I'm trying to prove that every extension of $\mathbb Q$ is separable.

I take an extension $E$ of $\mathbb Q$. Let $\alpha\in E$ be algebraic over $\mathbb Q$ and $p(x)$ be its minimal polynomial over $\mathbb Q$.

Suppose that the multiplicity of $\alpha$ is $m \gt 1$.

Taking the derivative $p'(x)$, we have $\alpha$ a root of $p'(x)$ which is of lower degree than $p(x)$, contradiction. Then $E$ is a separable extension of $\mathbb Q$.

I have a felling we can prove this to every field, when I'm using the fact the field is $\mathbb Q$?

Thanks a lot.

  • 2
    $\begingroup$ What if $p'$ is the zero polynomial? $\endgroup$ – Zhen Lin Dec 18 '12 at 6:52
  • $\begingroup$ @ZhenLin it can't because the characteristic is 0 (see the answer below). Thank you for your commentary. $\endgroup$ – user42912 Dec 18 '12 at 7:40

In characteristic $0$, indeed every irreducible polynomial is separabele, because your intuition that $p'$ is a nontrivial polynomial of lower degree is fine. However, in characteristic $\ne 0$, the derivative may be the zero polynomial and thus no contradiction arises. For example consider $X^6+aX^3+b$ in characteristic $3$: It's derivative is $6X^5+3aX^2$, i.e. $0$.

  • $\begingroup$ Then every extension of every field of characteristic 0 is separable? Thank you very much for your answer, very helpful! $\endgroup$ – user42912 Dec 18 '12 at 7:46
  • $\begingroup$ Be careful, every algebraic extension of a field of characteristic $0$ is separable. The extension must be algebraic. This is also true for every finite field. (This is why they are called "perfect fields") $\endgroup$ – Asinus Dec 30 '13 at 16:34

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