How to formally prove that $\Bbb Q[i]$ is not isomorphic to any subring of $\Bbb R$?

Question

I am trying to prove that $\Bbb Q[i]$ is not isomorphic to any subring of $\Bbb R$. We know that the ring $\Bbb Q[i]$ is actually a field. So if it is isomorphic to some subring of $\Bbb R$, $\Bbb R$ then becomes an extension of $\Bbb Q[i].$ So do we just prove that there is no ring homomorphism from $\Bbb Q[i]$ to $\Bbb R$?

Answer 1

Suppose for the sake of contradiction that $\mathbb{Q}[i]$ is isomorphic to a subring of $\mathbb{R}$, and call the isomorphism from the former to the latter, $\varphi$.

Consider $\varphi(i^2 + 1) = \varphi(i^2) + \varphi(1) = [\varphi(i)]^2 + 1$, where the first equality follows because homomorphisms respect addition, and the second follows because homomorphisms respect multiplication (and because $1$ is mapped to $1$, an exercise I presume you have seen or completed).

And so now we have $[\varphi(i)]^2 + 1 \in \mathbb{R}$.

But we can evaluate the earlier expression in a different way: $\varphi(i^2 + 1) = \varphi(0) = 0$, where the first equality follows because $i^2 + 1 = 0$, and the second equality follows because $0$ is mapped to $0$ (again, an exercise I presume you have seen or completed).

Put together, we have: $[\varphi(i)]^2 + 1 = 0$, for some real number $\varphi(i)$. But this is problematic, since it would mean we could square the real number $\varphi(i)$ and get $-1$. Contradiction, since no real number can be squared to yield $-1$. So our supposition was wrong, and no such isomorphism exists.