Could someone tell me the definition of the separable closure of a field $K$?
Furthermore, I would like to know whether it is a Galois extension of $K$. Also, why is this construction useful?
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Let us consider a field $K$ and let us say we fix an algebraic closure $K^{\text{alg}}$ of it.
Now, $K^{\text{alg}}$ might contain elements that are not separable (over $K$), that is, their minimal polynomial is not a separable polynomial (i.e., it has roots of multiplicity greater $1$ in $K^{\text{alg}}$ or equivalently it is not co-prime with its derivative).
Non-separable extensions and elements are not so nice in some ways, in particular recall that an extension is Galois if it is normal and separable. So one might consider only considering all separable (over $K$) elements in $K^{\text{alg}}$. The collection of all these forms again a field and is called the separable closure of $K$.
And, yes, this extension is a Galois extension; in fact it is the maximal Galois extension of $K$.
An extension $L/K$ is called normal if each irreducible polynomial in $K[X]$ that has a zero in $L$ can be decomposed into linear factors in $L$. Or put differently, if $L$ contains one of the zeros of a polynomial $P$ it contains all the zeros of $P$. Now, an element is separable if its minimal polynomial is a separable polynomial. Thus, either all the zeros of $P$ are separable and P decomposes in linear factors, or non of them is separable and it has no zero at all. In any case, each polynomial that has a zero in the separable closure will also decompose in linear factors; thus ext. is normal.
Also, note that for some fields such as the rationals or any field of characteristic $0$ but also for finite fields, the separable closure is nothing but the algebraic closure. The point being that these fields are perfect fields and thus every algebraic extension is separable.
