# Let $F$ be a field, when is the quotient ring $F[x]/(x^2+1)F[x]$ an integral domain?

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?

• What could happen? Either the polynomial splits or not. – Wuestenfux Nov 6 '18 at 12:28

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.
$$(x^2+1)$$ is a prime ideal iff $$x^2+1$$ is a prime element.
$$x^2+1$$ is a prime element iff $$x^2+1$$ is irreducible, since $$F[x]$$ is a PID and so a UFD.
Note that this argument is not specific to $$x^2+1$$.