In this answer and other's that I've read, finding the units in $\mathbb{Z}[\sqrt{2}]$ (elements of the form $a + b\sqrt{2}$ for integers $a$ and $b$) is equivalent to finding the solutions to $a^2-2b^2 = \pm 1$. So from my understanding, for a ring $R$, $u \in R$ is a unit if there exists a $v \in R$ such that $uv = vu = 1_R$, with $1_R$ being the multiplicative identity in the ring.

So I have 2 questions:

  1. My understanding is that we arrive at this because $(a+b\sqrt{2})(a-b\sqrt{2}) = a^2-2b^2$, so units can only come in conjugate pairs. I could intuitively reason that this is because $(a+b\sqrt{2})x$ would have a $\sqrt{2}$ term if $x$ is not conjugate, but I'm not sure if this is strictly the reason why the units are only conjugates.
  2. Why can they multiply to $-1$ and still be a unit? Is it because if $(a+b\sqrt{2})(a-b\sqrt{2}) = a^2-2b^2 = -1$, then $(a+b\sqrt{2})^2(a-b\sqrt{2})^2 = (-1)^2 = 1$, and thus the inverse of $(a+b\sqrt{2})$ is $(a+b\sqrt{2})(a-b\sqrt{2})^2$?
  • $\begingroup$ Are you familiar with the norm $N(x)$ defined on $\mathbb Z[\sqrt 2]$? $\endgroup$ – Frpzzd Oct 23 '18 at 23:31
  • $\begingroup$ @Frpzzd I am not - this question came up in one of my first lectures on the topic of rings, so I'm very new to it. $\endgroup$ – rb612 Oct 23 '18 at 23:32
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    $\begingroup$ Okay. I will define it for you in my answer. $\endgroup$ – Frpzzd Oct 23 '18 at 23:33
  • $\begingroup$ Have I explained by answer sufficiently? Please let me know if anything in my answer is confusing or worded badly so that I can clarify. $\endgroup$ – Frpzzd Oct 23 '18 at 23:43

Define the norm $N:\mathbb Z[\sqrt 2]\mapsto \mathbb Z$ as follows: $$N(a+b\sqrt 2)=a^2-2b^2\tag{1}$$ or $$N(x)=x\bar{x}\tag{2}$$ Then notice that for all $x,y\in\mathbb Z[\sqrt 2]$, we have that $$N(xy)=N(x)N(y)\tag{3}$$ Thus, suppose that $x\in\mathbb Z[\sqrt 2]$ is a unit. Then let $x^{-1}$ be its inverse. We have by $(3)$ that $$1=N(1)=N(xx^{-1})=N(x)N(x^{-1})$$ and so $N(x)$, which is an integer, is a unit, with multiplicative inverse $N(x^{-1})$. Since the only units in $\mathbb Z$ are $\pm 1$, we have that $N(x)=a^2-2b^2$ must equal $+1$ or $-1$.

This proves that $a^2-2b^2$ must equal $1$ or $-1$ if $x$ is a unit. Now we must prove the converse.

Suppose $N(x)=a^2-2b^2=\pm 1$. Then, by $(2)$, $x\bar{x}=1$ or $x\bar{x}=-1$. If the former is true, then $\bar{x}$ is the inverse of $x$ and $x$ is a unit; if the latter is true, then $-\bar{x}$ is the inverse of $x$ and $x$ is a unit.

Proof complete!

Now I shall address a generalized form of your first question. Suppose that $x,y\in\mathbb Z[\sqrt 2]$, and that $xy\in\mathbb Z$. Write $x=a+b\sqrt 2$ and $y=c+d\sqrt 2$. Then $$xy=(ac+2bd)+(ad+bc)\sqrt{2}$$ For $xy\in\mathbb Z$ to hold, it must be true that the coefficient $ad+bc$ of $\sqrt 2$ vanishes. Thus, we have $$ad+bc=0$$ or $$\frac{a}{b}=-\frac{c}{d}$$ ...which implies that the ratios of the rational and irrational parts of $x$ and $y$ are equal and opposite. This is equivalent to stating the existence of $z\in\mathbb Z[\sqrt 2]$ such that $z|x$ and $\bar{z}|y$.

If you apply this to units, you can see that it is indeed true that all units must come in conjugate pairs.

  • $\begingroup$ Thanks so much for the thorough response! This is great. My main question from this though is why can we/do define the norm as such and then proceed from there? In other words, did we arbitrarily define that norm to proceed with the proof, or is it the only option for the norm of an element in this ring? $\endgroup$ – rb612 Oct 23 '18 at 23:51
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    $\begingroup$ @rb612 We choose that norm mostly because of its nice multiplicative property $N(xy)=N(x)N(y)$. It's not really that arbitrary; I've just assigned the value $a^2-2b^2$ mentioned in your question to a function $N(x)$, giving it a name. I could have done the proof without defining $N(x)$ at all and instead referring to $a^2-2b^2$; it's just the convention that I am used to using. $\endgroup$ – Frpzzd Oct 23 '18 at 23:53
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    $\begingroup$ @rb612 One more thing... once I've answered all of your questions, don't forget to $\checkmark$! :D $\endgroup$ – Frpzzd Oct 23 '18 at 23:55

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