Galois Group of an Inseparable Polynomial

Question

The following question has arisen as part of my revision and I want to seek clarity on how it should be answered:

If $f$ is a separable polynomial over a field $K$ and $L$ is its splitting field, state a formula for the order of the Galois group of $f$ in terms of $[L:K]=n$. Give an example to show that, if $f$ is not separable, this formula fails.

Source: Paper B9a 2009, Oxford University Maths Papers

Note that the definition of 'separable' we use is:

"$f$ is separable if each of its irreducible factors are; that is to say they each have $d_i$ distinct roots in a splitting field (where $d_i$ is the degree of the irreducible factor)"

So, I know that if we suppose that $\text{deg}(f)=d$, then $[L:K]|d!$ and that for finite extensions we have that $|\Gamma(L:K)| =[L:K]$

In this case, I do not know how to generate a counter-example which shows this cannot hold for an inseparable polynomial. I know that if we take $K = \mathbb{F}_p(t)$, then this is an infinite field of characteristic $p$ and the polynomial $f(x) = x^p − t ∈ K[x]$ is inseparable. Then the Galois group of $f$ turn out to be trivial, but this doesn't go against what we've said for the separable case, since $1$ will divide any degree of extension.

http://www.proofwiki.org/wiki/Order_of_Galois_Group_Equals_Degree_of_Extension — Caleb Stanford, May 04 '14 at 18:48
@Goos I've already included that result in my answer - I'll make it a little clearer. It's the inseparable case I'm interested in — AlgebraicallyMinded, May 04 '14 at 18:49
In regards to the last paragraph, you didn't merely say that $|\Gamma(L : K)| \mid [L : K]$ in the separable case. Rather, you stated equality. And as you have shown, this no longer holds in the inseparable case. — RghtHndSd, May 04 '14 at 19:02

score 1 · Answer 1 · edited Apr 13 '17 at 12:20

I think you are confused about what the question is asking.

First of all, the statement $[L : K] | n!$ is true for any field extension, not just a separable polynomial. So you will not be able to find a counterexample to this fact even if you allow $f$ to be inseparable. See here.

Second of all, when the question says "formula", it means an exact formula. Thus, the formula you want is merely "$n$". The order of the Galois group is $n$. That's the answer.

You have already provided a counterexample to this fact when the polynomial is separable; $x^p - t$ is inseparable and the degree of the extension is not equal to the order of the Galois group.

Galois Group of an Inseparable Polynomial

1 Answers1