I'm trying to figure out the prime ideals of these rings: $C:=\mathbb{C}[x]/(x^2+1)$ and $R:=\mathbb{R}[x]/(x^2+1)$.
What I know is: any prime ideal of $C$ must contain $I:=(x-i)/(x^2+1)$ or $J:=(x+i)/(x^2+1)$, which are maximal because $C/I\simeq \mathbb{C}[x]/(x-i)\simeq \mathbb{C}\simeq C/J$, and $\mathbb{C}$ is a field. Then $I$ and $J$ are the only prime and maximal ideals of $C$, since $(0)$ is not prime here.
Is this correct?
And also I want to know:
Is there a unit in $C$?
Now $R$ is confusing me, I know $x^2+1$ is irreducible, then $(x^2+1)$ is prime in $\mathbb{R}[x]$ and $(0)$ is prime in $R$. More questions appeared:
Is $\mathbb{R}[x]$ a PID? If yes, then I can conclude $(0)$ is the only prime ideal of $R$, right?
And also:
Is it true that $R\simeq\mathbb{C}$?
I tried setting $\phi:\mathbb{R}[x]\rightarrow \mathbb{C}$ s.t. $\phi (p(x))=p(i)$, but I got confused trying to convince myself that $\ker(\phi) = (x^2+1)$.
You are correct about $C$. As for the unit, when you have a ring $R$ with a unit $1$, and an ideal $I$, then $R/I$ is also a ring with a unit, $1+I$.
Now for $R$, it's easy to see that your mapping is well defined, and also a ring homomorphism. Moreover, it's surjective: for every $α+βi$, $α, β$ real, there is $p(x)=α+βx\in \mathbb{R}[x]$ such that $φ(p(x))=α+βi$.
It remains to show that $\ker φ=(x^2+1)$. Obviously, $(x^2+1)$ is a subset of $\ker φ$. Moreover, $0\in\ker φ$, while every other real number does not belong to the kernel. Nor do real polynomials of degree $1$, since that would mean they have $i$ as a root. So now let's take a polynomial $p(x)\in \ker φ$, $\deg p(x)\ge 2$. Then the division algorithm would imply $p(x)=(x^2+1)π(x)+r(x)$, with $\deg r(x)<2$ or $r(x)=0$. But then we should have $p(i)=0$ or $r(i)=0$, which holds if and only if $r(x)=0$. This shows that $p(x)\in (x^2+1)$ or $\ker φ$ is a subset of $(x^2+1)$. This concludes the proof that $\ker φ=(x^2+1)$.