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Thread: Polynomial Roots

  1. #1
    Senior Member vincisonfire's Avatar
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    Polynomial Roots

    Consider the case where $\displaystyle F =
    Z_p $ and the polynomial $\displaystyle X^{p^k} - X $
    splits over K containing F . Show that the set
    $\displaystyle GF_{p^k} =
    \{c \in K : c^{p^k}
    = c \} $ has exactly $\displaystyle p^k $ elements.
    I tried showing $\displaystyle X^{p^k} - X $
    has $\displaystyle p^k $ distinct roots but did not succeed.
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  2. #2
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    Hi

    Since $\displaystyle D(X^{p^k} - X)=-1,\ X^{p^k} - X$ has only distinct roots
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  3. #3
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    Here is another way to prove this that does not use derivatives.

    Let $\displaystyle F$ be a field with charachteristic $\displaystyle p$ then the polynomial $\displaystyle x^n - 1$ has $\displaystyle n$ roots in a splitting field if $\displaystyle p\not |n$.

    We can write, if $\displaystyle a$ is a zero then we can write $\displaystyle x^n - 1 = (x-a)(x^{n-1}+x^{n-2}a+...+xa^{n-2}+a^{n-1})$. But the second polynomial, $\displaystyle x^{n-1}+...+a^{n-1}$, when evaluated at $\displaystyle a$ is $\displaystyle a^{n-1}+a^{n-1}+...+a^{n-1} = n\cdot a^{n-1} \not = 0$ since $\displaystyle p\not | n$.
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