abstract algebra help

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April 1st 2009, 02:40 PM
dori1123

abstract algebra help

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

I think I need to define $K$ , but I'm not sure how to do it. Can anyone help?
April 1st 2009, 07:16 PM
NonCommAlg

Quote:

Originally Posted by dori1123

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

I think I need to define $K$ , but I'm not sure how to do it. Can anyone help?

are you sure the assumption is not that "the set $\{\alpha_1, \alpha_2, \cdots , \alpha_n \}$ is algebraically independent over $F$ ?" this is very different from "each $\alpha_i$ being transcendental (= not algebraic) over $F$ ."
April 1st 2009, 07:56 PM
dori1123

Quote:

Originally Posted by NonCommAlg

are you sure the assumption is not that "the set $\{\alpha_1, \alpha_2, \cdots , \alpha_n \}$ is algebraically independent over $F$ ?" this is very different from "each $\alpha_i$ being transcendental (= not algebraic) over $F$ ."

Yes, it says "do not assume the $\alpha$ 's are algebraic over $F$
April 1st 2009, 09:58 PM
NonCommAlg

Quote:

Originally Posted by dori1123

Yes, it says "do not assume the $\alpha$ 's are algebraic over $F$

every element of $K$ is in the form $u=\frac{P(\alpha_1, \cdots , \alpha_n)}{Q(\alpha_1, \cdots , \alpha_n)},$ where $P, Q$ are polynomials in $\alpha_1, \cdots , \alpha_n$ with coefficients in $F$ and $Q(\alpha_1, \cdots , \alpha_n) \neq 0.$ if $\alpha_i$ were algebraic over $F,$ then elements of $K$

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