Root of a polynomial is a separable element

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]$. - 1 Answer Your argument is correct. Here is an alternative phrasing of it (only difference really is that it doesn't use an argument by contradiction): Let$\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