# Short exact sequence of algebras implies bimodule structure

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).

• Are your algebras unital? – darij grinberg Apr 17 at 22:14

## 2 Answers

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.

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 $$a \cdot m = jq(a) \cdot m$$

• My bad, the sequence does not split. However the algebras are on the same field, and are associative algebras. At some point one needs to use that the product in M is zero I guess. – Gericault Apr 17 at 21:28
• @Gericault What do you mean by "the product in M is zero"? – Auclair Apr 17 at 21:47
• @Auclair That means that $m n = 0 \in A$ for each $m, n \in M$, or in other words $M^2 = 0$ as an ideal. – Dean Young Apr 17 at 21:49
• @DeanYoung Ah, thanks for clearing that up. – Auclair Apr 17 at 21:51