Let $k$ be a field, and $$0\longrightarrow k\stackrel{\alpha}{\longrightarrow} A \longrightarrow k\longrightarrow 0$$ where $A = k[x]/(x^2)$ and $\alpha$ is given by $\alpha(t) = tx$.
I'd like to show this short exact sequence does not split.
$A$ has a nilpotent element, but $k \oplus k$ does not.
But this requires that one should see them as rings.
How could I prove this?
Interpreting this as a sequence of $A$ modules (as we are all suspecting was intended) one would note that $A=k[x]/(x^2)$ is directly indecomposable (actually local even.). So no mapping at all like that is going to split it properly.
It can't have a nontrivial pair of summands. Each half would have to be contained in the unique maximal ideal, and could never generate all of $A_A$.