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I was browsing through Wikipedia for some reason or another, and it says that there is a way to do Galois theory without worrying about whether or not polynomials are irreducible. That's a bit surprising to me.

Is Wikipedia wrong (as is often the case when I read something surprising)? If not, where can I hear more about it?

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Can you provide a link? Or, keeping with the spirit of Wikipedia, [citation needed]. – Douglas S. Stones Oct 31 '10 at 3:46 – user126 Oct 31 '10 at 3:54
" In this connection, the concept of separability is of lesser importance if P is not assumed irreducible, since repeated roots may then just reflect that P is not square-free." – user126 Oct 31 '10 at 3:57
up vote 5 down vote accepted

There is a very big difference between "you can do Galois theory without worrying about whether or not a polynomial is irreducible" and "one can do Galois theory while working with polynomials that are not irreducible". The former implies that the results will hold regardless of whether the polynomial is irreducible or not; the latter merely says that one can adjust the work so as to deal with reducible polynomials.

If $f(x)$ is not irreducible, but factors as $f(x)=p_1(x)^{a_1}\cdots p_k(x)^{a_k}$ with $p_j(x)$ irreducible and separable, then you can certainly still talk about the Galois group of the splitting field of $f(x)$, and it will have all the usual properties: it would simply be compositum of the splitting fields of the different $p_i$, and so the Galois group of $f$ would have normal subgroups corresponding to the splitting fields of each $p_i$.

When the polynomials are inseparable one runs into greater difficulties; the usual technique is to separate the extension into a separable extension and a purely inseparable extension. For the separable ones, the usual techniques of Galois Theory work; for the purely inseparable ones, we have special results that apply.

That said, the quote from Wikipedia is rather poor, since separability is a property of the irreducible factors of the polynomials. Repeated factors do not affect separability or lack thereof. But even if we were to take it at face value, it does not claim that one can do Galois theory "without worrying"; it would say that if you are doing Galois theory with non-irreducible polynomials (in the way that one usually does: by splitting the polynomial into irreducible factors), then you would not worry about whether the polynomial itself has repeated roots or not, because repeated roots could come from repeated separable irreducible factors.

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Ah, I was guessing (perhaps incorrectly!) that you could use some crazy techniques like looking at étale fundamental groups and somehow this gives us more information for extensions where ordinary Galois theory fails. – user126 Oct 31 '10 at 6:32
your convention of only having the separability of a polynomial depend on its irreducible factors is a common one, but I have come to much prefer the other convention. Namely, the following definition feels more natural: a polynomial $P(t) \in k[t]$ is separable iff the algebra $k[t]/(P(t))$ is separable, i.e., is geometrically reduced. This definition has the advantage of being invariant under base extension. What advantages does your definition have? (Not a rhetorical question.) – Pete L. Clark Oct 31 '10 at 16:58
I just looked at the wikipedia article in question, and I see that it gives "my" definition of separability as an alternative to Arturo's. – Pete L. Clark Oct 31 '10 at 17:06
@Pete: Let me clear: I was quoting the definition Wikipedia's page gives. I strongly suspect that the sentence in question is left-over from a previous edit in which the "criterion" refered to was checking the gcd with the formal derivative (I haven't bothered to check); in that case, the criterion would indeed not be very good if you do not assume your polynomial to be irreducible. I don't have a personal preference for a definition of irreducibility, I was quoting Wikipedia's since the question was in the context of the statement there. – Arturo Magidin Oct 31 '10 at 22:20

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