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The title already states my question: Is $(X^2+1)$ a prime ideal in $\mathbb{R}[X,Y]$?

I know that if this is true, Then $\mathbb{R}[X,Y]/(X^2+1)$ must be a domain. And if that is a domain then $(X^2+1)$ must be the kernel of a homomorphism from $\mathbb{R}[X,Y]$ to itself.

However, how do I prove that $(X^2+1)$ is the kernel of such homomorphism, or that this does not exist?

Thanks!

jbuser430
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  • The "right" homomorphism is $\mathbb{R}[X,Y] \to \mathbb{R}[X,Y]/(X^2+1)$, not an endomorphism of $\mathbb{R}[X,Y]$. In general, an ideal is prime if and only if it is the kernel of a homomorphism to some domain. I don't understand your claim that there must be such an endomorphism of $\mathbb{R}[X,Y]$. – 57Jimmy Apr 10 '18 at 08:35

1 Answers1

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We have \begin{align*} {\bf{R}}[X,Y]/(X^{2}+1)\cong({\bf{R}}[X])[Y]/(X^{2}+1)\cong({\bf{R}}[X]/(X^{2}+1))[Y]\cong{\bf{C}}[Y]. \end{align*}

user284331
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