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

Question

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?

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$ $ For $1.$, here are some comments:

• This is true for $\Bbb C$, see Can I embed $\Bbb{C}(x)$ into $\Bbb{C}$?. The key point here is the fact that $\Bbb C$ is algebraically closed, of cardinality $2^{\aleph_0}$, and of characteristic $0$. The same argument doesn't work for $\Bbb R$.

• We can actually find a subfield $F$ of $\Bbb R$ such that $F \cong F(X) \cong F(X,Y) \cong \cdots$, see Is there a subfield $F$ of $\Bbb R$ such that there is an embedding $F(x) \hookrightarrow F$?.

• Notice that $\Bbb Q(X)$ does not embed in $\Bbb Q$ (as a field).

• This is equivalent to: Is there an injective ring morphism $\Bbb R[X] \hookrightarrow \Bbb R$ ? (By taking the fraction fields).

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.

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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 $\mathscr B$.

Answer 1

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 $r$.