$(2, 1 + \sqrt[]{−17})$ is a prime ideal in $\Bbb Z[\sqrt[]{-17}]$ How can we see that $(2, 1 + \sqrt[]{−17})$ is prime ideal in $\Bbb Z[\sqrt[]{−17}]$?
We have \begin{align*}
\frac{\mathbb{Z}\left[\sqrt{-17}\right]}{\left(2, 1 + \sqrt{-17}\right)} &\cong \frac{\mathbb{Z}[x]/(x^2 + 17)}{(2, 1 + x, x^2 + 17)/(x^2 + 17)} \cong \frac{\mathbb{Z}[x]}{(2, 1+x, x^2 + 17)}.
\end{align*} 
I don't see why this is an integral domain? 
Thank you for your help.
 A: $$\frac{\Bbb Z[\sqrt{-17}]}{(2,1+\sqrt{-17})}\simeq\frac{\frac{\Bbb Z[\sqrt{-17}]}{(2)}}{\frac{(2,1+\sqrt{-17})}{(2)}}\simeq\frac{\Bbb Z_2[i]}{(1+i)}\simeq\Bbb Z_2.$$
Here I have used, Third Isomorphism theorem and reduction modulo $2$.
A: A general element of $\Bbb{Z}[\sqrt{-17}]$ is $a + b\sqrt{-17}$ for $a,b \in \Bbb{Z}$.  When you mod out by $(1+\sqrt{-17}) \subset (2,1+\sqrt{-17})$, you impose the relation $\sqrt{-17} = -1$, so 
$$  a + b\sqrt{-17} \cong a-b \pmod{(1+\sqrt{-17})}  \text{.}  $$
This makes $\frac{\Bbb{Z}[-17]}{(1 + \sqrt{-17})}$ (a subring of) $\Bbb{Z}$.  (There's a reasonable chance that $(1 + \sqrt{-17})$ contains an integer, perhaps something near $18$, so we may actually have a much smaller ring than $\Bbb{Z}$.)  Then when we quotient by $(2)$, we get either $\{0\}$ (a.k.a., the one ring) or $\Bbb{Z}/2\Bbb{Z}$.  The former happens when we actually have $\Bbb{Z}$ mod an odd number (because having both $\text{"some odd integer"} \cong 0$ and $2 \cong 0$ sends all of $\Bbb{Z}$ to zero) and the latter when we actually have $\Bbb{Z}$ or $\Bbb{Z}$ mod an even number after the first quotient.
If you can show $(2, 1+\sqrt{-17})$ is a proper ideal, you do not get the one ring.  So you get a field, showing that $(2, 1+\sqrt{-17})$ is a maximal ideal.  Since maximal ideals are prime ideals, $(2, 1+\sqrt{-17})$ is a prime ideal of $\Bbb{Z}$.
Can show $(2, 1+\sqrt{-17})$ is a proper ideal?
A: Modding out by $x+1$ is the same as setting $x = -1$. Or more precisely, using the Third Isomorphism Theorem
\begin{align*}
\frac{\mathbb{Z}[x]}{(2, 1+x, x^2 + 17)} & \cong \frac{\mathbb{Z}[x]/(1+x)}{(2, 1+x, x^2 + 17)/(1+x)} \cong \frac{\mathbb{Z}[-1]}{(2, (-1)^2 + 17)} = \frac{\mathbb{Z}}{(2, 18)} = \frac{\mathbb{Z}}{(2)} \, .
\end{align*}
