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On page 2 of Humphreys' book "Linear algebraic groups" he presents the "extension theorem". I will copy it below:

Extension theorem. Let $R/S$ be an integral extension, $K$ an algebraically closed field. Then any homomorphism $\varphi:S\longrightarrow K$ extends to a homomorphism $\varphi^\prime:R\longrightarrow K$. If $x\in R$, $a\in K$, $\varphi$ can first be extended to a homomorphism $R[x]\longrightarrow K$ sending $x$ to $a$ (then be further extended to $R$, $R$ being integral over $R[x]$), provided $f(x)=0$ implies $f_\varphi(a)=0$ for $f(T)\in S[T]$ ($f_\varphi(T)$ the polynomial over $K$ gotten by applying $\varphi$ to each coefficient of $f(T)$).

I have two questions.

The first is if there is a misprint here and the two occurrences of $R[x]$ should be in fact $S[x]$ (or I am missing something here)? Isn't $R[x]=R$?

The next question is could the inductive procedure described here by extending from $S$ to $S[x]$ etc.., eventually reaching $R$ be infinite? That is, maybe we may need infinitely many $x_i$ such that $R=S[x_1,x_2,...]$? If so, is the number of generators $x_i$ countably infinite or possibly uncountable?

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Yes, $R[x]$ is certainly a typo for $S[x]$.

I suspect that there is an implicit maximality argument here: first replace $S$ by a maximal extension inside $R$ to which $\varphi$ extends (guaranteed by Zorn), then show that, if $S$ is a proper subring of $R$, you can extend a little further.

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  • $\begingroup$ Thank you for your answer, I understand the argument now. So, it is possible to have an integral extension $R/S$ such that for any countable subset $I$ of $R$ that $S[I]$ is a proper subring of $R$? $\endgroup$
    – Mr Dumas
    Commented Dec 29, 2023 at 3:16
  • $\begingroup$ @MrDumas, re, sure; take $S$ to be a finite field and $R$ to be an uncountable direct sum of copies of $S$. (I don't know how many of those restrictions are necessary, but they make it easy to check that it's really an example.) $\endgroup$
    – LSpice
    Commented Dec 29, 2023 at 3:56
  • $\begingroup$ Oh ok, if $\Lambda$ is an uncountable index set and $R=\oplus_{\lambda\in\Lambda}\mathbb F_p$ and $S$ is the diagonal embedding of $\mathbb F_p$ in $R$ then for any $x=(x_\lambda)\in R$ an equation of integral dependence of $x$ over $S$ is given by the polynomial $t^p-t\in S[t]$. Thanks for the help. $\endgroup$
    – Mr Dumas
    Commented Dec 29, 2023 at 8:06

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