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Thread: Separable extension..

  1. #1
    ynj
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    Separable extension..

    Let $\displaystyle F$ be a field with characteristic 0.$\displaystyle E$is a finite field extension of $\displaystyle F$. Prove that $\displaystyle E$is a separable extension...
    I know that for an $\displaystyle \alpha$, if the minimal polynomial $\displaystyle f(x)$ splits on $\displaystyle E$, then $\displaystyle f(x)$is separable on $\displaystyle E$. But why $\displaystyle f(x)$splits?
    Last edited by ynj; Sep 2nd 2009 at 10:24 PM.
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    Quote Originally Posted by ynj View Post
    Let $\displaystyle F$ be a field with characteristic 0.$\displaystyle E$is a finite field extension of $\displaystyle F$. Prove that $\displaystyle E$is a separable extension...
    I know that for an $\displaystyle \alpha$, if the minimal polynomial $\displaystyle f(x)$ splits on $\displaystyle E$, then $\displaystyle f(x)$is separable on $\displaystyle E$. But why $\displaystyle f(x)$splits?
    You are misunderstanding what "seperable" means. An irreducible polynomial $\displaystyle f(x)\in F[x]$ is "seperable" over $\displaystyle F$ iff $\displaystyle f(x)$ has no repeated roots in its splitting field. Now $\displaystyle \alpha \in E$ is seperable iff the minimal polynomial for $\displaystyle \alpha$ is seperable over $\displaystyle F$. This does not mean that the miniminal polynomial must split over $\displaystyle E$.
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