Let $R=\Bbb{Z}[\sqrt2]=\{{a+b\sqrt2:a,b\in\Bbb{Z}}\}$ and the principal ideal $I=(2-\sqrt2)$ in $R$. Is $(2-\sqrt2)$ a proper ideal of $R$?

Question

The concept of proper ideals is confusing me just a little bit. Do I need to prove that $I\neq R$?

There's a second part as well:

b) Rewrite the coset $5 + 3\sqrt2 + I$ from $R/I$ in the form $a + I$, where $a$ is an integer. Try to use the smallest non-negative integer $a$ that you can.

Answer 1

$I = (\sqrt{2})(\sqrt{2}-1)$ the second factor is a unit, so the ideal is just $(\sqrt{2})$ which is indeed a proper ideal. Now for the second part note that $\sqrt{2}|2j$ for all $j\in\Bbb Z$ so with $5+3\sqrt{2}=2\cdot 2 + 1 +3\sqrt{2}$ what can you conclude?

Answer 2

This is most easily checked using the norm on $\mathbf Z[\sqrt2]$: by definition, $$N(a+b\sqrt 2)=a^2-2b^2.$$ This norm is multiplicative, i.e. $N(xy)=N(x)N(y)$ for all $x,y\in\mathbf Z[\sqrt2]$, and $N(1)=1$.

There results an element $x$ is a unit if and only if its norm is a unit in $\mathbf Z$. Here $N(2-\sqrt2)=4-2=2$, hence this element has no inverse, and it generates a non-trivial ideal.

An example of a unit is $3-2\sqrt2$, which has norm $1$. Indeed $$(3-2\sqrt2)(3+2\sqrt2)=9-8=1.$$