Every finite ring with identity $a+a = 0$ is subring of $Mat_{n\times n} (\mathbb{F}_2)$ for some $n$? Is it true, that every finite ring $R$ with identity $\forall a \in R (a+a=0)$ is subring of $Mat_{n \times n}(\mathbb{F}_2)$ for some $n$?
 A: Yes. First, note that the hypothesis shows that the ring has a natural structure of a vector space over the field $\mathbb{F}_2$. Let's call the ring $A$. When we write $\text{End}_{\mathbb{F}_2}(A)$, we mean endomorphisms of that $\mathbb{F}_2$ space, not endomorphisms of the ring $A$ (any ring endomorphism is a vector space endomorphism, but the converse isn't true).
The ring $A$ then maps to $\text{End}_{\mathbb{F}_2}(A)$ by the rule $a \mapsto L_a$, where $L_a$  is the "left-multiplication by $a$." You have to check that this map is a ring homomorphism, i.e. that $L_{a}L_{b} = L_{ab}$ and that $L_{a+b} = L_a + L_b$. (The non-commutativity is tricky; if we'd used right-multiplication, this wouldn't be true. Don't forget the convention for composing functions.)
Finally, since the underlying set of $A$ is finite, certainly $A$ is a finite dimensional vector space; pick a basis. This choice (by the usual argument from linear algebra) gives an isomorphism of rings
$$
\text{End}_{\mathbb{F}_2}(A) \to \text{Mat}_{n \times n}(\mathbb{F}_2)
$$
where $n$ is the dimension, and the composition of these maps is the embedding that you want.
A: Hint: The additive group of the ring is a vector space of finite dimension over $\mathbb F_2$.
