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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?

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

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There are no weaker conditions.

$(x^2+1)$ 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 $x^2+1$.

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