Determine if the following ideals are prime.

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Determine if the following ideals are prime.

  1. $I=\langle x^2+2\rangle\subseteq \mathbb{Z}[x]$
  2. $I=\langle x^2+2\rangle\subseteq \mathbb{Z}_3[x].$

By the definition, a proper ideal $I$ of $R$ is prime if whenever $a,b\in R$ such that $ab\in I,$ then either $a\in I$ or $b\in I.$

1) I know $I$ is a proper subset of $\mathbb{Z}[x]$ as $1\in \mathbb{Z}[x]$ but $1\not\in I.$ But I can't find values $a,b\in \mathbb{Z}[x]$ such that if $ab\in I,$ then $a\in I$ or $b\in I.$ Is this one prime?

2) Is this not prime? I'm having a hard time thinking of an example. How is $x^2+2$ even factorable?

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We claim that $I=\langle x^2+2\rangle\subseteq \mathbb{Z}[x]$ is prime. Suppose $a,b\in\mathbb{Z}[x]$ such that $ab\in I.$ Then $ab = r(x^2+2), r\in\mathbb{Z}[x].$ We have $\deg (ab) \geq 2$ or $ab=0\Rightarrow 0=a\in I\vee 0=b\in I,$ so we just need to consider when $\deg(ab)\geq 2.$ We show that $x^2+2$ is irreducible in $\mathbb{Z}[x].$ Suppose $\exists a,b\in\mathbb{Z}[x]$ such that $ab = x^2+2$ and $a$ and $b$ are not units of $\mathbb{Z}[x].$ Note that since $ab\neq 0, 0<\deg(a),\deg(b)<2.$ Then since the leading coefficient is $1,$ $a=x+c$ and $b=x+d,$ where $c,d\in\mathbb{Z}.$ But then, $ab=(x+c)(x+d)=x^2+(c+d)x+cd.$ This means $c+d=0\Rightarrow c=-d$ and $cd = 2\Rightarrow -d^2=2.$ The latter equation is unsolvable in $\mathbb{Z}[x];$ hence either $a$ or $b$ are units of $\mathbb{Z}[x]$. Thus, since $x^2+2$ is irreducible, either $a$ or $b$ is a multiple of $x^2+2.$ Thus $a\in I$ or $b\in I.$

We claim that $I=\langle x^2+2\rangle\subseteq \mathbb{Z}_3[x]$ is not prime. Consider $x+2,x+1\in \mathbb{Z}[x].$ We have $(x+2)(x+1)=x^2+2\in\mathbb{Z}_3[x],$ but $x+1,x+2\not\in I.$ Thus, this ideal is not prime.