Representations of superalgebras If I have a superalgebra $A$ and consider the category of finite-dimensional $A$-modules.
Is this category the same as the category of finite-dimensional $A$-modules which are super-vector spaces? 
I.e. is every representation of $A$ a super-vector space?

Let $A=k[x]$ graded so that a monomial $x^i$ is even or odd according to the partity of $i$. Let $V=k^2$ and view $V$ as an $A$-module in such a way that multiplying by $x$ has matrix $\begin{pmatrix}1&1\\0&1\end{pmatrix}$. It is easy to see that $V$ does not have a nontrivial direct sum decomposition preserved by $x^2$. It follows that if it is in some way a supermodule over $A$ then it is all odd or all even. But that is not possible, since $x$ changes parity and is an isomorphism. 

A finite dimensional example: let $A=k[x]/(x^9)$ graded again as before, let $V=k^3$, and let $A$ act on $V$ so that $x$ acts as $\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}$. Unless I messed up, the only decompositions of $V$ as nontrivial direct sums of subspaces invariant under $x^2$ are


*

*$\langle(a,b,0)\rangle\oplus\langle(1,0,0),(d,e,f)\rangle$ with $(e,d)\neq(0,0)$ and $bf\neq0$.

*$\langle(\alpha,\beta,0)\rangle\oplus\langle(a,b,c),(d,b,c)\rangle$ with $a\neq d$ and $c\beta\neq0$.
In each of this cases, the image under $x$ of the $2$-dimensional summand is not contained in the other summand. It follows that if $V$ is a graded $A$-module, then it is all odd or all even. But that is impossible, since $x$ changes parity and acts not by zero.
