7

My problem is to prove that the splitting field of $f(X)=X^5+X+1$ is, in fact, a separable extension of a field $F$ with characteristic zero.

After some research, I found that every extension of a field with characteristic zero is separable.

The idea is that, if $\alpha$ is a root of $f$ with multiplicity $>1$, then $\alpha$ is a root of $f'$ too. Why does this prove that, in the case where $F$ has characteristic zero, the extension is separable?

Robin
  • 6,201
Mateus Rocha
  • 2,746
  • 1
    Is it not enough to note that the formal derivative of $f$ cannot be zero because the leading term doesn't vanish? https://en.wikipedia.org/wiki/Separable_extension#Separable_and_inseparable_polynomials – Greg Martin Nov 04 '19 at 16:43

1 Answers1

11

An extension $K/F$ is separable iff every element of $K$ has a separable minimal polynomial over $F$. Minimal polynomials are irreducible, and in characteristic $0$ all irreducible polynomials are separable. Thus the result follows.

So why are minimal polynomials separable? A polynomial $f$ is separable iff $f$ and $f'$ have no nontrivial factors in common. And in characteristic $0$, if $f$ has degree at least $1$ and is irredicuble (since it's the minimal polynomial of some element in $K$), it has only trivial factors in common with $f'$.

In positive characteristic, however, even non-constant polynomials can have a derivative which is $0$ (say, for instance, $x^5 - 1$ in characteristic $5$), which is why there are non-separable extensions in those cases. Although for any finite field it turns out any finite extension is still separable. So the simplest examples of non-separable extensions are things like adjoining a fifth root of $T$ to the field $\Bbb F_5(T)$ (i.e. a root of the irreducible, non-separable polynomial $x^5 - T$).

Arthur
  • 204,511
  • I have a confusion on "all polynomials are separable in characteristic zero" ? How is it true? Over $\mathbb{R}$ or $\mathbb{C}$ the polynomial $(x-1)^2$ is not separable. Does it require to be minimal polynomial ? – MAS Mar 20 '21 at 03:57
  • 2
    @Why I said all irreducible polynomials are separable in characteristic $0$. Minimal polynomials are always irreducible polynomials. – Arthur Mar 20 '21 at 04:37
  • Thank you very much – MAS Mar 20 '21 at 13:02
  • 2
    I had a hard time following your deductions (though they're correct). Maybe the easier explanation would be: If $f$ is irreducible and has a multiple root, then $f'$ annihilates that root. But $f'$ is a polynomial with degree less than $f$, hence must be $0$. If characteristic is $0$, this cannot happen. Hence, $f$ doesn't have multiple roots. – toxic Jun 27 '22 at 20:58