How many elements are there in this quotient ring? So we are having this undergraduate course in my department of commutative algebra and there is a problem sheet that we have to submit. The second problem goes like this:
Let $R$ be the ring $R=\mathbb{Z}[\sqrt2]=\{a+b\sqrt2:a,b\in\mathbb{Z}\}$ and let $J=17R\subseteq R.$ How many elements are there in the quotient ring $R/J$? Is $J$ a prime ideal?
Now, I know a theorem that states that $\mathbb{Z}[\sqrt n]/p\mathbb{Z}[\sqrt n]$ (p: prime) is a field (hence $p\mathbb{Z}[\sqrt n]$ is a maximal and therefore a prime ideal) if and only if there is no $a\in\mathbb{Z}$ such that $$a^2\equiv n\mod p$$ and in that case, $\mathbb{Z}[\sqrt n]/p\mathbb{Z}[\sqrt n]$ has $p^2$ elements.
In my case the previous theorem does not work because for $6\in\mathbb{Z}$ I got $6^2\equiv 2\mod 17$ so am a bit confused.
ps. This is my first post, so accept my apologies for any mistakes i've made.
ps2. Sorry for my english.

You can do it by hand very easily: two elements $a+b\sqrt{2}$ and $c+d\sqrt{2}$ lie in the same coset iff their difference belongs to $17R$. Since $17R=\{17x+17y\sqrt{2}\colon x,y\in \mathbb Z\}$, you have that $a+b\sqrt{2}$ and $c+d\sqrt{2}$ lie in the same coset iff $a\equiv b\bmod 17$ and $c\equiv d\bmod 17$. So you see that $R/17R$ has $17^2$ elements and it is not a field for the reason you pointed out. In fact, $R/17R$ is isomorphic as a ring to $\mathbb F_{17}\times \mathbb F_{17}$.
