# Calculating Separable Closures

In my study of fields, the notion of the separability of an algebraic field extension is one of the more slippery concepts I have encountered thusfar. What is particularly vexing to me is the notion of a separable closure. Let $E/F$ be an algebraic field extension. The separable closure of $F$ in $E$ is defined to be the subfield

$E_{sep}=\{\alpha \in E \; | \; \alpha$ is separable over $F \}$.

While the concept of a separable closure is easy enough to understand based on the definition, the actual calculation of a separable closure $E_{sep}$ for a given algebraic field extension $E/F$ is not obvious to me.

For example, I was recently assigned the following problem:

Let $k=\mathbb{F}_2(t)$, $\alpha$ be a root of $X^4+tX^2+t\in k[X]$ in an algebraic closure of $k$ and $K=k(\alpha)$, where $\mathbb{F_2}=\mathbb{Z}/2\mathbb{Z}$. Determine $K_{sep}$ and $[K_{sep}:K]$.

My first thought is that since the separability of an element $a$ of the algebraic field extension $E/F$ is defined based on the structure of its minimal polynomial $m_{a,F}(X) \in F[X]$, we can use this to determine the elements of $E_{sep}$. An element $e \in E$ is inseparable if and only if $m_{e,F}(X)=g(X^p)$ for some separable $g(X) \in F[X]$. So the separable elements of $E$ are those whose minimal polynomial is not an element of $F[X^p]$. This, however, does not seems to be a tractable idea.

My questions are

How does one approach the problem of calculation the separable closure of an algebraic extension, like, for example, the one above?

If there is no general, one-size-fits-all method, what are some examples of techniques used to deduce the separable closure?

Note: any questions asked herein are not current homework questions.

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I don't know of any general approaches, but I do see that for this particular question, $\alpha^2$ is separable over $k$ because its minimal polynomial is $X^2+tX+t$, hence $k(\alpha^2)/k$ is a separable extension, and because $[K:k(\alpha^2)]=2$ is prime the only extension larger than $k(\alpha^2)$ is $K$ itself, so there is no larger extension that is separable over $k$ than $k(\alpha^2)$ because $K/k$ is inseparable. –  Zev Chonoles May 21 '12 at 16:45
Dear Holdsworth88, a minor terminological point : the field $E_{sep}$ is called the separable closure of $F$* in $E$ and not the separable closure of $E$ (which would be, logically enough, a field *larger than $E$). –  Georges Elencwajg May 21 '12 at 17:02
Georges, I have fixed the notation. Thank you for pointing out the inconsistency. –  Holdsworth88 May 21 '12 at 17:10

Here are a few details, completing Zev's completely correct, succinct comment (which I upvoted, of course).

An important first remark is that the polynomial $f(X)=X^4+tX^2+t\in k[X]$ is irreducible over $k=\mathbb F_2(t)=Frac (\mathbb F_2[t])$ : this is Eisenstein's criterion in all its splendour, applied with the prime $t\in \mathbb F_2[t].\:$ Hence $[K:k]=4$.
The polynomial $g(X)=X^2+tX+t\in k[X]$ is then also irreducible , so that $\alpha^2$ is of degree $2$ over $k$, since it is killed by $g$.
Moreover since $g$ is not a polynomial in $X^2$ and is irreducible ( an important, often forgotten condition!) , it is separable over $k$ and so is $\alpha ^2$, one of its roots.

So we have the tower of quadratic extensions $k\subset k(\alpha ^2)\subset K$ with $k\subset k(\alpha ^2)$ separable and $k(\alpha ^2)\subset K=k(\alpha )$ obviously purely inseparable.
Only one intermediate field of $k\subset K$ can perform both these feats: $K_{sep}$ .
Hence $K_{sep}=k(\alpha^2)$.

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This is why I like algebra so much. Thank you for expanding upon that. –  Holdsworth88 May 21 '12 at 18:34
I am happy to learn that you like algebra so much, Holdsworth88: we sure are in a club to be proud of! –  Georges Elencwajg May 21 '12 at 18:45