# abstract algebra help

• Apr 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,
$
, 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)
$
. 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?
• Apr 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,
$
, 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)
$
. 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 $F$."
• Apr 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 $F$
• Apr 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.