Structure and Automorphisms of $\mathcal{O}_F/p\mathcal{O}_F$ as $\mathbb{F}_p$-algebra Let $F/\mathbb{Q}$ be a galois number field and $p \in \mathbb{Z}$ a prime, then $A=\mathcal{O}_F/p\mathcal{O}_F$ is an $\mathbb{F}_p$-algebra.
What can we say about the structure of $A$ and the group of $\mathbb{F}_p$-automorphisms Aut$(A)$, in terms of $F$ and the factorisation of $p\mathcal{O}_F$?
I know that $A$ is a finite $\mathbb{F}_p$-algebra since $\mathcal{O}_F$ is finitely generated over $\mathbb{Z}$ and $\dim A=[F:\mathbb{Q}]$. In addition we have the natural group homomorphism
$$Gal(F/\mathbb{Q})\to Aut(A)$$
but it seems to behave very differently depending on $F$ and $p$. By algebraic number theory this map is an isomorphism when $p$ is inert in $F$, but I also found that for
$F=\mathbb{Q}(\sqrt{d})$ with $d\equiv 2,3\mod{4}$ and $p=2$ this map is trivial.
 A: If $F/\mathbf{Q}$ is Galois, then
$$p = \prod_{i=1}^{r} \mathfrak{p}^{e}_i,$$
where $\mathcal{O}_F/\mathfrak{p} = A/\mathfrak{p} = k$ is an extension of $\mathbf{F}_p$ of degree $f$ and $fre=n=[F:\mathbf{Q}]$.
There is an isomorphism
$$\mathcal{O}_F/p = A = \prod_{i=1}^{r} k[x]/x^e,$$
Since the automorphism group must preserve the idempotents $(0,0,\ldots,0,1,0,\ldots 0)$, you get
$$\mathrm{Aut}(A) = \mathrm{Aut}(k[x]/x^e) \wr S_r = \mathrm{Aut}(k[x]/x^e)^r \ltimes S_r.$$
There is a surjection
$$\mathrm{Aut}(k[x]/x^e) \rightarrow \mathrm{Aut}(k) = \mathrm{Gal}(k/\mathbf{F}_p) = \mathbf{Z}/f \mathbf{Z},$$
which splits as a map of groups. The kernel $\Gamma$ consists of automorphisms of $k[x]/x^e$ which fix $k$. They are given by any polynomial map $x \mapsto p(x)$ where $p(x) = a x + \ldots $ and $a \ne 0$ and where the group operation is composition. If $e = 2$ this is isomorphic to $k^{\times}$, but when $e > 2$ it is not typically commutative. We have
$$\mathrm{Aut}(k[x]/x^e)  = \Gamma \ltimes \mathbf{Z}/f \mathbf{Z}.$$
Note that if $e \ge 2$ then $|\Gamma| = |k^{\times}| |k|^{e-2} = (p^f - 1) p^{f(e-2)}$.
Hence
$$|\mathrm{Aut}(A)| = \begin{cases} f^r r!, & e = 1, \\
(p^f - 1)^r p^{f(e-2)} f^r r!, & e \ge 2. \end{cases}$$
On the other hand, if $G$ is the Galois group of $F$, then $|G| = n = efr$. There is a map from $G$ to $\mathrm{Aut}(A)$, but in general the target will be much bigger except in a few exceptional cases one can work out by a case bash, including $e=r=1$ when they are both cyclic of degree $f$, and then a few degenerate cases like $(e,r,f) = (1,2,1)$ and $(2,1,1)$ and $(3,1,1)$.
As for whether the map is injective or not:
The stabilizer of $G$ acting on a single factor is just the decompostion group $D$ at that prime $\mathfrak{p}$. There is a map
$$D \rightarrow \mathrm{Aut}(k[x]/x^e),$$
While $D$ surjects onto $\mathrm{Aut}(k)$, it doesn't have to surject onto this group in general. But the kernel by definition is the higher ramification group $I_{e-1}$. The kernel of the entire action is the intersection of the stabilizers so the largest normal subgroup of $G$ contained inside $I_{e-1}$. Since $I_{e-1}$ is a pro-p group, the map is certainly injective under any of the following circumstances:

*

*$F$ is unramified at $p$, or more generally only tamely ramified at $p$.

*The group $G$ has no proper normal subgroups of degree $p$, for example when $[F:\mathbf{Q}]$ has degree prime to $p$.

Of course it can happen that $I_{e-1}$ contains a non-trivial normal subgroup, for example if $p=2$, $e=2$, $r=f=1$ then $G = I_{1}$ has order $2$.
