 Solution

The evaluation map $\phi :\mathbb{R}[x]\to \mathbb{C}$, $p\mapsto p(i)$, is a ring homomorphism and is onto because $a+bx$ maps to any $a+bi$. By polynomial division, writeThen $p(i)=0$ precisely when $a=b=0$. Thus $\ker \phi =(x^2+1)$, and the first isomorphism theorem givesSince the quotient ring is a field, $(x^2+1)$ is a maximal ideal.