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I am trying to prove that the ideal $J=<x+1>$ is prime in the ring $\mathbb{Z}[x]$. I know that if the generator is prime, then the ring modulo the generator is an integral domain. I can show that this is true in this case after a bit of work. However, I was wondering if there was a quicker way to do this. Any thoughts?

Thanks

jonan
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    It's the kernel of the map to $\Bbb Z$ which takes $x$ to $-1$. – Angina Seng Dec 09 '18 at 17:16
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    We can of course directly prove that domain $R,\Rightarrow, x!-!a,$ prime in $,R[x],$ via the remainder theorem:

    $$x!-!a\mid fg\iff \underbrace{ f(a)g(a)=0\iff f(a)=0\ \ {\rm or}\ \ g(a)=0}_{\large R\ \text{a domain}}\iff x!-!a\mid f\ \ {\rm or}\ \ x!-!a\mid g$$

     – Bill Dubuque Dec 09 '18 at 17:33

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