This is probably a very basic question in Hochschild theory.
Let $k$ be a field, and let $A$ be a $k$-algebra (which is not commutative).
If $M$ is an $A$-bimodule, then the $n$-th Hochschild cohomology of $A$ with coefficients in $M$ is given by
$Ext^n_{A^e}(A,M)$.
Why is this not an $A^e$-module? First, both $A$ and $M$ are left $A^e$-modules, so this $Ext$ is defined, and is an abeliean group. But can't I also view $A$ as a right $A^e$-module? so that this is has a structure of an $A^e$-module?
No, it doesn't have an $A^e$-module structure, because the left and right actions of $A^e$ on $A$ don't commute unless $A$ is commutative, so $A$ is not an $A^e-(A^e)^{op}$ bimodule.
That doesn't prove that there's not an $A^e$-module structure defined some other way, but to see that this isn't the case, consider the example $k=\mathbb{R}$ and $A=\mathbb{H}$, the quaternions, considered as an $\mathbb{R}$-algebra. Then $H^0(A,A)=\mathbb{R}$, the centre of $A$, which can't have an action of $A^e$, or even of $A$, as $A$ doesn't have any modules with real dimension 1.