Thread: finite field (and subfields thereof)

1. finite field (and subfields thereof)

$\displaystyle K$ is a finite field, $\displaystyle F$ is a subfield of $\displaystyle K$, and $\displaystyle m$ is a positive integer.

$\displaystyle L=\{a \in K \vert a^{p^m} \in F \}$.

Show that $\displaystyle L$ is a subfield of $\displaystyle K$ containing $\displaystyle F$. Moreover, show that $\displaystyle L=F$.

Easy enough to show $\displaystyle L$ is a subfield of $\displaystyle K$, and I have a proof that $\displaystyle L=F$ assuming that $\displaystyle L$ contains $\displaystyle F$, but I do not know why $\displaystyle L$ must contain $\displaystyle F$.

Thanks in advance (and afterwards as well, naturally).

2. Originally Posted by mylestone
$\displaystyle K$ is a finite field, $\displaystyle F$ is a subfield of $\displaystyle K$, and $\displaystyle m$ is a positive integer.

$\displaystyle L=\{a \in K \vert a^{p^m} \in F \}$.

Show that $\displaystyle L$ is a subfield of $\displaystyle K$ containing $\displaystyle F$. Moreover, show that $\displaystyle L=F$.

Easy enough to show $\displaystyle L$ is a subfield of $\displaystyle K$, and I have a proof that $\displaystyle L=F$ assuming that $\displaystyle L$ contains $\displaystyle F$, but I do not know why $\displaystyle L$ must contain $\displaystyle F$.
you're kidding, right? well, if $\displaystyle a \in F \subseteq K,$ then obviously $\displaystyle a^{p^m}$ will also be in $\displaystyle F$ (because F is closed under multiplication). thus $\displaystyle F \subseteq L.$

3. Originally Posted by NonCommAlg
you're kidding, right? well, if $\displaystyle a \in F \subseteq K,$ then obviously $\displaystyle a^{p^m}$ will also be in $\displaystyle F$ (because F is closed under multiplication). thus $\displaystyle F \subseteq L.$
Jeepers ...apologies, and thank you.