# Show that $M = \{a + b\sqrt{2}\mid 5\mid a, 5 \mid b\}$ is a maximal ideal in $R = \mathbb{Z}[\sqrt{2}]$ and find $R/M$.

I came across this question as part of my self-study of abstract algebra and so I would prefer answers suitable for a beginner.

First I established that the set $$M = \{a + b\sqrt{2}\mid a, b\in \mathbb{Z}, 5\mid a, 5 \mid b\}$$ is an ideal in the ring $R = \mathbb{Z}[\sqrt{2}]$ and this is more of a routine exercise. To prove that $M$ is a maximal ideal took me some time. The approach is to start with an ideal $N$ which properly contains $M$ i.e. $M \subset N$ and then show that $1 \in N$ so that $N = R$. Thus I start with an element $a + b\sqrt{2} \in N \setminus M$ so that at least one of $a, b$ is not divisible by $5$. Using this I show that norm $a^{2} - 2b^{2}$ is not divisible by $5$ and that is the only fact from number theory I used here.

Now it is easy to show that there is an integer $k$ such that $k(a^{2} - 2b^{2}) \equiv 1\pmod{5}$. Thus we can see that $1 = k(a - b\sqrt{2})(a + b\sqrt{2}) - 5n \in N$ for some integer $n$ (note that $5n \in M \subset N$).

Next $R$ is commutative with unity and hence $R/M$ is a field as $M$ is a maximal ideal of $R$. The field $R/M$ consists of all the cosets of type $a + b\sqrt{2} + M$ with $a, b$ taking values $0, 1, 2, 3, 4$ so that it has 25 elements.

There is another way to look at this field and that is to consider the field $\mathbb{Z}_{5}$ and the polynomial $x^{2} - 2$ which is irreducible in $\mathbb{Z}_{5}[x]$ so that the quotient $\mathbb{Z}_{5}[x]/(x^{2} - 2)$ is a field and looking at its elements we see that it is isomorphic to $R/M$ considered earlier.

My question is whether the second viewpoint can be used to directly infer the fact that $M$ is a maximal ideal in $R$?

Yes, it can - and it would have been easier like that most likely. Your ideal $M$ is the principle ideal generated by $5$. We have $R \cong \mathbb{Z}[x]/(x^2-2)$ and thus $R/M \cong \left( \mathbb{Z}[x]/(x^2-2) \right)/(5)$. Now an important isomorphism theorem shows that you can rearrange the order in which you factor out. Namely: $\left( \mathbb{Z}[x]/(x^2-2) \right)/(5) \cong \left( \mathbb{Z}[x]/(5) \right)/(x^2-2) \cong \mathbb{Z}[x]/(x^2-2,5)$.
Then we can see it as $\mathbb{Z}_5[x]/(x^2-2)$ which is a field, yes.
edit: Here is the theorem: https://en.wikipedia.org/wiki/Isomorphism_theorems#Third_isomorphism_theorem_2 Look at the last point for rings. Taking $R = \mathbb{Z}[x]$, $A = (x^2-2,5)$ and $B = (x^2-2)$ should give you the right results to use above.