In today's lecture, we learned about a new theorem:
Let $f\subseteq f'$ be an algebraic field extension, $o:f\hookrightarrow L=\overline L$ a field extension and $L$ algebraically closed. Then we have a continuation of $o$ called $o'$ (which is an $f$-algebra-homomorphism $o':f'\rightarrow L$). Also, if $f'$ is algebraically closed and $L$ is algebraic over $o(f)\subseteq L$, then every $o'$ is an isomorphism.
So I was going through the proof, but there is something I don't quite understand. Why do we need to define $f'$ as an algebraic field extension and why does $L$ need to be algebraically closed. I tried to find counterexamples as to why this has to be so. Can someone give me a counterexample for each of the two properties, please?
You have an extension $i:Q\rightarrow R$ where $Q$ is the field of rational numbers and $R$ the field of real numbers. and a morphism $f:Q\rightarrow \bar Q$ where $\bar Q$ is the algebraic closure of $Q$, you cannot extend $f$ to $R$ since the cardinal of $R$ is strictly greater than the cardinal of $\bar Q$ and a morphism of fields is injective. Remark that $i$ is not algebraic.
http://planetmath.org/cardinalityofalgebraicclosure