Let $E$ be an extension of $\mathbb Z_p$ contained in an algebraic closure $\bar{\mathbb Z_p}$. Let $f$ be an irreducible polynomial in $\mathbb Z_p[x]$ and let $a, b \in \bar{\mathbb Z_p}$ be roots of $f$. If $a\in E$, show that $b\in E$.
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4Do you mean $,\Bbb Z_p=\Bbb F_p=\Bbb Z/p\Bbb Z;$ ? – DonAntonio Jun 10 '13 at 16:39
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This is true if $E/\mathbb Z_p$ is Galois. See my answer here. – Antonio Vargas Jun 10 '13 at 16:56
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@AntonioVargas, it is true always: any finite extension of a finite field is always Galois. – DonAntonio Jun 10 '13 at 17:22
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Hints:
Any finite field $\,\Bbb F_{p^n}\;$ is the splitting field of $\,p_n(x):=x^{p^n}-x\in\Bbb F_p\,$ , which means that field is the set of all the roots (in some algebraic closure of the prime field $\,\Bbb F_p\,$) of $\,p_n(x)\,$ .
Now, suppose $\,E=\Bbb F_{p^m}\;$ and suppose further $\;\deg f=n\;$ :
$$a\in \Bbb F_p(a)\cong\Bbb F_{p^n} \;$$
and since this last is the minimal field containing both $\,\Bbb F_p\,$ and $\,a\,$ then $\,F_{p^n}\le E\,$ .
But then both $\,p_n(x)\;,\;f(x)\;$ have $\,a\,$ as a root , so
$$f(x)\mid p_n(x)\;\;(\text{why?})\implies p_n(x)=f(x)g(x)$$
and the above means all the roots of $\,f(x)\,$ are also roots of $\,p_n(x)\,$ , so also $\,b\in \Bbb F_{p^n}\le E\;$ and we're done.
Note the above is far from being true for general fields...
DonAntonio
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Thanks. But I don't know why $f(x)\mid p_n(x)$. Would you mind giving me other hints? – Yeyeye Jun 10 '13 at 17:52
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@Rose, remember that if the minimal pol. of an algebraic element over some field always divides any other polynomial that vanishes on that element...and your $,f,$ is a minimal pol. of $,a,$ , though it must be monic to be the minimal pol. – DonAntonio Jun 10 '13 at 18:25