# isomorphism extension theorem

• Apr 13th 2009, 06:35 PM
ziggychick
isomorphism extension theorem
Here's the problem:

Let K be an algebraically closed field. Show any isomorphism $\displaystyle \sigma$ of K onto a subfield of K such that K is algebraic over $\displaystyle \sigma[K]$ is an automorphism of K, that is show $\displaystyle \sigma[K]=K$.

I know $\displaystyle \sigma^{-1}:\sigma[K] \rightarrow K$ can be extended to an isomorphism $\displaystyle \mu:K\rightarrow K'$ where K'<=K. And since K<=K'<=K we know $\displaystyle \sigma^{-1}$ can only be extended to an automorphism of K. But does this help me? I don't see how to make the connection with $\displaystyle \sigma[K]$.

Any advice would be great! :-)
• Apr 13th 2009, 08:48 PM
NonCommAlg
Quote:

Originally Posted by ziggychick
Here's the problem:

Let K be an algebraically closed field. Show any isomorphism $\displaystyle \sigma$ of K onto a subfield of K such that K is algebraic over $\displaystyle \sigma[K]$ is an automorphism of K, that is show $\displaystyle \sigma[K]=K$.

I know $\displaystyle \sigma^{-1}:\sigma[K] \rightarrow K$ can be extended to an isomorphism $\displaystyle \mu:K\rightarrow K'$ where K'<=K. And since K<=K'<=K we know $\displaystyle \sigma^{-1}$ can only be extended to an automorphism of K. But does this help me? I don't see how to make the connection with $\displaystyle \sigma[K]$.

Any advice would be great! :-)

Hint: show that $\displaystyle \sigma(K)$ is algebraically closed too and thus, since $\displaystyle K$ is algebraic over $\displaystyle \sigma(K) \subseteq K,$ we must have $\displaystyle \sigma(K)=K.$