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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

$$\beta_k=a_1\alpha_1+...+a_n\alpha_n$$

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.

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    Let $\alpha \in E$. An extension $E/F$ is separable if the minimal polynomial $m(X)=\text{min}_F(\alpha)$ is separable.2012-07-14
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    @dylanmoreland the question is copied almost verbatim; I haven't missed any hypotheses. I will post the verbatim problem in the question.2012-07-14
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    I see, Isaacs uses a less common definition of separability. It's the second one [on Wikipedia](http://en.wikipedia.org/wiki/Separable_polynomial).2012-07-14
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    This is a little overkill, but couldn't you just use Theorem 18.13?2012-07-14
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    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.2012-07-14
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    @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."2012-07-14
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    @Matt How do you mean $E$ does not occur? It sounds like you are reading it as "Suppose $F \subset E$ is the splitting field..."2012-07-14
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    @DylanMoreland Ah. I see. I misinterpreted "Suppose $E\supset F$ is the splitting field for both $f$ and $g$" as meaning $E$ is the splitting field of $\{f, g\}$ and not that the splitting field of $f$ equals the splitting field of $g$. Thanks.2012-07-14
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    @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?2012-07-14
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    @arturomagidan the definition of separability that I first learned is that an extension $E/F$ is separable if for each algebraic element $\alpha \in E$, $m(X)=\text{min}_F(\alpha)$ is separable, meaning it has distinct roots. I know this is equivalent to the statement $\text{gcd}(m(X),m'(X))=1$, where $m'(X)$ is the formal derivative of $m(X)$.2012-07-15

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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.

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    But the problem is about separability over $F$, no?2012-07-20
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    @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.2012-07-20
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    Did you look at the Theorem I suggested? I think the resulting proof is quite short.2012-07-20
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    @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.2012-07-20
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    In any case, a splitting field of a polynomial will always be finite. It's a finitely generated algebraic extension.2012-07-20
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    @dylanmoreland I just had one of the duh moments. Thank you for pointing that out. I was wondering why it was assumed that $E$ is the splitting field for both $f$ and $g$ if I didn't make real use of it, but I see that that is what allows us just to prove it one way (i.e. $f$ separable implies $g$ separable) instead of both. Thank you.2012-07-20