebra of $A$. The module category of $B$ is isomorphic to the module category of $A$-bimodules.

Take $M=A$ as a bimodule and $S$, the unique simple module of $B$.

Now a bimodule $N$ is symmetric if and only if $Hom_{A^e}(A,N) \cong N$.

$A$ and $S$ are symmetric. Now my computer tells me that there is a unique up to isomorphism non-split exact sequence $0 \rightarrow A \rightarrow W \rightarrow S \rightarrow 0$ (meaning that the dimension of $Ext_{A^e}^1(S,A)$ is equal to 1), where $W$ is isomorphic to the Jacobson radical of $B$.

We have that $Hom_{A^e}(A,W)$ is 2-dimensional but $W$ is 3-dimensional and thus $W$ can not be symmetric.

Answer 3

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.