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.

  • $\begingroup$ The cubic root of 2 comes to mind. $\endgroup$
    – Asaf Karagila
    May 19, 2017 at 21:54
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    $\begingroup$ @AsafKaragila : What do you mean ? I think it follows from $F(\alpha)$ be contained in $K/F$ such that $F = K^G$ (the fixed field of $G$) where $G = Aut(K/F)$ $\endgroup$
    – reuns
    May 19, 2017 at 22:04
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    $\begingroup$ It's true. See section V.4 of Lang's Algebra. $\endgroup$
    – user49640
    May 20, 2017 at 0:27
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    $\begingroup$ The result is true, and even more: $F(\alpha_1,\ldots \alpha_n)$ is a separable extension iff each $\alpha_i$ is separable over $F$. See: math.uconn.edu/~kconrad/blurbs/galoistheory/separable1.pdf $\endgroup$
    – Xam
    May 20, 2017 at 3:34
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    $\begingroup$ @AsafKaragila I presume you were talking about $\mathbb Q(\sqrt[3]{2})$ over $\mathbb Q$. Note that any field of characteristic $0$ is perfect, so by default any extension of a field of characteristic $0$ is separable. $\endgroup$
    – Cauchy
    Aug 18, 2017 at 16:40

1 Answer 1


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:

  1. $K$ is Galois over $F$.

  2. $K$ is normal and separable over $F$.

  3. $K$ 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 $F$.


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