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Let $F$ be a field, let $f \in F[x]$, let $E$ be the splitting field of $f$, and let $e \in E$ be written in terms of the roots of $f$.

I'm looking for a simple way to establish if $e \in F$.

If $E/F$ is separable, then $E/F$ is Galois and so $e \in F$ if and only if $\sigma(e) = e$ for every $\sigma$ in the Galois group of $E/F$. Equivalently, we have that $e \in F$ if and only if $e$ is a symmetric function of the roots of $f$.

If $E/F$ is inseparable, then I do not know a simple way to establish if $e \in F$. For sure, the method for separable $E/F$ does not work. In fact, if we consider the classic example of $F = \mathbb{F}_p(t)$ for a prime $p$ and $f(x) = x^p - t$, then $E = F(t^{1/p})$ and $e = t^{1/p}$ is a symmetric function of the (single) roots of $f$ but $e \notin F$.

Thanks for any help

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    $\begingroup$ Your first statement about the separable case isn't quite accurate, e.g., $f(x)=(x^2-2)(x^2-3)$ and $F=\mathbb Q$. But you probably meant to insist that $f(x)$ be irreducible in $F[x]$. $\endgroup$
    – Joe Silverman
    Commented Dec 8 at 12:17
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    $\begingroup$ Probably something with derivations ("all derivations vanish") instead of automorphisms ("all automorphisms fix"). There has been a recent (very technical) work by Brantner and Waldron on inseparable Galois theory: people.maths.ox.ac.uk/brantner/… $\endgroup$
    – darij grinberg
    Commented Dec 8 at 21:12

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Of course, inseparability is only a problem if the characteristic $p$ of $F$ is positive, so let's suppose that it is.

I don't know how to answer this question without changing $f$. Specifically, replace $f$ by the minimal polynomial of $e$ over $F$, hence $E$ by the normal closure of $F[e]$ over $F$. Then $e$ belongs to $F$ if and only if (1) every $F$-linear automorphism of $E$ fixes $e$, and (2) every $F$-linear derivation of $E$ annihilates $e$. (@darijgrinberg has also commented on the importance of derivations, and given a pointer to Brantner and Waldron - Purely inseparable Galois theory I.)

The "only if" direction is obvious. For the "if" direction, condition (1) guarantees that $F[e]/F$ is purely inseparable, so that $E$ equals $F[e]$. In particular, either $e$ belongs to $F$, or $F[e^p]$ is a proper subfield of $E$. By Lang - Algebra, Proposition VIII.5.4, condition (2) guarantees that $e$ belongs to $E^p F = F[e^p]$, so that $F[e^p]$ equals $F[e] = E$, so that $e$ belongs to $F$.

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    $\begingroup$ NB: Lang's Proposition VIII.5.4 (sic) is about single-root extensions, not their normal closures. And I'm not sure if any derivation can be extended to the normal closure, so I don't know whether this applies here. $\endgroup$
    – darij grinberg
    Commented Dec 8 at 21:41
  • $\begingroup$ @darijgrinberg, re, thanks for the comment (and reminder that I missed the chapter information in my citation; now fixed). Lang's notation $K = k(x)$ seems, confusingly, to mean $K = k((x))$, where $(x)$ is a tuple of generating elements (as on p. 369), not necessarily a singleton. Otherwise, it seems strange to ask whether $K = k(x)$ is finitely generated! (To be sure, it seems strange anyway, since the tuple is finite.) Indeed, the proof proceeds by adjoining one generator at a time. … $\endgroup$
    – LSpice
    Commented Dec 8 at 21:50
  • $\begingroup$ … But anyway I think that it's no big deal, because I only need to pass to the normal closure to speak of $F$-automorphisms of $E$ rather than of $F$-embeddings of $E$ into an algebraic closure of $F$. We could speak of such embeddings instead; or we can note that, once $e$ is fixed by all such automorphisms, so that $F[e]/F$ is purely inseparable, it is therefore also normal. Right? I edited to emphasise this point. $\endgroup$
    – LSpice
    Commented Dec 8 at 21:51

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