Let $F$ be a field, when is the quotient ring $F[x]/(x^2+1)F[x]$ an integral domain?
We know that for general rings, $R$ that $R/I$ is an integral domain if and only if $I$ is a prime ideal of $R$.
Thus $F[x]/(x^2 +1)F[x]$ is an integral domain if and only if $(x^2+1)F[x]$ is a prime ideal of $F[x]$.
One easy way that assure that $(x^2+1)F[x] = (x^2+1)$ is a prime ideal of $F[x]$ is if $x^2+1$ is irreducible over $F$ (for example in the case when $F = \mathbb{R}$), then $(x^2+1)$ is a maximal ideal of $F[x]$ and thus a prime ideal.
Now are there any weaker conditions than irreducibility of $x^2+1$ from which we can conclude that $(x^2+1)$ is a prime ideal?
Best Answer
The ideal $(x^2+1)$ is a prime ideal if and only if $x^2+1$ is irreducible in $F[x]$. This happens if and only if there is no $\lambda\in F$ such that $\lambda^2=-1$. For instance, if $F=\mathbb C$ or if $F=\mathbb{F}_5$, then $(x^2+1)$ is not a prime ideal.