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Give an example of an extension $L : K$ such that $char(K) = p > 0$, $L : K$ is finite but not normal.

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Take $K=\mathbb{Z}_2(X)$ and $L=K(\sqrt[3]{X})$. Then the polynomial $f(T)=T^3-X\in K[T]$ is irreducible (because it has no roots in $K$), has one root in $L$, namely $\sqrt[3]{X}$, and I leave you the pleasure to prove that this is the only root of $f$ in $L$.

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but what do you mean by $Z2(X)$ here? the field of rational expressions in X?? – Rahman Nov 13 '12 at 21:00
@user49295 Yes, you are right. – user26857 Nov 13 '12 at 21:01
Thanks a lot indeed :) – Rahman Nov 13 '12 at 21:06
@ navigetor23 did i do? – Rahman Nov 13 '12 at 21:12
i need 15 reputation maybe – Rahman Nov 13 '12 at 21:15

Here's a hint/general outline: We know that finite extensions of finite fields are normal, so we're going to need something a little more complicated. The simplest example here would be to take an extension of $\mathbb{F}_p(T)$, where $T$ is transcendental.

Over the rationals, the canonical example is to adjoin an $n^{\text th}$ root ($n>2$) of some integer $a$, by adding a root of the equation $X^n-a=0$. In short, this will fail to be normal because $\mathbb{Q}$ does not include the $n^{\text th}$ roots of $1$.

Try something similar with taking an $n^{\text th}$ root of $T$, where the $n^{\text th}$ roots of $1$ are not in $\mathbb{F}_p$.

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@ Brett Frankel thanks a lot – Rahman Nov 13 '12 at 21:28

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