When does induced module functor preserve indecomposables Let $A$ and $B$ be unital algebras over an algebraically closed field. Let $f:A\twoheadrightarrow  B$ be a surjective algebra homomorphism. Then for a right $A$-module $M$ we can define the right $B$-module $M\otimes_{A} B$, the induced module. My question is: if $M$ is indecomposable, does it follow that $M\otimes_{A} B$ is also indecomposable? 
My feeling is that $A$ being a "bigger" algebra since $f$ is surjective should make the answer yes. I tried to use the Hom-tensor adjunction to show that $End_B(M\otimes_{A} B, M\otimes_{A} B)$ but I didn't succeed...
If the answer is negative and there is a simple, well-known additional assumption that guarantees indecomposability of the induced module I would be happy to know it. Thanks!
 A: Here are two counterexamples where $A$ is commutative. Let $k$ be the ground field. In both of the two examples below, the $A$-module $M$ is indecomposable since $M$ is torsion-free and $M \otimes_{A} \mathrm{Frac}(A)$ has dimension $1$ as a $\mathrm{Frac}(A)$-vector space.


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*Set $A := k[t]$ and $B := k \times k$ and let $A \to B$ be the $k$-algebra map sending $t \mapsto (0,1)$. Take $M := A$. Then $M \otimes_{A} B \simeq B$ and $B \simeq (k \times 0) \oplus (0 \times k)$.

*Set $A := k[s,t]$ and $B := k$ and let $A \to B$ be the $k$-algebra map sending $s,t$ to $0$. Take $M := (s,t)$ (i.e. the ideal of $A$). Then $M$ has the presentation $M \simeq (Ax \oplus Ay)/(tx-sy)$ and $M \otimes_{A} B \simeq k \oplus k$.
On the other hand, if $A$ is semisimple, then indecomposable $A$-modules are also irreducible so the canonical surjective $A$-linear map $M \to M \otimes_{A} B$ is either the zero map or an isomorphism, so if $M \otimes_{A} B$ is nonzero then it is indecomposable as a $B$-module.
