# Separable extension characterization

Let $$F$$ be a field of characteristic $$\operatorname{char}F=p\neq 0$$. It is well-known that a simple extension $$F is separable if and only if $$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?

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 $$\alpha$$.

• 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. May 18, 2023 at 5:56
• @user1104937 okay, I modified my 2nd paragraph to account for your edit.
– KCd
May 18, 2023 at 6:09