# Show that $I = \{a +\sqrt2b \in \Bbb Z[\sqrt2] : a$ and $b$ are both multiple of $5\}$ is a maximal ideal.

Let $I = \{a +\sqrt2b \in \Bbb Z[\sqrt2] : a\text{ and } b\text{ are both multiple of }5\} \subset \mathbb Z[\sqrt 2]$. I have shown that $I$ is an ideal.

Now I want to show that it is maximal. Since I can write

$$I = 5\mathbb Z[\sqrt 2] = \{ 5(x+y\sqrt 2): x, y, \in \mathbb Z\},$$

I guess $\mathbb Z[\sqrt 2] / I\cong \mathbb Z/5\mathbb Z$. This is a field and thus $I$ is maximal. I am not sure about if the $\cong$ is true though.

For $a+b\sqrt 2\notin I$, show that there exists $c+d\sqrt 2$ such that $(a+b\sqrt 2)(c+d\sqrt 2)\in 1+I$. This means that every nonzero element of $\mathbb Z[\sqrt 2]/I$ is invertible, i.e., $\mathbb Z[\sqrt 2]/I$ is a field, i.e., $I$ is a maximal ideal.

• First consider the case that $5\mid b$ (and hence $5\nmid a$), say $b=5b'$. Since $\mathbb Z/5\mathbb Z$ is a field, we can find $c$ such that $ac\equiv 1\pmod 5$, say $ac=1+5e$. Then $(a+b\sqrt 2)(c+0\sqrt 2)=ac+bc\sqrt 2=1+5(e+b'c\sqrt 2)$.
• Next consider the case $5\mid a$ (and hence $5\nmid b$), say $a=5a'$. Then multiplication with $\sqrt 2$ gives $2b+a\sqrt 2$ and we are back in the first case.
• Finally consider the case that $5\nmid a$ and $5\nmid b$. Then $a^2\equiv \pm 1\pmod 5$ and $2b^2\equiv \pm 2\pmod 5$, so certainly $5\nmid a^2-2b^2$. Multiplication with $a-b\sqrt 2$ gives $(a+b\sqrt 2)(a-b\sqrt 2)=a^2-2b^2+0\sqrt 2$, i.e., again the first case.

One may follow this route:

$\mathbb{Z}[\sqrt{2}]=\frac{\mathbb{Z}[x]}{(x^2-2)}$

The pullback ideal of $I=5\cdot\mathbb{Z}[\sqrt{2}]$ in $\mathbb{Z}[x]$ is $(x^2-2,5)$, hence $\frac{\mathbb{Z}[\sqrt{2}]}{I}=\frac{\mathbb{Z}[x]}{(5,x^2-2)}$

But the latter can (see [$\star$]) be proved to be $\frac{\mathbb{F}_5[x]}{(x^2-\overline{2})}$, which is a field because $x^2-\overline{2}$ is irreducible in $\mathbb{F}_5[x]$

[$\star$] The map $\ a_dx^d+\ldots+a_0\mapsto \overline{a_d}x^d+\ldots+\overline{a_0}$ sends $(5,x^2-2)\rightarrow(x^2-\overline{2})$...