# Thread:

1. ## Multiple root

Let $P$ be an arbitrary field, while $C$ is the complex field. $f(x)$ is a polynomial in $P$ with degree $>0$. Suppose $f$ has no multiple factor in $P$, prove then $f$ has no multiple root in $C.$

Here, we mean by $g(x)$ is a multiple factor of $f(x)$, if $g(x)$ is irreducible, and for some $k>1$, $g^k(x)\mid f(x).$

2. ## Re: Multiple root

This doesn't make any sense as written? I'm going to assume that $P$ is a subfield of $\mathbb{C}$. Recall then that $f$ having no multiple roots is equivalent to $(f,f')=1$, which is trivially independent of what field you are discussing. Once again, this question dreally doesn't make sense.
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