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We know that if F is a field which ch(F)=0,p(x) is a polynomial with coefficient of F,then p(x)root solvablity if and only if the Galois group of p(x) is solvablity .Here I want to know if the character of F is not zero ,how to judge the root solvablity ?Is there a analogue theorem?

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You need separability conditions, but given thos, the theory is much the same in characteristics other than $2$ and $3$. – Geoff Robinson Mar 16 '13 at 7:32
When you say "root solvability", perhaps you mean "solvable by radicals". In prime characteristic $p$, you will have to replace (for example) the radical $\root p\of$ by the "Artin-Schreier radical" $\wp_p^{-1}(\ )$, where $\wp_p^{-1}(a)$ stands for a root of $T^p-T-a$, etc. Apart from such obvious changes, the theory should be the same as in characteristic $0$. See for example… – Chandan Singh Dalawat Mar 16 '13 at 8:22
@Geoff Robinson: what is different about characteristics 2 and 3? – user30035 Mar 16 '13 at 17:48

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