Existence of field injections between $\mathbb{R}(x)$ and $\mathbb{C}$.

Question

This came with a set of questions:

Find a field injection between the following fields or prove that such injection doesn't exist:

• $\mathbb{Q}[i]\subset\mathbb{C}\to\mathbb{R}$
• $\mathbb{Q}(x)\to\mathbb{C}$
• $\mathbb{R}(x)\to\mathbb{C}$
• $\mathbb{Q}[\alpha]\subset\mathbb{C}\to\mathbb{R}$, where$\alpha=2^{\frac{1}{3}}e^{2\pi i/3}$

I have worked with the finite case before, (for example $\mathbb{F}_{29}\to \mathbb{F}_{57}$), in which case I could use the cardinality of the image to show that no such field injection exists. But I don't think I could continue with this strategy when everything is infinite.

I think I have tackled all other parts, my proof goes something like this:

Suppose $\phi:\mathbb{Q}[i]\to\mathbb{R}$ is an injection. Suppose $\phi(i)=r\in \mathbb{R}$. Then $\phi(-1)=\phi(i\cdot i)=\phi(i)\cdot\phi(i)=r^2$. But since $$\phi(0)=0=\phi(1)+\phi(-1)=1+\phi(-1)$$, $\phi(-1)=-1=r^2$, and this is a contradiction since there is no such real number whose square is $-1$.

For the second part, I think $\phi:\mathbb{Q}(x)\to\mathbb{R}:p(x)\to p(e)$ would be an injection, based off the fact that $e$ is transcendental over $\mathbb{Q}$. But I have no idea about the third part. I only have a vague feeling that there might be some argument based on cardinality.

For the last part,the map $\phi:\mathbb{Q}[\alpha]\to\mathbb{R}:(a_0+a_1\alpha+a_2\alpha^2)\to(a_0+a_12^\frac{1}{3}+a_22^\frac{2}{3})$ seems to be valid.Many thanks to @reuns for pointing this out.

I also found this question, which seems to be more or less relate：https://math.stackexchange.com/questions/935877/how-to-prove-that-the-evaluation-map-is-a-ring-homomorphism

Any help is appreciated, thank you in advance.

Answer 1

Part $(3)$ is a bit subtle.

There's one key difference here between $\mathbb{R}$ and $\mathbb{Q}$. From a field perspective, $\mathbb{Q}$ is very "explicit:" if $F$ is a field, there is only ever at most one field homomorphism $\mathbb{Q}\rightarrow F$. This is basically because every element of $\mathbb{Q}$ is "definable" in a particularly nice way, namely by combining the additive and multiplicative units ($0$ and $1$) via addition, multiplication, subtraction, and division. Since all of that structure is preserved by field homomorphisms, this gives the uniqueness above.

By contrast, $\mathbb{R}$ does not have this property. In particular, a homomorphism $h:\mathbb{R}(x)\rightarrow\mathbb{C}$ is allowed to do terrible things to the "$\mathbb{R}$-part" of $\mathbb{R}(x)$: it need not be the case that $h$ restricted to $\mathbb{R}$ is just the inclusion map $\mathbb{R}\rightarrow\mathbb{C}$.

At this point it's a good exercise to prove that there is no field embedding $h:\mathbb{R}(x)\rightarrow\mathbb{C}$ which restricts to the inclusion on $\mathbb{R}$ - that is, such that $h(r)=r$ for all $r\in\mathbb{R}\subset\mathbb{R}(x)$. Basically, think about what happens if we try to whip up such an $h$ which sends $x\in\mathbb{R}(x)$ to $a+bi\in\mathbb{C}$.

So when we look at $\mathbb{R}(x)\rightarrow\mathbb{C}$ we're in the realm of very "flexible" objects. In fact, the fields in question are now pathologically flexible - for example, the field of complex numbers has $2^{2^{\aleph_0}}$-many nontrivial automorphisms.

Here, now, is a rather indirect proof that there is a field homomorphism from $\mathbb{R}(x)$ to $\mathbb{C}$:

Every field has an algebraic closure. Let $F$ be an algebraic closure of $\mathbb{R}(x)$ and $i:\mathbb{R}(x)\rightarrow F$ be a field homomorphism. Then $F$ is an algebraically closed field of characteristic $0$ (immediate) of transcendence degree $2^{\aleph_0}$ (by thinking about cardinality). But up to isomorphism, there is only one algebraically closed field of characteristic $0$ and transcendence degree $\kappa$ for any $\kappa$ - and in the particular case of $\kappa=2^{\aleph_0}$, it's just $\mathbb{C}$ itself.

The bolded fact above is not trivial. It's an extension of the more familiar result that any two vector spaces of the same dimension are isomorphic - here, "algebraically closed field of characteristic $0$" (or indeed any fixed characteristic) replaces "vector space" and "transcendence degree" replaces "dimension" - but more work needs to be done to prove it.

That's not even the worst bit, though. The above argument uses the axiom of choice in two ways: to build an algebraic closure of $\mathbb{R}(x)$, and to build transcendence bases of $F$ and $\mathbb{C}$ in order to get an isomorphism between them. This is not in general something we can avoid, and I am not certain whether $\mathsf{ZF}$ (= set theory without the axiom of choice) alone can prove that there is a field homomorphism from $\mathbb{R}(x)$ to $\mathbb{C}$.