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Thread: finite field (and subfields thereof)

  1. #1
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    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).
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  2. #2
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    Quote Originally Posted by mylestone View Post
    $\displaystyle K$ 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 $\displaystyle F \subseteq L.$
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    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.
    Last edited by mr fantastic; Jul 10th 2011 at 04:29 PM.
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