Isomorphism between $\Bbb R$ and $\Bbb R(X)$?

My questions are:

$1.$ Is there a field morphism $\Bbb R(X) \hookrightarrow \Bbb R$ ?

$2.$ If the answer to $1.$ is "yes", are $\Bbb R$ and $\Bbb R(X)$ isomorphic as fields?



 For $1.$, here are some comments:

As for 2. :

• They are isomorphic as abelian groups, and actually as $\Bbb Q$-vector spaces. They are not isomorphic as $\Bbb R$-algebras, because they haven't the same transcendence degree over $\Bbb R$.

• Let's try to see how to construct such a field isomorphism. We could try$^{[1]}$ to find a subset $B$ of $\Bbb R$ such that $\Bbb R = \Bbb Q(B)$ and $C \subsetneq B \implies \Bbb R \neq \Bbb Q(C)$. The set $B$ must contain transcendental elements, pick such an $x_0 \in B$.

Then $\Bbb R = \Bbb Q(B) = \Bbb Q(B \setminus \{x_0\})(x_0)$ and we could try$^{[2]}$ to find a field isomorphism $\Bbb Q(B \setminus \{x_0\}) \cong \Bbb Q(B)$, so that $\Bbb Q(B \setminus \{x_0\})(x_0) \cong \Bbb Q(B)(x_0) \cong \Bbb R(X)$.

For ${[1]}$, I tried to use Zorn's lemma on $\mathscr B = \{B \subset \Bbb R \mid \Bbb R = \Bbb Q(B)\}$ and the partial order $B_1 ≤ B_2 \iff B_1 \supset B_2$. But this is not clear to me why this should be an inductive set.

For ${[2]}$, my idea was to define $f : \Bbb Q(B \setminus \{x_0\}) \cong \Bbb Q(B)$ by choosing a bijection $b : B \setminus \{x_0\} \to B$ (these sets should be uncountable), and then trying to extend $f\vert_{B \setminus \{x_0\}} = b$ to a field morphism.

Edit for $^{[1]}$: in this article (MINIMAL GENERATING SETS OF GROUPS, RINGS, AND FIELDS, by LORENZ HALBEISEN, MARTIN HAMILTON, AND PAVEL RUZICKA) the theorem 2.4. proves that $\Bbb R$ has no minimal generating set as field (or $\Bbb Q$-algebra, i.e. as ring). The argument uses Artin-Schreier theorem. Minimal generating sets of modules are also discussed here.

I think that $B$ is a minimal generating set of a subfield $K \subset \Bbb R$ over $\Bbb Q$ if and only if any element of $B$ cannot be written as a rational fraction of other elements of $B$. Under these conditions and using (transfinite) induction, we can probably prove (and this was my first idea) that any function $f: B \to \Bbb C$ such that $b$ and $f(b)$ are conjugate over $\Bbb Q$ (i.e. have the same minimal polynomial) extends to a (unique) field morphism $\sigma : K = \Bbb Q(B) \to \Bbb C$.

Actually, the collection $\mathscr B$ defined above doesn't have to be an inductive set, at least if we replace $\Bbb R$ by $K =\Bbb Q(\sqrt 2, \sqrt[4]{2},\sqrt[8]{2},\dots)$, because the descending chain $\left(E_n = \{\sqrt[2^m]{2} \mid m \geq n\}\right)_{n \geq 1}$ satisfy $K = \Bbb Q(E_n)$, but $K \neq \Bbb Q\left(\bigcap_{n≥1} E_n\right)=\Bbb Q$, so the chain has no upper bound w.r.t. $≤$ in $\mathscr B$.

• Wouldn't $\mathbb{R}$ and $\mathbb{R}(x)$ being isomorphic imply $\mathbb{C}(x)$ is algebraically closed? – Steve D Aug 19 '16 at 17:44
• @SteveD: in general the algebraic closure $\overline{F(x)}$ of $F(x)$ is not the same as $\overline{F}(x)$. – Watson Aug 19 '16 at 17:57
• Right, but $\mathbb{R}$ has index $2$ in its algebraic closure, so the isomorphism would suggest so does $\mathbb{R}(x)$. – Steve D Aug 19 '16 at 18:38
• @SteveD: ok and then by Artin–Schreier's theorem, it would follow that $\overline{\Bbb R(x)} \cong \Bbb R(X,i) \cong \Bbb C(x)$. – Watson Aug 19 '16 at 19:44
• For my first question, a field morphism $\Bbb R(X) \to \Bbb R$ would induce a field morphism $\overline{\Bbb R(X)} \to \Bbb C$, but there is no contradiction since $\overline{\Bbb R(X)} \cong \Bbb C$ as it is an algebraically closed field of characteristic $0$ and of cardinality $2^{\aleph_0}$. – Watson Aug 19 '16 at 19:46

There is no homomorphism $\mathbb{R}(X)\to\mathbb{R}$. Indeed, the only homomorphism $\mathbb{R}\to\mathbb{R}$ is the identity, so a homomorphism $\mathbb{R}(X)\to\mathbb{R}$ would have to restrict to the identity on constants and then there is nowhere $X$ could go.
Here's a proof that the only homomorphism $f:\mathbb{R}\to\mathbb{R}$ is the identity. We know $f$ must be the identity on $\mathbb{Q}$. Now note that for any $r\in\mathbb{R}$ and any rational $q<r$, we must have $$f(\sqrt{r-q})^2=f(r-q)=f(r)-q,$$ so $f(r)-q\geq 0$ since it is a square. This means $f(r)\geq r$ for all $r$. By a similar argument using $\sqrt{q-r}$ instead of $\sqrt{r-q}$, you can also show that $f(r)\leq r$. Thus $f(r)=r$ for all $r$.
• Is there a description of the field homomorphisms $\Bbb R \to \Bbb C$ ? I believe that this is more difficult because we could restrict to $\Bbb R$ any wild automorphism of $\Bbb C$ . – Watson Aug 20 '16 at 10:26
• Yeah, there are lots and lots of homomorphisms $\mathbb{R}\to\mathbb{C}$. They come not just from automorphisms of $\mathbb{C}$ but from endomorphisms of $\mathbb{C}$, which need not be surjective. – Eric Wofsey Aug 20 '16 at 15:11