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If $F$ is a field with characteristic $p$ and $\alpha$ is element of an extension of $F$, $F(\alpha)$ is also a field with characteristic $p$?

At the first glance, it seems obvious, because $F(\alpha)$ is the field generated by $F$ and $\alpha$. The problem is this element $\alpha$ which doesn't need to be in an extension of characteristic $p$.

I'm a little confused, I need help.

Thanks

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why downvoted?? –  user42912 Dec 19 '12 at 0:24
    
I upvoted to counter the downvote which I also don't understand. –  lentic catachresis Dec 19 '12 at 0:48
    
@BrunoStonek yes, thank you –  user42912 Dec 19 '12 at 0:53

2 Answers 2

Hint: You have that $F$ is a subfield of $F(\alpha)$ and the additive order of $1$ is the same whether you are thinking about $1$ in $F$ or $F(\alpha)$.

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I didn't understand very well your hint, please can you explain a little bit more? thank you for your answer! –  user42912 Dec 19 '12 at 4:43

A field $F$ has characteristic $p$ exactly when the kernel of the canonical ring map $$ \phi:\Bbb Z\longrightarrow F, \qquad\text{$\phi(n)=n\cdot1_F$ if $n\geq0$} $$ is the ideal $\Bbb Zp$. This makes obvious that if $F\subseteq F^\prime$ is an inclusion of fields, then the characteristics of $F$ and $F^\prime$ are the same.

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