Prove $\mathbb{R}$ does not contain a subring isomorphic to $\mathbb{C}$

Question

I'm trying to prove that the quaternions ring $\mathbb{H}$ is not a $\mathbb{C}$-algebra, so I assume $\mathbb{H}$ actually is a complex algebra and that implies that there exists an injective ring homomorphism from $\mathbb{C}$ to the center of $\mathbb{H}$ which is $\mathbb{R}$. This seems to be a contradiction to me, but I can't produce a formal proof of this fact, so I would appreciate any hint. Thanks in advance.

Answer 1

An embedding $\Bbb C\hookrightarrow \Bbb R$ must send $i$ to a root of $x^2+1$.

Answer 2

HINT: if $f$ is an injective ring homomorphism from $\mathbb{C}$ to $\mathbb{R}$, what can $f(i)$ be?

---

There's another approach (which is essentially the same argument but in different language): any subfield of $\mathbb{R}$ has a compatible ordering, so it's enough to show that $\mathbb{C}$ does not have any compatible ordering. The reason this is the same argument in disguise is that (to the best of my knowledge) the only way to show that $\mathbb{C}$ has no compatible ordering . . . is to look at $i$! Still, I think it's a neat observation.

Note that we can show by general topology that there is no injective continuous map from $\mathbb{C}$ to $\mathbb{R}$, without ever talking about the special properties of $i$; unfortunately, there's no reason a ring homomorphism from $\mathbb{C}$ has to be continuous. Interestingly, any ring homomorphism from $\mathbb{R}$ is continuous, essentially because the topology on $\mathbb{R}$ comes from the ordering, which in turn is definable in a ring-theoretic way ($a<b\iff \exists c(c^2+a=b)$).

Answer 3

If such a subring exists, then the group of units of complex numbers will be a subgroup of the group of units of reals. The former contains elements of any finite order (the roots of unity), whereas in the latter the only torsion elements are $\pm1$. The hint by Noah Schweber is this for torsion elements of order 4.