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Let $F$ be a field with characteristic $0$, $f \in F[t]$ the polynomial ring over $F$. Show that $f$ is square free implies $ f, f'$ are relatively prime.

I know this is actually an if and only if statement, but the other direction is easy. I get stuck on this direction, and I did some search but a lot of materials using extension fields and separable polynomials to get this. But our course haven't reached that far. So can anyone help me with an elementary proof? Thanks a lot.

user112564
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Suppose that $f$ and $f'$ have a common irreducible factor $p$. Let $f(t)=p(t)q(t)$. Then $f'(t) = p(t)q'(t) + p'(t)q(t)$.

Since $p$ is irreducible and $F$ has characteristic zero, $p$ and $p'$ do not have a common factor, so $p(t)$ divides $q(t)$. In other words, $p^2 \mid f$.

Andrew Dudzik
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  • Thanks! But how did you get $p, p'$ do not have common factor since $F$ has characteristic zero? – user112564 Mar 13 '14 at 23:59
  • Irreducible polynomials do not have factors, forget common factors with another polynomials! – P Vanchinathan Mar 14 '14 at 00:34
  • @PVanchinathan I think the right question for me to ask is why do we need $F$ to have characteristic zero? Isn't it true that $p,p'$ are relatively prime if $p$ is irreducible in any polynomial ring? – user112564 Mar 14 '14 at 00:35
  • This is the pure inseparability arising in prime characteristic. Formal derivative $x^p$ is $px^{p-1}$ which is identically zero hence has a common root with every polynomial that has a root in that field. – P Vanchinathan Mar 14 '14 at 00:42
  • @user112564 Almost. We know that the degree of $p'$ is less than the degree of $p$, which means they share no common factor—unless $p'=0$. This can happen in nonzero characteristic (specifically, over fields that are not "perfect", so don't try finding counterexamples over finite fields). For example, let $F=(\mathbb{Z}/q)(X)$. Then the polynomial $p(t)=t^q-X$ is irreducible, but $p'(t) = qt^{q-1}=0$. – Andrew Dudzik Mar 14 '14 at 00:42
  • @User-33433 Oh, that's the point! I got it, thank you so much! – user112564 Mar 14 '14 at 01:11
  • @Slade, regarding your counterexample: could you explain, please, why the polynomial $p(t) = t^q - X$ is irreducible in $(\mathbb{Z}/q)(X)[t]$? – Alexey Jan 01 '19 at 23:07
  • @Alexey Probably the most straightforward proof is to apply Eisenstein's criterion to the prime $X$ in the PID $(\mathbb{Z}/q)[X]$. There is also a standard result that says that $X^q-\alpha$ is irreducible over $\mathbb{Z}/q$ whenever $\alpha$ is not a $q$-th power. – Andrew Dudzik Jan 02 '19 at 03:44