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In this lecture on Frobenius endomorphisms, we imagine a Galois extension over $\mathbb{Q}$ given as $M=\mathbb{Q}/(f(x))$ for an irreducible monic polynomial $f(x)\in\mathbb{Z}[x]$. We then consider $R = \mathbb{Z}[x]/(f(x))$ and $R/pR$. If $f(x)=f_1(x)\dots f_k(x)\in (\mathbb{Z}/p\mathbb{Z})[x]$, then $R/pR$ splits into a product of fields $F_i: = (\mathbb{Z}/p\mathbb{Z})[x]/(f_i(x))$.

Any automorphism of $R$ will give an automorphism of $R/pR$, and this will permute the roots of $f(x)$. This will either permute the fields $F_i$ in the decomposition of $R/pR$, or permute the roots within each field.

The claim he makes that confuses me (at around 9:27) is that if an automorphism maps $F_1$ to itself, then this automorphism is the identity on $R$ if and only if it induces the identity on $F_1$. I can see that a non-identity automorphism on $R$ must give a non-identity automorphism on $R/pR$, but I don't see why I couldn't have an automorphism that acts as the identity on $F_1$, but has a non-trivial action on some other field $F_i$ in the decomposition of $R/pR$.

As an example, suppose I have a polynomial $f(x)$ over the rationals that has real and complex roots. If $f(x)=f_1(x)f_2(x)\mod p$, and all the real roots end up as roots of $f_1(x)$, then complex conjugation would be the identity on $F_1$ but not the identity on $F_2$. So this can't happen?

Sam Jaques
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It is not the case that $F_1$ contains just some of the roots of (the reduction of) $f$ while $F_2,\dots$ contain the others. Here, every field $F_i$ contains all roots of $f$: We have the composition of maps $$R\to R/p\cong \prod_{i=1}^k F_i\to F_1.$$ Since $f$ splits over $R$ into linear factors (this was an assumption, see around 5:00. An example that shows that this is not the case automatically when $R$ is a non-maximal order is given here), it will also do so over $F_1$ due to the projection map above. Thus the argument around 10:30 works: if $\sigma$ is a non-trivial automorphism of $M$ it permutes the roots of $f$ non-trivially. If $\sigma$ induces an automorphism of $F_1$ it will also induce a non-trivial permutation of the roots of $f$ in $F_1$ (since $f$ is separable mod $p$ and has all of its roots contained in $F_1$ by the above), hence $\sigma$ will not restrict to the identity on $F_1$.

leoli1
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  • Ah, so a big portion of the work is done by the fact that $\mathbb{Q}[x]/(f(x))$ is a Galois extension? My "counterexample" of real and complex roots is not really a counterexample because it will not have this property. – Sam Jaques Sep 21 '21 at 16:50
  • Yes, we need that $M$ (actually $R$) contains the roots of $f$. Since $f$ is irreducible this means that $M$ is the splitting field of $f$ over $\Bbb Q$ and hence Galois. It is not possible for a such a polynomial $f$ to have both real and non-real roots: If $\alpha$ is a real root, then $M=\Bbb Q(\alpha)\subseteq \Bbb R$ and as all roots of $f$ are in $M$ all of them are real. – leoli1 Sep 21 '21 at 17:51
  • Or more generally: if $\sigma$ is an automorphism of $M$ that fixes one root of $f$, it has to be the identity. This is because $M$ is generated by a single root of $f$. The statement about complex/real roots is then a special case when $\sigma=$ complex conjugation. – leoli1 Sep 21 '21 at 17:54