Are the Complex Numbers Isomorphic to the Polynomials, mod $x^2+1$? My friend told me that the Complex Numbers are Isomorphic to the Polynomials mod $x^2+1$, is this so? And how can this be proved?
 A: Let $\phi : \mathbb R[x] \to \mathbb C$ be defined by $\phi(P)=P(i)$. 
This can be proven to be ring/group morphism, which is onto and $\ker(\phi)=\langle x^2+1 \rangle$. Apply the first isomorphism theorem.
A: $\Bbb R[x]$ is a PID so every prime ideal is maximal. $x^2+1$ is irreducible over $\Bbb R[x]$, and therefore $(x^2+1)$ is a maximal ideal, so $\Bbb R[x]/(x^2+1)$ is a field.
Now, consider the evaluation homomorphism $\phi: \Bbb R[x]\rightarrow\Bbb C$, such that $\phi(f)=f(i)$.
It is clear that $(x^2+1)\subseteq\ker(\phi)$, and equality follows since $(x^2+1)$ is maximal. From the 1st isomorphism theorem, we have $$\Bbb R[x]/(x^2+1)\cong\Bbb C$$


Let $R$ be a PID, then every nonzero prime ideal is maximal

Proof: Let $P\subset R$ be a nonzero prime ideal, then since $R$ is a PID we have $P=(p)$ for some $p\in R$. Suppose there is an ideal $I=(x)$ such that $$(p)\subseteq (x)\subseteq R$$
Then $p\in(x)$ so that $p=kx$ for some $k\in R$. Since $(p)$ is a prime ideal, either $x\in (p)$ or $k\in (p)$:

*

*If $x\in (p)$, then $(x)=(p)$.

*If $k\in (p)$, then $k=py\Rightarrow p=pyx\Rightarrow p(1-yx)=0\Rightarrow yx=1\Rightarrow x$ is a unit $(x)=R$
Thus $P$ is maximal.


$x^2+1\in\mathbb R[x]$ is irreducible.

Proof: If $x^2+1$ were reducible, then we could write:
$$x^2+1=(ax+b)(cx+d)$$
Expanding, we get $$x^2+1=acx^2+(ad+bc)x+bd$$ So:

*

*$ac=1$

*$bd=1$

*$ad+bc=0$
This leads to $$ad=-bc\\\Rightarrow (ac)d=-bc^2\\\Rightarrow d=-bc^2\\\Rightarrow (bd)=-(bc)^2\\\Rightarrow 1=-(bc)^2$$ Which is a contradiction, so $x^2+1$ is irreducible.
