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It is known that $\mathbb{C}$ has proper subfields which are isomorphic to $\mathbb{C}$, see this question; let $K$ be such subfield of $\mathbb{C}$.

Let $\iota: K \to \mathbb{C}$, $\iota(k)=k$ for all $k \in K$.

Now, I am confused:

(1) If $K \subseteq \mathbb{C}$ is an algebraic field extension, then since $K$ is algebraically closed we obtain that $K=\mathbb{C}$.

(2) If $K \subseteq \mathbb{C}$ is non-algebraic, then $\mathbb{C}$ should be isomorphic to $K(T_1,\ldots,T_n)$, where the $T_j$'s are variables over $K$, and $n \geq 1$ ($n$ is the transcendence degree of the extension). But this seems impossible, since, for example, algebraically closedness of $\mathbb{C}$ would imply that a root of $f(T)=T^2-T_1$, denote it by $w$, belongs to $K(T_1,\ldots,T_n)$, but the square root of $T_1$ does not belong to $K(T_1,\ldots,T_n)$.

Question: Where is my error?

(A Remark: Perhaps the construction of such $K$ does not imply that there exists $\iota: K \to \mathbb{C}$ with $\iota(k)=k$? Maybe $1_K \neq 1_\mathbb{C}$ and the usual ring theory results are not relevant here?).

Thank you very much!

user26857
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user237522
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    You always have case 1, $K\cong \Bbb C$, but the image of $i$ is not necessarily the whole of the copy of $\Bbb C$ that you started with. Think about $(\Bbb Z,+)$ as a group and the many subgroups isomorphic to itself that it contains – Alessandro Codenotti Apr 21 '21 at 07:48
  • @AlessandroCodenotti, thank you, interesting. Could you please elaborate your explanation in an answer? – user237522 Apr 21 '21 at 07:49
  • Please, so such $K \subseteq \mathbb{C}$ is algebraic of finite or infinite degree? I am still confused... – user237522 Apr 21 '21 at 07:53
  • Not algebraic does not imply purely transcendental. – Thorgott Apr 21 '21 at 08:03
  • Thank you. In the second case I thought that all the algebraic elements over $K$ are already contained in $K$, and hence only added the transcendental part. Perhaps I was missing something. – user237522 Apr 21 '21 at 08:07
  • The algebraic part lies over the purely transcendental part, not the other way around. Consider $\mathbb{C}(T)$, the function field in one variable over $\mathbb{C}$. Then $\overline{\mathbb{C}(T)}\cong\mathbb{C}$ (since both are algebraically closed of uncountable transcendence degree over their prime field), but the obvious inclusion $\mathbb{C}\rightarrow\overline{\mathbb{C}(T)}$ is neither algebraic nor purely transcendental. Indeed, first comes the purely transcendental part $\mathbb{C}\rightarrow\mathbb{C}(T)$ and then the algebraic part $\mathbb{C}(T)\rightarrow\overline{\mathbb{C}(T)}$ – Thorgott Apr 21 '21 at 08:09
  • The situation you're considering looks exactly the same (except with perhaps more variables), but what is $\overline{\mathbb{C}(T)}$ in my situation is $\mathbb{C}$ in yours and the embedded copy of $\mathbb{C}$ in $\overline{\mathbb{C}(T)}$ corresponds to your $K$. – Thorgott Apr 21 '21 at 08:10
  • So my error was considering first the algebraic part and then the transcendental part, instead of vice versa. So case 2 always holds. Nice. You can make your comments into an answer, if you like. – user237522 Apr 21 '21 at 08:15
  • @AlessandroCodenotti, please see Thorgott's answer. Perhaps in your argument it is not possible to move from groups to fields? – user237522 Apr 25 '21 at 00:35

1 Answers1

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Your error lies in the fact that an extension can be non-algebraic and still not purely transcendental, even if the base field is algebraically closed. Any extension $L/K$ can be decomposed as $L/K(B)/K$, where $B$ is a transcendence basis, i.e. $K(B)/K$ is a purely transcendental extension and $L/K(B)$ is an algebraic extension. $K$ is algebraically closed, but $K(B)$ is not, so $L/K(B)$ may very well be non-trivial.

Indeed, let's explicitly look at the purely transcendental extension $\mathbb{C}(T)/\mathbb{C}$. Then, we also have an algebraic extension $\overline{\mathbb{C}(T)}/\mathbb{C}(T)$. The composite $\overline{\mathbb{C}(T)}/\mathbb{C}$ is an extension of $\mathbb{C}$ that is neither algebraic nor purely transcendental. On the other hand, if $B$ is a transcendence basis for $\mathbb{C}$ over $\mathbb{Q}$, then $B\cup\{T\}$ is a transcendence basis for $\overline{\mathbb{C}(T)}$ over $\mathbb{Q}$, so that by the decomposition from the first paragraph and the fact that if an algebraically closed field is algebraic over some other field, it is its algebraic closure (up to isomorphism, of course), we obtain that $\mathbb{C}\cong\overline{\mathbb{Q}(B)}\cong\overline{\mathbb{Q}(B\cup\{T\})}\cong\overline{\mathbb{C}(T)}$ (the middle isomorphism comes from the fact that $B$ is infinite, more precisely uncountable). This is the prototypical example of the phenomenon that is also observed in your scenario.

Indeed, if we fix an isomorphism $\overline{\mathbb{C}(T)}\cong\mathbb{C}$, then the copy of $\mathbb{C}$ in $\overline{\mathbb{C}(T)}$ now corresponds to a proper subfield $K$ of $\mathbb{C}$ such that $\mathbb{C}/K$ is a non-algebraic, non-purely transcendental extension of transcendence degree $1$. In fact, for any proper subfield $K$ of $\mathbb{C}$ isomorphic to $\mathbb{C}$, the extension $\mathbb{C}/K$ looks like this up isomorphism, only perhaps with more variables. For we can find a transcendence basis $B$ of $\mathbb{C}$ over $K$, so that $\mathbb{C}/K(B)$ is agebraic, whence $\mathbb{C}\cong\overline{K(B)}\cong\overline{\mathbb{C}(B)}$ is the algebraic closure of a function field over $\mathbb{C}$ up to isomorphism and the copy of $K$ in $\mathbb{C}$ looks like the canonical copy of $\mathbb{C}$ in the algebraic closure of that function field. The number of algebraically independent elements here can be anything from $0$ to continuum.

Thorgott
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    Thank you very much! Nice answer. – user237522 Apr 21 '21 at 09:21
  • Please, sorry for the following perhaps trivial question, but is such extension of $\mathbb{C}$, namely, $\mathbb{C} \subset \overline{\mathbb{C}(T)}$ separable? (Should I ask this as a separate question?). – user237522 Apr 21 '21 at 21:50
  • What is a separable transcendental extension? My motivation for all this is the following question, where we consider $A/m \subseteq B/n$: https://mathoverflow.net/questions/390649/local-rings-r-subsetneq-s-with-r-regular-and-s-cohen-macaulay-non-regula?noredirect=1&lq=1 – user237522 Apr 21 '21 at 21:57
  • I don't know what a separable transcendental extension is. To me, the concept of separability only makes sense for algebraic extensions. – Thorgott Apr 21 '21 at 22:30
  • Thank you for the comment; nice of you trying to help. Actually, immediately after writing to you the last comment an hour ago, I realized that in the cases I am interested, namely, $A/m \subseteq B/n$ where $(A,m)$ and $(B,n)$ are local rings, with some finitenss condition or other nice conditions (for example, $S$ is finitely generated as an $R$-module or $S$ is algebraic over $R$), then the field extension would be algebraic. Please see (if this interests you) my recent edit of https://mathoverflow.net/questions/390831/is-the-following-local-map-unramified – user237522 Apr 21 '21 at 23:22