# If $F$ is a field with characteristic $p$, $F(\alpha)$ is also a field with characteristic $p$?

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

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

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 $F(\alpha)$.
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.