$A \subseteq B \subseteq C$, with $A$ and $C$ simple rings, but $B$ is not a simple ring

Question

Let $A \subseteq B \subseteq C$ be three associative $k$-algebras, where $k$ is a field.

Assume that $A$ and $C$ are simple rings. (Recall that a ring $R$ is simple, if it has no two-sided ideals other than $0$ and $R$).

Is it true that $B$ must also be a simple ring? If not, it would be nice to see a counterexample. Does the characteristic of $k$ relevant to the answer?

It would be nice to see a few counterexamples, according to the commutativity or non-commutativity of the rings concerned. (But just one counterexample will also be ok).

Perhaps it is also relevant to know if $C$ is finitely generated as a $B$-module, etc.

Thank you very much!

Answer 1

No, $B$ need not be simple: Let $A=k$, $B=k[X]$, and $C=k(X)$.

Answer 2

For a finite-dimensional counterexample, let $A=K$ be a field, $C$ a matrix ring over $k$. Then $A$ and $C$ are simple but there is no reason that $B$ must be.

For a simple example, a nonzero nilpotent matrix generates a subring isomorphic to $k[X]/(X^n)$, which is not simple.

Note that all finite-dimensional counterexamples over a field are non-abelian.