I am trying to solve the following problem in I. Martin Isaacs' Algebra: A graduate course, p.290:

Let $f(X),g(X) \in F[X]$ and suppose $E \supseteq F$ is the splitting field both for $f(X)$ and for $g(X)$ over $F$. Show that $f(X)$ is separable over $F$ if and only if $g(X)$ is separable over $F$.

To prove this, one only needs to show one direction since $f(X)$ and $g(X)$ are interchangeable. To be honest, I have no idea where to begin. By definition, $E$ would have to be the smallest field containing all the roots of both $f(X)$ and simultaneously $E$ would have to be the smallest field containing all the roots of both $g(X)$. I cannot, however, see how I can relate repeated roots in $f(X)$ to repeated roots in $g(X)$ with this information.

Here is my first attempt:

Let $\alpha_1,...,\alpha_n$ and $\beta_1,...,\beta_m$ be the roots of $f(X)$ and $g(X)$ in the algebraic closure of $F[X]$. Using Isaacs' definition of a splitting field we see that $E=F(\alpha_1,...,\alpha_n)$ and $E=F(\beta_1,...,\beta_m)$, and so $F(\alpha_1,...,\alpha_n)=F(\beta_1,...,\beta_m)$. Assume $f(X)$ is separable over $F$. Then every irreducible component of $f(X)$ has distinct roots. This implies that the minimal polynomial of the roots of $f(X)$ are separable. Let $g_i(X)$ be an irreducible component of $g(X)$ and assume $g_i(X)$ has a multiple root, say $\beta_k$. What can we say about $\beta_k$? Well, since $\beta_k \in F(\alpha_1,...,\alpha_n)$ we know


for some $a_1,...,a_n \in F$.

I'm thinking that there must be some way to obtain a contradiction about the non separability of $g_i(X)$ from the fact that the minimal polynomials of all the $\alpha_j$'s are separable.

  • $\begingroup$ Let $\alpha \in E$. An extension $E/F$ is separable if the minimal polynomial $m(X)=\text{min}_F(\alpha)$ is separable. $\endgroup$ – Holdsworth88 Jul 14 '12 at 13:19
  • $\begingroup$ @dylanmoreland the question is copied almost verbatim; I haven't missed any hypotheses. I will post the verbatim problem in the question. $\endgroup$ – Holdsworth88 Jul 14 '12 at 13:22
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    $\begingroup$ Something is wrong in the question. "Suppose $E\supset F$ is the splitting field ..." But then $E$ doesn't occur anywhere in the problem. It is irrelevant data, so surely take $f$ to be separable and $g$ to not be and you have a counterexample. $\endgroup$ – Matt Jul 14 '12 at 16:26
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    $\begingroup$ @Matt I'm confused: the fact that they have the same splitting field is a piece of data, right? The argument I have in mind is, "If $f$ is separable then $E/F$ is Galois, hence separable, and it follows that $g$ is separable." $\endgroup$ – Dylan Moreland Jul 14 '12 at 19:36
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    $\begingroup$ @Holdsworth88: As with a lot of field-extension-theoretic definitions, there are many equivalent ways of defining "separability". Which definition and what equivalences do you already know? $\endgroup$ – Arturo Magidin Jul 14 '12 at 23:21

Using Dylan Moreland's suggestion, I have the following proof.

Assume $f(X)$ is separable over $F$. Since $E$ is the splitting field for $f(X)$ over $F$ and $f(X)$ is separable, we see that $E/F$ is Galois. This is equivalent to $E$ being a normal, separable extension. Let $g_i(X)$ be an irreducible component of $g(X)$. Then $g_i(X)$ is the minimal polynomial of its roots over $F$. Since $E$ is separable, $g_i(X)$ must have unique roots, hence $g_i(X)$ is separable.

  • $\begingroup$ But the problem is about separability over $F$, no? $\endgroup$ – Dylan Moreland Jul 20 '12 at 8:56
  • $\begingroup$ @dylanmoreland I am trying to fix it now. I have posted the beginnings the fixed proof in the question, but, of course, I am stuck. Hopefully I can figure it out. $\endgroup$ – Holdsworth88 Jul 20 '12 at 9:45
  • $\begingroup$ Did you look at the Theorem I suggested? I think the resulting proof is quite short. $\endgroup$ – Dylan Moreland Jul 20 '12 at 10:16
  • $\begingroup$ @dylanmoreland Sorry, I had overlooked it because $E/F$ is not assumed to be finite, but, looking again at the relevant parts, I see that finite degree is not needed for part of 18.13 I need. I will write it up now. $\endgroup$ – Holdsworth88 Jul 20 '12 at 10:23
  • $\begingroup$ In any case, a splitting field of a polynomial will always be finite. It's a finitely generated algebraic extension. $\endgroup$ – Dylan Moreland Jul 20 '12 at 10:52

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