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Thread: abstract algebra help

  1. #1
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    abstract algebra help

    Show that if the field $\displaystyle K$ is generated over $\displaystyle F$ by the elements $\displaystyle \alpha_1, \alpha_2, ..., \alpha_n,
    $, where each $\displaystyle \alpha_i$ is not algebraic over $\displaystyle F$, then an automorphism $\displaystyle \sigma$ of $\displaystyle K$ fixing $\displaystyle F$ is uniquely determined by $\displaystyle \sigma(\alpha_1), \sigma(\alpha_2), ..., \sigma(\alpha_n)
    $. In particular, show that an automorphism fixes $\displaystyle K $ if and only if it fixes a set of generators for $\displaystyle K$.

    I think I need to define $\displaystyle K$, but I'm not sure how to do it. Can anyone help?
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  2. #2
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    Quote Originally Posted by dori1123 View Post
    Show that if the field $\displaystyle K$ is generated over $\displaystyle F$ by the elements $\displaystyle \alpha_1, \alpha_2, ..., \alpha_n,
    $, where each $\displaystyle \alpha_i$ is not algebraic over $\displaystyle F$, then an automorphism $\displaystyle \sigma$ of $\displaystyle K$ fixing $\displaystyle F$ is uniquely determined by $\displaystyle \sigma(\alpha_1), \sigma(\alpha_2), ..., \sigma(\alpha_n)
    $. In particular, show that an automorphism fixes $\displaystyle K $ if and only if it fixes a set of generators for $\displaystyle K$.

    I think I need to define $\displaystyle K$, but I'm not sure how to do it. Can anyone help?
    are you sure the assumption is not that "the set $\displaystyle \{\alpha_1, \alpha_2, \cdots , \alpha_n \}$ is algebraically independent over $\displaystyle F$ ?" this is very different from "each $\displaystyle \alpha_i$ being transcendental (= not algebraic) over $\displaystyle F$."
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    are you sure the assumption is not that "the set $\displaystyle \{\alpha_1, \alpha_2, \cdots , \alpha_n \}$ is algebraically independent over $\displaystyle F$ ?" this is very different from "each $\displaystyle \alpha_i$ being transcendental (= not algebraic) over $\displaystyle F$."
    Yes, it says "do not assume the $\displaystyle \alpha$'s are algebraic over $\displaystyle F$
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  4. #4
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    Quote Originally Posted by dori1123 View Post

    Yes, it says "do not assume the $\displaystyle \alpha$'s are algebraic over $\displaystyle F$
    every element of $\displaystyle K$ is in the form $\displaystyle u=\frac{P(\alpha_1, \cdots , \alpha_n)}{Q(\alpha_1, \cdots , \alpha_n)},$ where $\displaystyle P, Q$ are polynomials in $\displaystyle \alpha_1, \cdots , \alpha_n$ with coefficients in $\displaystyle F$ and $\displaystyle Q(\alpha_1, \cdots , \alpha_n) \neq 0.$ if $\displaystyle \alpha_i$ were algebraic over $\displaystyle F,$ then elements of $\displaystyle K$

    would have simpler forms, that is in the form $\displaystyle P(\alpha_1, \cdots , \alpha_n).$ anyway, if $\displaystyle \sigma$ is an $\displaystyle F$ automorphism of $\displaystyle K,$ then $\displaystyle \sigma(u)=\frac{P(\sigma(\alpha_1), \cdots , \sigma(\alpha_n))}{Q(\sigma(\alpha_1), \cdots , \sigma(\alpha_n))}.$ now you should be able to solve your problem easily.
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