# abstract algebra help

• Apr 1st 2009, 02:40 PM
dori1123
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?
• Apr 1st 2009, 07:16 PM
NonCommAlg
Quote:

Originally Posted by dori1123
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 $\displaystyle F$."
• Apr 1st 2009, 07:56 PM
dori1123
Quote:

Originally Posted by NonCommAlg
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 $\displaystyle F$
• Apr 1st 2009, 09:58 PM
NonCommAlg
Quote:

Originally Posted by dori1123

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.