Let R a ring with subring $\mathbb{C}$. Show there is no ring homomorphism $\phi:R\to\mathbb{R}$.
My attempt: every ring homomorphism from a field to a non-zero ring is injective. So we can consider the restriction of $\phi$ to $\mathbb{C}$.
If $\phi(i)=r$, then $r^4=1\implies r^2=\pm1$.
$r^3=-r\implies r(r^2-1)=0\implies r=0 $ or $r=\pm 1$. In any case, $\phi$ is not injective. QED
Is this right?
Best Answer
You were on the right track, but you went a little off course.
Starting as you did, let $r=\phi(i)$.
Since $\phi$ is a ring homomorphism, we must have $\phi(1) = 1$, hence $$i^2+1 = 0 \implies \phi(i^2+1) = 0 \implies r^2 + 1 = 0$$ contradiction, since for any $r \in \mathbb{R}, r^2+1 > 0$.
Hence there is no such homomorphism.