Let $I:=\langle x^2+1 \rangle$ be an ideal of the polynomial ring $R:=\mathbb{Z}[x]$.
How to show that $I$ is a prime ideal?
I tried this:
$I$ is a prime ideal if for all $a,b \in \mathbb{Z}[x]$ with $ab \in I$, it's $a \in I$ or $b \in I$.
So let $a,b \in \mathbb{Z}[x]$ with $ab \in I$. Then:
$ab=r_1(x^2+1) \in I$, with $r_1 \in R$.
Now it should be $a=x^2+1$ and $b=r_1$.
Since $a \in I$ if $b=1$ it follows that $I$ is a prime ideal.
I'm not sure if this way to show it is correct. Or is there another method to show that it's a prime ideal?
Perhaps an easier way:
Define $\;\phi:\Bbb Z[x]\to\Bbb Z[i]\;,\;\;\phi(f(x)):=f(i)\;$ , with $\;i=\sqrt{-1}\;$ . Show this is a surjective homomorphism of rings and, of course, $\;\Bbb Z[i]\;$ is an integral domain. Of course, this happens iff the homomorphism's kernel is a prime ideal...
In fact, it is pretty common also to see $\;\Bbb Z[\sqrt{-1}]=\Bbb Z[i]\;$