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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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    @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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    @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