Galois group of a non-separable polynomial

In my notes, I am given the definition of the Galois group of a polynomial only in the case when the polynomial is separable (if $f$ is a separable polynomial over $K$ with splitting field $L$, then $\mathrm{Gal}(f) = \mathrm{Gal}(L/K)$). This makes sense, since $L$ is a splitting field implies normality, and an extension generated by separable elements is separable, so $L$ is necessarily a Galois extension of $K$.

I should probably tell you that my definition of separable is the one which involves the irreducible factors of the polynomial having distinct roots in a splitting field (I'm aware there's another definition, and that they coincide when defining a separable extension).

When I look around online, I see that the Galois group is defined for any polynomial as the Galois group of its splitting field. Why is the splitting field necessarily Galois?

Thanks

• To any field extension you can attach its automorphism group. The standard convention is to say that this group is the Galois group of the extension only if the extension is Galois, i.e. is normal and separable. This is just a matter of convention. There is nothing to understand here. – Pierre-Yves Gaillard Jan 9 '12 at 12:00
• You can define the Galois group of any polynomial in $K[x]$ to be the group of $K$-automorphisms of its splitting field. The reason why we concentrate on separable polynomials is that this group is otherwise smaller than expected. For example, if $a\notin K^p$ and $p=char K$, then the Galois group of the irreducible polynomial $x^p-a$ is trivial. – Jyrki Lahtonen Jan 9 '12 at 12:07
• Dear Stephen, a polynomial is separable iff it has distinct roots in some splitting field. The other definition, saying that its irreducible factors should have distinct roots in a splitting field, is dreadful and should never be used: I cannot imagine an algebraist who wouldn't recoil in horror at the assertion that $X^2$ is a separable polynomial or that the algebra $k[X]/(X^2)$ , $k$ a field, is separable (or equivalently: étale). – Georges Elencwajg Jan 9 '12 at 12:46
• @JyrkiLahtonen, how does one show that the Galois group of the irreducible polynomial x^p - a is trivial? Thanks! – Eric Jun 25 '12 at 0:02
• @Eric: The splitting field of that polynomial is $k(a^{1/p})$. Here $$x^p-a=(x-a^{1/p})^p,$$ so this polynomial has a single root of multiplicity $p$. An automorphism has to map a zero of this polynomial to another one, so ... – Jyrki Lahtonen Jun 25 '12 at 12:02

Only when infinite fields of characteristic $p>0$ are involved you can encounter inseparable extensions.