Subring of $M_7(\mathbb{Z}_2)$ isomorphic to $\mathbb{F}_{128}$? Let $A \subset M_7(\mathbb{Z}_2)$ be a subring such that no proper nonzero subgroup $V \subset \mathbb{Z}_2^7$ is invariant under all matrices in $A$. I suspect that $A \cong \mathbb{F}_{128}$, but I'm at a loss on how to show it... could anyone give me a hint or help me out?
 A: The first sentence of the problem's statement says that $(\mathbb{Z}/2\mathbb{Z})^7$ is a simple $A$-module; hence, by Schur's Lemma, the ring $A' = \text{End}_A((\mathbb{Z}/2\mathbb{Z})^7)$ is a skew field. By the Artin-Wedderburn Theorem, $A'$ is a field, thus $A' \cong \mathbb{F}_{2^n}$ for some $n$. Since $A'$ acts on $\mathbb{Z}/2^7\mathbb{Z}$, the latter has a structure of an $\mathbb{F}_{2^d}$ vector space, hence $2^7 = (2^n)^d$ (where $d$ is the dimension of this vector space), so $n\,\vert\,7$. Thus, either $n = 1$ and $A' = \mathbb{Z}/2\mathbb{Z}$, or $n = 7$ and $A' = \mathbb{F}_{128}$, $d = 1$.
By the Jacobson Density Theorem, $A = \text{End}_{A'}((\mathbb{Z}/2\mathbb{Z})^7)$. If $A' = \mathbb{Z}/2\mathbb{Z}$, then $A = \text{Mat}_7(\mathbb{Z}/2\mathbb{Z})$, but this is excluded by $A$ being a strict subset of $\text{Mat}_7(\mathbb{Z}/2\mathbb{Z})$. Thus, $A' = \mathbb{F}_{128}$, and $(\mathbb{Z}/2\mathbb{Z})^7$ is a dimension $1$ vector space over $\mathbb{F}_{128}$, so $A \cong \text{End}_{\mathbb{F}_{128}}(\mathbb{F}_{128}) = \mathbb{F}_{128}$.
