Solution verification: Describe homomorphisms from $A=\mathbb Q[x]/(x^2+1)$ to $\mathbb C$ and $\mathbb R$
So this is a watered down version of a question I encountered in a test, and I want to see if I got the principle correct.
So $(x^2+1)=(x-i)(x+i)$. Roots in the range field correspond with homomorphisms. Since $i, -i \in \mathbb C$ there are two homomorphisms from $A$ to $\mathbb C$. But they aren't in $\mathbb R$, so there are no homomorphisms to $\mathbb R$. Is the last bit correct?
The fact that $+/- i$ are not in $\mathbb R$ does not necessarily say anything about homomorphisms into $\mathbb R$, since we can have homomorphisms of rings which consist of entirely different elements.
However you have $(\varphi (i))^2=\varphi (i^2)=\varphi(-1)$ if $\varphi$ is a homomorphism of $\mathbb Q / (x^2+1)$ into $\mathbb R$. Since homomorphisms map identity to identity, you have $\varphi(-1)=-\varphi(1)=-1$. Thus we see that $i$ must map to a term in $\mathbb R$ whose square is $-1$. But there is no element in $\mathbb R$ whose square is $-1$.
Using the same reasoning, it is clear that a homomorphism from $\mathbb Q / (x^2+1)$ into $\mathbb C$ must map $i$ to $+/-i$.