$\mathbb{F}_{p}A$-module Yesterday I was introduced to the definition of a module in a course: "Homological Algebra". I'm doing a project involving $p-$groups and in the text I got from my supervisor they use the word: $\mathbb{F}_{p}A$-module. 
What does $\mathbb{F}_{p}A$-module mean? The problem is that the text does not define it and it is difficult to search for an answer using only the notation.
Note: In this case $A$ is a $p$-group. 
 A: I don't know what $A$ is, but hopefully it is a group. If that's not true, it's safe to ignore everything I write below! Even though $A$ is probably abelian, I'll write it multiplicatively to make the below easier. This isn't a rigorous definition but is hopefully enough to let you search.
The ring $\mathbb{F}_pA$, sometimes written $\mathbb{F}_p[A]$, is the ring of finite sums
$$
\sum_i \lambda_i a_i
$$
where $\lambda_i$ belongs to the finite field $\mathbb{F}_p$ and $a_i$ belongs to $A$. To add and multiply these, just "treat them like polynomials," remembering that we're writing $A$ multiplicatively. In other words, if $a_i \neq a_j$ then $a_i + a_j$ doesn't simplify, but $a_ia_j = a_k$ using the group law on $A$. (If this informal explanation doesn't make sense, the phrase to google is "group ring" or "group algebra.")
Since this is a ring, it makes sense to talk about a module over it. (The phrase to google is "module" or "R-module.")
A: When $A$ is a group, $\mathbb{F}_pA$ does just mean the group ring $\mathbb{F}_p[A]$ over the field $\mathbb{F}_p$. That is as a set
$$\mathbb{F}_p[A] = \{\sum_{g\in A} m_g \cdot g | m_g \in \mathbb{F}_p \}$$
EDIT: Since $A$ is supposed to be a $p$-group, an $\Bbb F_pA$-module $M$ is just an abelian group with a scalar multiplication by $\Bbb F_pA$. Since $1 \in M$, you can just view $\Bbb F_pA$ as a submodule of $M$. 
Another way of thinking about a $\Bbb F_pA$-module $M$ is that $M$ is a linear representation of the $p$-group $A$ over the field $\Bbb F_p$, since every element $a\in A$ gives rise to an action $GL_n(M)$ for some $n \geq 0$.
