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Let $F$ be a field of characteristic $\operatorname{char}F=p\neq 0$. It is well-known that a simple extension $F<F(\alpha)$ is separable if and only if $F(\alpha^{p^k})=F(\alpha)$, for any $k\geq 1$, if and only if $F(\alpha^p)=F(\alpha)$. Is $\min(\alpha, F)=\min(\alpha^p, F)$ in this case?

My proof is: define $\phi: F(\alpha)\to F(\alpha^p)$ to be $\phi|_F=1_F$(the identity map on $F$), and $\phi(\alpha^n)=(\alpha^p)^n,\;n\geq 1$, and extend $\phi$ to $F(\alpha)$ by linearity. Then $\phi$ is a surjective field homomorphism, hence an isomorphism. Furthermore, $\phi$ is an $F$-isomorphism by definition, with $\phi(\alpha)=\alpha^p$. Hence, $\min(\alpha, F)=\min(\alpha^p, F)$ by an answer given here: Algebraic extensions $F(\alpha), F(\beta)$ with a $F$-isomorphism yield the same minimal polynomial for $\alpha, \beta$.

Am I missing anything?

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You are missing something: your $\phi$ need not be a field homomorphism since it need not be multiplicative. I wonder if you checked your argument on an example where $F \not= \mathbf F_p$.

(You implicitly intend to have $\phi$ be $F$-linear, and should have said that directly in your definition of $\phi$. EDIT: the definition of $\phi$ was originally less explicit than it is now.)

Example. The polynomial $x^2 - x - 1$ is irreducible over $\mathbf F_2$. Set $F = \mathbf F_2(\gamma)$, where $\gamma$ is a root of $x^2 - x - 1$, so $|F| = 4$ and $\gamma^2 = \gamma + 1$. The polynomial $f(x) = x^2 - x - \gamma$ is irreducible over $F$. Let $\alpha$ be a root of $f(x)$, so $\alpha^2 = \alpha + \gamma$ and $F(\alpha)/F$ is separable with $$ F(\alpha) = F + F\alpha. $$

Your description suggests that you want to define $\phi \colon F(\alpha) \to F(\alpha)$ by $$ \phi(a + b\alpha) = a + b\alpha^2 = a + b(\alpha + \gamma) = (a+b\gamma) + b\alpha $$ for $a, b \in F$. In particular, $\phi(\alpha) = \alpha^2 = \gamma + \alpha$. Then $$ \phi(\alpha^2) = \phi(\gamma + \alpha) = \gamma + \alpha^2 = \alpha $$ while $$ \phi(\alpha)^2 = (\gamma + \alpha)^2 = \gamma^2 + \alpha^2 = (\gamma+1) + (\alpha+\gamma) = 1+\alpha, $$ so $\phi(\alpha^2) \not= \phi(\alpha)^2$.

The answer to your question ("do $\alpha$ and $\alpha^p$ have the same minimal polynomial over $F$?") is NO in general. Indeed, when $F$ is a finite field of order $q$ and $\alpha$ is algebraic over $F$, the roots of the minimal polynomial of $\alpha$ over $F$ are $\alpha, \alpha^{q}, \alpha^{q^2}, \ldots$, which need not include $\alpha^p$ if $q \not= p$. In the example above, where $p = 2$ and $q = 4$ (I needed $q \not= p$ to get a counterexample), the roots of $x^2 - x-\gamma$ are $\alpha$ and $\alpha^4$, where the second root is $\alpha+1$, while $\alpha^2 = \alpha+\gamma$ is different from $\alpha+1$ and $\alpha$.

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  • $\begingroup$ I edited the definition of $\phi$, as you suggested, thank you - sorry if my new definition makes your answer slightly less clear. With my new definition, your example shows that $\phi$ is not well-defined - indeed, $\phi(\alpha^2):=\phi(\alpha)^2$ by my definition(this is the definition I had in mind), but also $\phi(\alpha^2)=\alpha\neq \phi(\alpha)^2=1+\alpha$. I am pretty sure the same logic applies for infinite fields - I am just very new to the subject, and cannot supply my own examples sometimes. $\endgroup$ May 18, 2023 at 5:56
  • $\begingroup$ @user1104937 okay, I modified my 2nd paragraph to account for your edit. $\endgroup$
    – KCd
    May 18, 2023 at 6:09

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