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Problem: Let $f$ be an irreducible polynomial in $F[x]$. Prove that $f$ has no multiple root in any field extension of $F$ unless the derivative $f'(x)$ is the zero polynomial.

My work so far: I'm assuming that $f'(x)=0$ and I want to should that $f$ has a multiple root. I know that if $f$ has a multiple root $\alpha$, then $\alpha$ is a common root of $f$ and $f'$. In general, for $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0, f'(x)=na_nx^n{n-1}+\cdots +a_1$ and $deg{f}\geq def{f'}$.

If $f$ has a multiple root, then there is a factor $(x-\alpha)^k$ s.t. $k\leq n$ and $(x-\alpha)|f$ and $(x-\alpha)|f'$. I know I want to do something s.t. I can say $deg{f}\leq deg{f'}$ to get their degrees to be equal. In the textbook I'm using (Algebra by Artin, pg 458), it says "Since it is irreducible, $f$ will have a nonconstant factor in common with another polynomial $g$ only if $f|g$." I'm struggling to wrap my mind around how I can see $f|g$. In general, $a|b$ and $a|c$ doesn't imply $b|c$.

Thanks in advance!

cheeseboardqueen
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  • compute the gcd of $f,f'$ over the given field. It can't be $1$ since it has a root, namely $\alpha$. Thus it would be a non-trivial divisor of $f$, which was assumed to be irreducible. – lulu Apr 20 '21 at 17:04
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    By $f$ irreducible, the gcd $(f,f') = f$ or $1$. If $,(f,f') = f,$ then $,f\mid f'$ so $f'=0.,$ Else $(f,f')=1$, which can't occur, since by the linked dupe the gcd $(f,f')=1$ persists over the extension field, so they can't have a nonunit common factor $,x-\alpha.,$ Said more conceptually: persistance of gcds $\Rightarrow$ persistence of double root test. – Bill Dubuque Apr 20 '21 at 18:57

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