Let $R$ be a commutative ring and consider the ideal $I=(x^2+1)$ of the polynomial ring $R[x]$ generated by $x^2+1$. If $R=\mathbb{C}$ prove that $I$ is not a prime ideal.
In order to prove that, I thought of proving that the quotient ring $R[x]/I$ is not an integral domain, but I am having some difficulties finding $a$ and $b$ such that $ab=0$. Is this approach correct? If so, how should try and show it?
It's not necessary to prove that $R[x]/I$ is not a domain. You can prove directly that $I$ is not prime by finding $a,b\in R[x]$ such that $ab\in I$ but neither $a$ nor $b$ are in $I$. (Hints have already been given in the comments.) This is essentially the same as finding zero divisors in $R[x]/I$, but it's more direct.
Generally, it's easier to prove that $R/I$ is an integral domain when you already know something about the structure of $R/I$. If you don't know very much about $\mathbb{C}[x]/(x^2+1)$, then you may have an easier time examining the ideal $(x^2+1)$ instead.
(It's also more common to prove that an ideal $I$ is prime by showing that $R/I$ is a domain. Proving that $I$ is not prime in my experience is often done directly.)