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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.

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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.

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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 $\mathbb F_{17}\times \mathbb F_{17}$.

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We have $$(5-2\sqrt{2} )\cdot (5+2\sqrt{2} )=17 \in J$$ but $5-2\sqrt{2}\notin J$ and $5+2\sqrt{2}\notin J$ hence $J$ is not prime ideal.

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OP already has $\ 17\mid 6^2-\sqrt{2}^2 = (6-\sqrt 2)(6+\sqrt 2)\ \ $ – Bill Dubuque Feb 14 '14 at 14:32

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