Proper subfields of $\mathbb{C}$ isomorphic to $\mathbb{C}$ 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!
 A: 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.
