Short exact sequence of algebras implies bimodule structure

Question

I read a statement in "Algebraic Operads" which I don't understand :

Let $$0 \rightarrow M \rightarrow A' \rightarrow A \rightarrow 0 $$

be a short exact sequence of associative algebras over the same field, such that the product in M is 0. Then M is a bimodule over A.

I don't see how M inherits any bimodule structure (or even say left-module).

Answer 1

Are they algebras over the same field? Is there any relationship between $A$ and $A'$?

You don't get $A' \cong A \oplus M$ unless the sequence splits. If it happens to split, then you have maps \begin{align*} j: A' & \rightarrow M\\ q: A &\rightarrow A' \end{align*} such that $ji = id_{M}$ and $pq = id_{A}$. Then I suppose you could define for $m \in M$ and $a \in A$ $$m\cdot a = m \cdot jq(a)$$ and $$a \cdot m = jq(a) \cdot m$$

Answer 2

Let $\phi : A \rightarrow B$ be a surjection of associtive algebras over a ring $R$ with kernel $M$, such that $M^2 = 0$. Then $M$ has a canonical $B$-bimodule structure. (note: taking $R = \mathbb{Z}$ we get the case for rings).

We define $\mu : B \times M \rightarrow M$ where $\mu((b,m)) = a m$ where $a$ is taken such that $\phi(a) = b$. If $a_1, a_2$ are such that $\phi(a_1) = b = \phi(a_2)$, then $a_1 - a_2 \in M$, so that $(a_1 - a_2)m = 0$ (we have assumed that $M^2 = 0$), so that $a_1 m = a_2 m$. So this is well defined. $\mu$ is $R$-balanced, or in other words, for $a \in A$, $\mu((b\phi(a), m)= b \phi(a) m = \mu(b, \phi(a) m)$. We get a map $\mu : B \otimes_R M \rightarrow M$ making $M$ into a left $B$-module.

A symmetrical construction gives a map $\mu : M \otimes_R B \rightarrow M$, making $M$ into a right $B$-module.