Let k be a field, E an algebraic extension of k,and $\sigma : k \rightarrow L$ an embedding of k into algebraically closed field L.

Let $S$ be the set of all pairs $(F,\tau )$ where $F$ is subfield of E containing k, and $\tau $ is an extension of $\sigma $ to an embedding of $F$ in L. If $(F,\tau ) \leq (F',\tau ')$ if $F\subset F' $ and $\tau' |F=\tau $. Now S is non empty which has chain(totally ordered subset) $\{(F_i,\tau_i)\}$ and $F=\cup F_i $ and $\tau |F_i= \tau_i $. Then $(F,\tau)$ is an upperbound and hence Zorn's lemma implies existence of maximal element $(K,g)$ in S. where $K=E$ (if K $\neq $ E then it breaks maximality of K as we can a construct a new extension by attaching a element of E to K) and $g$ is an extension of $\sigma $.

I need clarification where did we make use of the extension being algebraic but I guess this was required to prove that the maximal element is $E$ itself ?

And what if I remove the requirement of extension being algebraic ?

  • $\begingroup$ You need the assumption $E/k$ to be algebraic to guarantee the ability to extend $\sigma$ at any step. Or, as @Tsemo said, to prove that maximal element is an embedding of $E$ rather than some intermediate field. See also here. A duplicate I think, but given that I answered tat version I should not lead the rush to close this as a dupe. $\endgroup$ – Jyrki Lahtonen Apr 12 '18 at 14:18

It when you construct the new extension which contains $K$ if $K\neq E$, you take $x\in E-K$, it is algebraic. This enables to extend $\sigma$ to $K(x)$. I believed you have shown before for that if $x$ is algebraic over $k$ you can extend $\sigma$ to $k(x)$ where $k(x)$ is the field generated by $k$ and $x$.


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