Let $F \subseteq E$ be an extension field.
A polynomial $f\in F[x]$ is separable if every irreducible factor of $f$ has distinct roots. An element $\alpha \in E$ is separable if the minimal polynomial of $\alpha$ over $F$ is separable. An extension field $E$ is separable if every element in $E$ is separable.
My question is as follows: Let $\alpha \in E$ be a separable element. I want to prove or disprove that $F[\alpha]$ is a separable extension.
One way (but perhaps not the best way) to prove this is using Galois theory. First, one proves the following theorem (see page 42 of Patrick Morandi's Field and Galois Theory):
Theorem 4.9. Let $K$ be an algebraic extension of $F$. Then the following statements are equivalent:
$K$ is Galois over $F$.
$K$ is normal and separable over $F$.
$K$ is a splitting field of a set of separable polynomials over $F$.
Patrick Morandi defines $K$ to be Galois over $F$ if $F = \mathcal{F}(\operatorname{Gal}(K/F))$, where $\operatorname{Gal}(K/F)$ is the set of all $F$-automorphisms of $K$.
Then, as a corollary (see page 44) we get the required result. Here's a proof of the result assuming the above theorem:
Let $E/F$ be an extension of fields and let $\alpha \in E$ be separable over $F$. Then, $\min(F,\alpha)$ is separable. If $K$ is a splitting field of $F$ contained in an algebraic closure of $E$, then $K/F$ is normal and separable by the above theorem. Since $F \subset F(\alpha) \subset K$, $F(\alpha)$ is also separable over $F$.
(Remark: since $\alpha$ is algebraic over $F$, $F[\alpha] = F(\alpha)$.)
In fact, the same method of proof allows one to show that if $\alpha_1,\dots,\alpha_n \in E$ are separable over $F$, then $F(\alpha_1,\dots,\alpha_n)$ is separable over $F$. Simply let $K$ be the splitting field of the set of separable polynomials $\{ \min(F,\alpha_i) : 1 \leq i \leq n \}$. Then, $K$ is normal and separable over $F$ by the theorem, and so, $F(\alpha_1,\dots,\alpha_n)$ is separable over $F$.
(1) Galois theoretic approach (discussed more below)
(2) "From first principles" field embedding counting,
KConrad1 Theorem 3.5. Let $L/K$ be a finite extension and write $L = K(α_1, . . . , α_r)$. Then $L/K$ is separable if and only if each $α_i$ is separable over $K$
Proven by a fairly fiddly looking argument counting field extensions.
(3) Focusing on char $p$ ($F$ field, $\alpha$ separable on $F$. Is $F(\alpha)$ a separable extension of $F$?, Finite extension of a characteristic p field is separable if and only if $E=F(E^p)$, If the elements $y_1,y_2,\dots,y_r \in E$ are linearly independent over $F$, show that $y_1^p,y_2^p,\dots,y_r^p$ are linearly independent over $F$.)
(4) More complicated approaches using traces, tensor products or derivations, see KCd answer If $\alpha$ separable over $F$ then $F(\alpha )/F$ is a separable extension.
Let me summarize/simplify the Galois theoretic proof/ideas (discussed in a previous answer)
I prove in my answer here (using nothing more than the "lifting lemma" for field homomorphisms) that an extension $L/K$ s.t. $L=K(\alpha_1,\ldots, \alpha_n)$ where the minimal polynomials $\text{Irr}(\alpha_i,K)\in K[t]$ all split and are separable in $L$ satisfies $L^G=K$ where $G:= \text{Aut}(L/K)$. Then, I mentioned that this MSE post (Normal field extension separable over its fixed field) shows that then $L/K$ is a separable extension.
This PDF recasts some of the Galois theory in terms of degree counting, so it's like halfway between (1) and (2). I like that it highlights the surprising "Buy one get everything else free" nature of this result.
Theorem 1.5 $E/F$ is a separable splitting field if and only if it is the splitting field of a separable polynomial (or finite set of irreducible separable polynomials).
Here again the “only if” is easy (we’ll omit the proof), whereas the “if” is highly non- obvious. It says that after adjoining the roots of a single separable polynomial, presto!—every element in the resulting field extension is separable.
Fix an algebraic closure $\overline{K[\alpha]}$ (or just fix the splitting field of $P_{min,\alpha,K}(X)$). As $\alpha$ is separable this means its minimal polynomial has degree $>0$ and has only simple roots, now consider the $K-$endomorphism $[\cdot\alpha]:K[\alpha]\longrightarrow K[\alpha]$ consisting of multiplying by $\alpha$ and consider $P_{min,[\cdot\alpha],K}(X)$ it is easy to show (if one decide to take such minimal polynomial their monic version) that $P_{min,[\cdot\alpha],K}(X)=P_{min,\alpha,K}(X)$, hence $P_{min,[\cdot\alpha],K}(X)$ has only simple roots, this means equivalently that $[\cdot\alpha]\in End_{K}(K[\alpha])$ is diagonalizable i.e. that there exist a basis $\mathcal{B}\in \overline{K[\alpha]}^{[K[\alpha]:K]}$ such that $mat_{\mathcal{B}}([\cdot\alpha])$ as a diagonal matrix representation in $Mat_{[K[\alpha]:K]}(\overline{K[\alpha]})$ i.e. $mat_{\mathcal{B}}([\cdot\alpha])=diag(\lambda_i\in\overline{K[\alpha]})$. Now Given $w\in K[\alpha]$ then $w=ev_{\alpha}(Q(X))=Q(\alpha)$ for some $Q(X)\in K[X]$ . We compute: $$mat_{\mathcal{B}}([\cdot w])=mat_{\mathcal{B}}([\cdot Q(\alpha)])=mat_{\mathcal{B}}(Q([\cdot\alpha]))=Q(mat_{\mathcal{B}}([\cdot\alpha]))$$ (since matrix representation of sum and product IS the sum and product of matrix representation). But as $mat_{\mathcal{B}}([\cdot\alpha])$ is diagonal this means $Q(mat_{\mathcal{B}}([\cdot\alpha]))$ is diagonal, hence $mat_{\mathcal{B}}([\cdot w])$ is diagonal which is equivalent to say $P_{min,[\cdot w],K}(X)=P_{min,w,K}(X)$ has only simple roots which shows $w$ is separable.
Now you can show by induction that $K[a_1,...,a_n]$ is separable iff a_1,...,a_n are separable elements.
