The field $\mathbb{R}[X]/(X^2+1)$

Homomorphism. $\varphi(P+Q)=P(i)+Q(i)=\varphi(P)+\varphi(Q)$ and $\varphi(PQ)=P(i)Q(i)=\varphi(P)\varphi(Q)$.

Kernel. $P(i)=0$ iff $X-i$ divides $P$ over $\mathbb{C}$. But $P\in\mathbb{R}[X]$, so if $i$ is a root, $-i$ is also a root (conjugate roots). So $(X-i)(X+i)=X^2+1$ divides $P$ over $\mathbb{R}[X]$. Kernel = $(X^2+1)$.

Surjective. $a+bi=\varphi(a+bX)$.

Conclusion. By the first isomorphism theorem, $\mathbb{R}[X]/(X^2+1)\cong\mathbb{C}$.