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?
 A: In every context I've ever worked, an $A$-(super)module would be a $\mathbb{Z}/2\mathbb{Z}$-graded vector space $V=V_{\bar{0}}\oplus V_{\bar{1}}$ such that $a.v\in V_{\bar{i}+\bar{j}}$ for $a\in A_{\bar{j}}$ and $v\in V_{\bar{i}}$. 
Equivalently, a representation of $A$ is an even homomorphism $\rho:A\to M_{m,n}(\mathbb{C})$ (or some other field if you want), where $\dim V_{\bar{0}}=m$ and $\dim V_{\bar{1}}=n$. 
The even part of $M_{m,n}(\mathbb{C})$ consists of the block matrices of the form
$$\begin{pmatrix}X&0\\0&Y\end{pmatrix}$$
where $X$ is $m\times m$ and $Y$ is $n\times n$. The odd part consists of matrices of the form
$$\begin{pmatrix}0&X'\\Y'&0\end{pmatrix}$$
where $X'$ is $m\times n$ and $Y'$ is $n\times m$.
In the special case $m=n$, there is an interesting subalgebra $Q_n(\mathbb{C})\subseteq M_{n,n}(\mathbb{C})$ consisting of matrices of the form
$$\begin{pmatrix}X&Y\\Y&X\end{pmatrix}.$$
The super vector space $V=\mathbb{C}^{n|n}$ is irreducible as a $Q_n(\mathbb{C})$-supermodule. But, if we forget the $\mathbb{Z}/2\mathbb{Z}$-grading, $V$ decomposes as the direct sum of two isomorphic ungradable submodules $W\oplus\overline{W}$.
A: 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.
A: It's a matter of definition. A super algebra is nothing but a particular case of a graded algebra; when we speak of modules over a graded ring we typically mean graded modules. This grading is given by the same group that gives the algebra it's grading (in this case $\mathbb{Z}/2\mathbb{Z}$), and the multiplication is required to respect that grading. 
So the answer is yes, you are correct by definition. 
