Prove that an isomorphism $\Phi : F[x] \to F’[x]$ preserves seperability.

Question

If we have a field isomorphism $\phi : F \to F’$. Then this induces a natural isomorphism $\Phi : F[x] \to F’[x]$. If we have a separable polynomial say f(x) in F[x] then is it true that $\Phi(f)=f’(x)$(say) is again seperable over $F’[x]$? Is the converse true?
My attempts: for field with characteristic 0 and for finite field a polynomial is seperable polynomial iff it is disjoint product of irreducibles.
So for this let $f=f_1f_2f_3..f_m$ then $\Phi(f)=\Phi(f_1) \Phi(f_2)…\Phi(f_m)$ which is obviously seperable by the above fact. But what should I do for infinite field but of characteristic finite case?

Answer 1

Let $K$ and $K'$ be algebraic closures of $F$ and $F'$, respectively. Show that $\phi$ can be extended to an isomorphism $\overline\phi\colon K\to K'$ (i.e., such that $\overline \phi|_F=\phi$) and that $\overline\phi$ bijects the roots of $f$ in $K$ with the roots of $\Phi(f)$ in $K'$.

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Let $\mathcal Z$ be the set of all field homomorphisms $\psi\colon E\to K'$ where $F\subseteq F\subseteq K$ and $\psi|_F=\phi$. We say $(\psi_1\colon E_1\to K')\preceq (\psi_2\colon E_2\to K')$ if $E_1\subseteq E_2$ and $\psi_2|_{E_1}=\psi_1$, thereby turning $\mathcal Z$ into a partially ordered set. If $\mathcal C=\{\,\psi_i\colon E_i\to K'\mid i\in I\,\}$ is non-empty and totally ordered by $\preceq$, let $\hat E=\bigcup_{i\in I}E_i$ and define $\hat\psi\colon \hat E\to K'$ by setting $\hat\psi(a)=\psi_i(a)$ if $a\in E_i$. Then $\hat\psi$ is well-defined, is $\in \mathcal Z$ and is an upper bound for $\mathcal C$. By Zorn's lemma, $\mathcal Z$ has a maximal element $\psi_\max\colon E_\max\to K'$.

Let $L=\psi_\max(E_\max)$.

Claim 1. $E_\max=K$.

Proof. Let $a\in K$. Then $a$ is algebraic over $E_\max$, hence there exists irreducible $g\in E_\max[X]$ with $g(a)=0$. $\psi_\max\colon E_\max\to K'$ induces $\Psi_\max\colon E_\max[X]\to L[X]\subseteq K'[X]$. Let $b\in K'$ be any root of $\Psi_\max(g)$. Verify that by setting $\psi(a)=b$, we obtain an extension of $\psi_\max$ to a homomorphism $\psi\colon E_\max[a]\to K'$. By maximality of $\psi_\max$, we must have $\psi=\psi_\max$, i.e., $a\in E_\max$. $\square$

Claim 2. $L=K'$.

Proof. Let $b\in K'$. Then $b$ is algebraic over $L$, hence there exists irreducible $h\in L[X]$ with $h(b)=0$. As $\Psi_\max\colon K[X]\to L[X]$ is an isomorphism, there is $g\in K[X]$ with $\Psi_\max(g)=h$. Let $a\in K$ be a root of $g$. Then $\psi_\max(a)\in L$ is a root of $h$. It follows that $X-\psi_\max(a)$ divides $h$ in $L[X]$. It follows that $h(X)=X-\psi_\max(a)$ and $b=\psi_\max(a)\in L$. $\square$

We conclude that $\overline \phi:=\psi_\max$ is indeed an isomorphism $K\to K'$ with $\overline\phi|_F=\phi$.

For any $f\in F[X]$ and $a\in K$ a root of $f$, it is clear that $\overline\phi(a)\in K'$ is a root of $\Phi(f)\in F'[X]$, i.e., $\overline\phi$ maps roots of $f$ to roots of $\Phi(f)$. Likewise, $\overline\phi^{-1}$ maps roots of $\Phi(f)$ to roots of $\Phi^{-1}(\Phi(f))=f$. These mappings between roots of $f$ and roots of $\Phi(f)$ are clearly inverses of each other, hence $\overline hi$ induces a bijection of the roots of $f$ with the roots of $\Phi(f)$. In particular, $f$ and $\Phi(f)$ have the same number of roots. As also clearly $\deg f=\deg\Phi(f)$, we conclude that the number of roots of $f$ (in an algebraic closure) corresponds to its degree if and only if the number of roots of $\Phi(f)$ corresponds to its degree, i.e., $\Phi$ preserves separability.

Answer 2

If $f \in F[x]$ is separable, then there are polynomials $a,b$ such that $af+bf’=1$. Apply $\Phi$ to thus equality to find out that $\Phi(f)$ is separable.