# Elements of $K$ which are separable over $F$ form a subfield of $K$?

I'm trying to prove the following statement:

If $K$ is an extension of $F$ prove that the set of elements in $K$ which are separable over $F$ forms a subfield of $K$.

I have a proof for the set of algebraic elements in $K$ forming a subfield, but I'm stuck with the separable elements.

I'm assuming I should be splitting this into cases of characteristic = 0 and characteristic $\neq$ 0. Any help? I'm at a loss.

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## 1 Answer

Characteristic $0$ is immediate, since all elements are separable (irreducible polynomial has no repeated roots).

For characteristic $p$, let $\alpha$ and $\beta$ be separable. The characteristic polynomial of $\beta$ over $F(\alpha)$ divides the characteristic polynomial of $\beta$ over $F$, hence has no repeated roots; so $\beta$ is separable over $F(\alpha)$. Now consider the tower $F(\alpha)\subseteq F(\alpha+\beta)\subseteq F(\alpha,\beta)$. Looking at the separable degrees, we have: $$[F(\alpha,\beta):F]_s = [F(\alpha,\beta),F(\alpha+\beta)]_s[F(\alpha+\beta):F]_s.$$ But $$[F(\alpha,\beta):F]_s = [F(\alpha,\beta):F(\alpha)]_s[F(\alpha):F]_s = [F(\alpha,\beta):F(\alpha)][F(\alpha):F] = [F(\alpha,\beta):F],$$ and $[F(\alpha,\beta):F(\alpha+\beta)]_s\leq [F(\alpha,\beta):F(\alpha+\beta)]$, $[F(\alpha+\beta):F]_s \leq [F(\alpha+\beta):F]$. So in order to get \begin{align*} [F(\alpha,\beta):F]&=[F(\alpha,\beta):F]_s \\ &= [F(\alpha,\beta),F(\alpha+\beta)]_s[F(\alpha+\beta):F]_s\\ &\leq [F(\alpha,\beta):F(\alpha+\beta)][F(\alpha+\beta):F]\\ &= [F(\alpha,\beta):F],\end{align*} we must have equality throughout, hence $\alpha+\beta$ is separable over $F$. The same argument holds for $\alpha\beta$.

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Thanks very much. I'm also wondering if there's a way to argue without using degree-arguments. This answer is very helpful though. – peter Nov 16 '11 at 22:45
Dear Arturo, I think there is some problem in your proof because $F(\alpha)\subseteq F(\alpha+\beta)$ doesn't hold in general (take $\beta=-\alpha$). – user18119 Nov 16 '11 at 22:56
@QiL: Oops; that should be $F$ itself. Thanks! – Arturo Magidin Nov 17 '11 at 5:25
@peter: First, there was a mistake, as noted by QiL; it's been fixed. As to your question: what precisely is your definition of separable? there are many equivalent ways of defining it. You could show that $\alpha+\beta$ is separable over $F(\alpha)$ (its irreducible polynomial is a translate of the irreducible of $\beta$) and then use the same constructive argument used to prove algebraic extension of algebraic extension is algebraic to show $\alpha+\beta$ satisfies a polynomial with no repeated roots and coefficients in $F$. Similarly with $\alpha\beta$. – Arturo Magidin Nov 17 '11 at 5:33