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Can you please help with the following problem?

Let $K$ be a field and let $f(x) \in K[x]$ be a separable polynomial. If $E/K$ is a splitting field of $f(x)$, prove that every root of $f(x)$ in $E$ is a separable element over $K$.

My attempt:

Let $z$ be a root of $f(x)$ in $E$ and suppose $z$ is not a separable element over $K$, then by definition the minimal polynomial, say $h(x)$, of $z$ over $K$ is not separable, so we can find a repeated root $q \in E$ of $h(x)$. Now since $f(z)=0$ and $h(x)$ is the minimal polynomial of $z$ over $K$ then $h(x)|f(x)$ so that $q$ is also a repeated root of $f(x)$. This implies then that $f(x)$ is not separable in $E[x]$, this contradicts the fact that $f(x)$ is separable over K[x]$.

For this problem I'm using the fact that if $E/K$ is a field extension and $f(x) \in K[x]$ is a separable polynomial then $f(x)$ is separable when considered as an element of $E[x]$.

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1 Answer 1

up vote 2 down vote accepted

Your argument is correct. Here is an alternative phrasing of it (only difference really is that it doesn't use an argument by contradiction):

Let $\alpha\in E$ be a root of $f$. Then $\text{Irr}(\alpha,K)\mid f$, where $\text{Irr}(\alpha,K)$ denotes the minimal polynomial for $\alpha$ over $K$. Therefore, because $f$ is a separable polynomial, we also have that $\text{Irr}(\alpha,K)$ is a separable polynomial. Thus $\alpha$ is a separable element over $K$.

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Thanks, essentially what I wrote right? (of course shorter heh) –  user6495 May 3 '11 at 0:51
    
@user6495: Yes, sorry, I kind of jumped to answer - this is exactly the same as your argument. –  Zev Chonoles May 3 '11 at 0:52
    
Thanks! sometimes I have trouble writing things clearly, thanks again. –  user6495 May 3 '11 at 0:52
    
@user6495: No problem! The way you phrased it is great actually, you should stick with it - it explains more than mine. –  Zev Chonoles May 3 '11 at 0:54

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