$(2, 1 + \sqrt[]{−17})$ is a prime ideal in $\Bbb Z[\sqrt[]{-17}]$

Question

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.

Answer 1

$$\frac{\Bbb Z[\sqrt{-17}]}{(2,1+\sqrt{-17})}\simeq\frac{\frac{\Bbb Z[\sqrt{-17}]}{(2)}}{\frac{(2,1+\sqrt{-17})}{(2)}}\simeq\frac{\Bbb Z_2[i]}{(1+i)}\simeq\Bbb Z_2.$$ Here I have used, Third Isomorphism theorem and reduction modulo $2$.

Answer 2

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?

Answer 3

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*}