$\mathbb R[x]/\langle x^2+1\rangle\simeq\mathbb C$ and $\mathbb R[x]/\langle x^2+1\rangle$ contains a zero of $x^2+1$

Question

• How to show that $\mathbb R[x]/\langle x^2+1\rangle\simeq\mathbb C$ and $\mathbb R[x]/\langle x^2+1\rangle$ contains a zero of $x^2+1$

Due to the division algorithm, $\mathbb R[x]/\langle x^2+1\rangle=\{ax+b+\langle x^2+1\rangle:a,b\in\mathbb R\}$

Indeed I'm having problem to interpret, $\mathbb R[x]/\langle x^2+1\rangle$ contains a zero of $x^2+1.$ Does it mean $\exists$ $ax+b+\langle x^2+1\rangle\in\mathbb R[x]/\langle x^2+1\rangle$ such that $$(ax+b+\langle x^2+1\rangle)^2+(1+\langle x^2+1\rangle)=\langle x^2+1\rangle?$$

Answer 1

Yes. Let $\bar x=x+\langle x^2+1\rangle$. What is $\bar x^2+1$ in the quotient?