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I believe that for any subfield $F$ of $\Bbb R$, with finite degree of transcendence over $\Bbb Q$, the field of rational fractions in one variable $F(x)$ can't be embedded in $F$. However the situation can be different for those with infinite degree of transcendence over $\Bbb Q$.

I know that $F(x)$ always embeds in $\Bbb C$. Actually $\Bbb C(x)$ embeds in $\Bbb C^{[1]}$. I only know that if $j:F(x) \hookrightarrow F$ denotes a field morphism, then $j(x)$ is transcendental over $\Bbb Q$.

Here is an idea for $F=\Bbb R$. Let's take a transcendence basis $T$ of $\Bbb R$ over $\Bbb Q$ (which has cardinality $2^{\aleph_0}$) and complete it into a (linear algebra) basis $B$ of $\Bbb R$ over $\Bbb Q$. Then $T$ and $T \cup \{x\}$ ($x$ is the indeterminate) have the same cardinality, we can pick a bijection $b:T \cup \{x\}\to T$. Then we could try to build some $j : \Bbb R(x)=\Bbb Q(B \cup \{x\}) \to \Bbb Q(B)=\Bbb R$...

Any hint would be appreciated!

$^{[1]}$ A nice example from heptagon shows the existence of a subfield $L \subset \Bbb C$ such that $L(x) \cong L$. Indeed, we can embed $L=\Bbb R(x_1,x_2,\dots)$ in $\Bbb C$.

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  • How can $F$ be equal to $\mathbb{R}$ if the trancsendence degree over $\mathbb{Q}$ is finite? – heptagon Aug 17 '16 at 14:51
  • @heptagon: I consider any subfield of $\Bbb R$. My "belief" is related to the ones with finite degree of transcendence, but my general question also deals with the ones with infinite degree of transcendence. – Watson Aug 17 '16 at 14:52
  • Maybe there is no subfield $F \subset \Bbb R$ such that $F(x) \hookrightarrow F$ (even the ones with infinite degree of transcendence over $\Bbb Q$), but I'm not sure how to (dis)prove it. – Watson Aug 17 '16 at 14:53
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    First of all, for those with finite transcendence degree embeddings are not possible: $F(x)$ has greater transc.degree than $F$, so there should be an algebraically independent subset of $F(x)$ sent to an algebraically dependent subset of $F$ under embedding. – heptagon Aug 17 '16 at 14:57
  • @heptagon: ok, just to be sure for instance we can never have $k(X_0,X_1) \cong k(X_1)$ (as fields) where $k$ is any field, because the transcendence degrees over $k$ are $2 ≠1$? Even for $k = \Bbb Q(X_2,X_3,X_4,\dots)$ ; doesn't the map fixing $\Bbb Q$ and sending $X_j$ to $X_{j-1}$ give a field isomorphism $k(X_1) \to k(X_0,X_1)$ ? – Watson Aug 17 '16 at 15:03
  • No, this is important that $k$ is rationals here, because in general $k$ must not be sent into $k$ under embeddings. – heptagon Aug 17 '16 at 15:04
  • @heptagon: Let us continue this discussion in chat, if you don't mind. I would have a few more questions related to transcendence degree. – Watson Aug 17 '16 at 15:06

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I think I've found an example, thanks to this answer. We pick a subset $\{x_i\}_{i\in \mathbb{N}}$ of pairwise distinct elements of a transcendence basis of $\mathbb{R}$ over $\mathbb{Q}$ (this is possible since a transcendence basis of $\Bbb R$ over $\Bbb Q$ has cardinality $2^{\aleph_0}$).

Then the subfield $F = \Bbb Q(x_1, \dots, x_n, \dots)$ satisfies $F \cong F(X) \cong F(X,Y) \cong \cdots$ (as fields), so in particular $F(X)$ embeds in $F$.

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