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Thread: Multiple root

  1. #1
    Junior Member
    Jun 2011

    Multiple root

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

    Here, we mean by $\displaystyle g(x)$ is a multiple factor of $\displaystyle f(x)$, if $\displaystyle g(x)$ is irreducible, and for some $\displaystyle k>1$, $\displaystyle g^k(x)\mid f(x).$
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  2. #2
    MHF Contributor Drexel28's Avatar
    Nov 2009
    Berkeley, California

    Re: Multiple root

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