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Let $f(X)$ be an irreducible polynomial in the polynomial ring $k[X]$ over the field $k$. Prove that the number of distinct roots of $f(X)$ divides the degree of $f(X)$.

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If f has a root isn't it either linear or reducible? – Gaffney Dec 26 '13 at 7:39
@Gaffney No. Consider $X^2+2$ over $\mathbb R$. (The roots don't have to be in $k$.) – Potato Dec 26 '13 at 7:40
Strictly speaking Gaffney is correct. The question should refer to roots in a splitting field. – Derek Holt Dec 26 '13 at 10:00
up vote 5 down vote accepted

Irreducible polynomials over fields of characteristic zero have distinct roots (in the splitting field), so it suffices to consider the case where $k$ has positive characteristic. Fix some $f$ over such a $k$. We may assume $f$ has repeated roots, or the result is trivial.

The multiple roots are roots of $\gcd(f,f')$. Since $f$ is irreducible, it must divide $f'$, and this is only possible if $f'=0$, so $f(X)=g(X^p)$, where $p$ is the characteristic of $k$. We may continue this process to write $f(X)=h(X^{{p^e}})$, where $e\ge1$ and $h$ is irreducible and separable. Hence every root of $f$ has the same multiplicity $p^e$, and the result follows.

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I don't understand "continue this process." Could you please explain? – Myath Feb 8 at 5:31

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