# If $K$ is a field of charateristic $0$ then every irriducible polynomial over $K$ is separable over $K$

Let the seguent propositions:

Lemma $1$:

A polynomial $f \not=0$ over a field $K$ has a multiple zero in a splitting field if and only if $f$ and $Df$ have a common factor of degree $\ge1$

Lemma $2$:

If $K$ is a field of charateristic $0$ then every irriducible polynomial over $K$ is separable over $K$

Proof Lemma $2$: An irredubicible polynomial $f$ over $K$ is inseparable if and only if Lemma $1$. If so, then since $f$ is irreducible and $Df$ has smaller degree than $f$. We must have $Df=0$

I don't understand last implication. If I take in $\mathbb{Q}$ the polynomial $x^2+1$ is irreducible. I know that $Df=2x$ and $HCF(f,Df)=1$ and $deg (HCF(f,DF))=deg(1)=0$. In this example I don't have that $Df=0$

Where is the mistake that i'm doing?

• What the proof tries to say is that if $K$ has characteristic $0$, then a polynomial in $K[x]$ that has a multiple zero in $\overline K[x]$ is reducible in $K[x]$. For instance, the polynomial $x^4 + 2x^2 + 1\in \Bbb R[x]$ has a multiple zero at $x = i$, if you see it as a polynomial in $\Bbb C[x]$. Therefore, the polynomial is reducible over the reals. – Arthur Feb 22 '15 at 12:26
• I find this proof of lemma 2 particularly badly stated. – Olórin Feb 22 '15 at 12:28
• @RobertGreen Me too. I'm studying from Galois Theory, Ian Stewart II edition – Skills Feb 22 '15 at 12:36
• @Skills In the third edition at least (I don't know about the second one) it says "...$\;f\,,\,f'\;$ have a common factor, and since $\;f\;$ is irreducible this factor has to be $\;f\;$ . But $\;\deg f'<\deg f\;$ , so $\;f\;$ being a factor of $\;f'\;$ means $\;f'=0;$ . I think this is nicely proved. – Timbuc Feb 22 '15 at 12:43

1) $\;f\;$ is irreducible, so it cannot have non-trivial factors.
2) If $\;f\,,\,\,f'\;$ have a common factor, then $\;f\;$ is reducible unless $\;f'=0\;$
Both points above mean, at least, that if $\;f'\neq 0\implies f\;$ is separable