Units of $\mathbb{Z[\sqrt2]}$

Question

If we take $a+b\sqrt2\in\mathbb{Z[\sqrt2]}$ such that this is a unit, is it necessary therefore that $1$ is the highest common divisor of both $a$ and $b$?

Is there a way to approach this question straight from definition of unit? Or in other words, not from the perspective of norms etc?

Answer 1

Suppose that $a + b\sqrt{2}$ is a unit. Let $c + d\sqrt{2}$ be its inverse. (Where $c$ and $d$ are some integers.)

Then, we have $$(a + b\sqrt{2})(c+d\sqrt{2}) = 1$$ or $$(ac + 2bd) + (ad + bc)\sqrt{2} = 1.$$

As $\sqrt{2}$ and $1$ are indepdendent over $\mathbb{Q}$, we get that $ac + 2bd = 1$ and $ad + bc = 0$. The latter is not important.

Now, note that if $p \in \mathbb{Z}$ is a common divisor of $a$ and $b$, then $p$ must divide $ac + 2bd = 1.$ This forces $p \in \{1, -1\}$.

Answer 2

Let $u = a + b\sqrt{2}$ be an unit (that means $u\cdot r=1$ for some $r$ in the ring). Let $d = \gcd(a,b) \in \mathbb Z$. Then $d$ divides both $a$ and $b$. So let us write $a = a'\cdot d$, $b = b' \cdot d$.

Since $ 1= (a + b\sqrt{2})r = d(a' + b'\sqrt{2})r$, we see that $d$ divides $1$. Therefore $d = \pm 1$.

Answer 3

Is there a way to approach this question straight from definition of unit? Or in other words, not from the perspective of norms etc?

Yes, units $u$ are the divisors of $1,\,$ so divisors of units remain units by transitivity of "divides", i.e. $\,c\mid u\mid 1\,\Rightarrow\, c\mid 1\Rightarrow\,c\,$ unit.

In particular if $\,c\mid a,b\,$ in $\,\Bbb Z\,$ then $\,c\mid a+b\sqrt 2= \,$ unit, so $\,c\,$ divides a unit $\,\Rightarrow\,c\,$ remains a unit.