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.
Best Answer
This is how I look at such problems. First of all, $\Bbb Z[\sqrt{2}]\cong \Bbb Z[x]/(x^2-2)$.
Then using isomorphism theorems, $\Bbb Z[\sqrt{2}]/(17)\cong \Bbb Z[x]/(x^2-2,17)\cong(\Bbb Z/(17))[x]/(x^2-2)=\Bbb F_{17}[x]/(x^2-2)$.
(I'm taking some liberties with the notation: the parenthesis around elements denote the ideal generated in the ring in context. That's why even though the $(17)$ all look alike, they're actually different sets as they are generated in their respective rings.)
So the question amounts to figuring out the structure of $\Bbb F_{17}[x]/(x^2-2)$, but quotients of polynomial rings over fields are pretty easy to analyze. The ideal $(x^2-2)$ is going to be prime iff $x^2-2$ is irreducible over $\Bbb F_{17}$, but you'll discover quickly that it has two distinct roots over this field, and is reducible.
Given the two roots $\alpha,\beta$, the Chinese remainder theorem says that $\Bbb F_{17}[x]/(x^2-2)\cong \Bbb F_{17}[x]/(x-\alpha)\times \Bbb F_{17}[x]/(x-\beta)\cong \Bbb F_{17}\times \Bbb F_{17}$, so the ring has $289$ elements.