Finding the units of $\mathbb{Z}\left[\frac{1+\sqrt{-19}}{2}\right]$

Question

I was trying to comprehend a simple exercise from my elementary number theory class.

Let $\theta=\frac{1+\sqrt{-19}}{2}$, and let $R$ be the ring $\mathbb{Z}[\theta] .$ Show that the units of $R$ are $\pm 1$, and that 2,3, and $\theta$ are irreducible in $\mathbb{Z}[\theta]$.

The official solution: If $z=a+b \theta$ is a unit of $\mathbb{Z}[\theta]$ then $z \bar{z}=1$, so we have $a^{2}+a b+5 b^{2}=1$ from which we conclude $a=\pm 1$, so the units are $\pm 1$. If $z=a+b \theta$ is a nonunit, non-associate divisor of 2 then $z \bar{z}=2$, but there are no solutions to $a^{2}+a b+5 b^{2}=2$. The arguments for 3 and $\theta$ are similar.

This solution is much too concise for me.

1. Firstly, why does $z=a+b \theta$ being a unit of $\mathbb{Z}[\theta]$ imply $z \bar{z}=1$? At this point in time, the norm for $\mathbb{Z}[\theta]$ has not been defined, so cannot be used. I was hypothesizing that for $z_1,z_2 \in \mathbb{Z}[\theta]$, $z_1 z_2 \in \mathbb{Z}$ implied (by comparing imaginary parts) that $z_1 = \lambda \overline{z_2}$ for some $\lambda \in \mathbb{Q}$, but this also is not easy to prove. I even failed to brute force this, I couldn't continue the calculation.

2. Why does $z=a+b \theta$ being a nonunit, non-associate divisor of 2 imply that $z \bar{z}=2$?

Any partial answers (answering only one of the questions) are very welcome! I will upvote them too. If the above problem has other solutions, not necessarily following the same reasoning as the official solution, then those are also very welcome!

Answer 1

Denote $z\bar z$ by $N(z)$. Then

• $N(\alpha\beta)=N(\alpha)N(\beta)$ where $\alpha,\beta\in\Bbb Z[\theta]$.
• $N(z)=a^2+ab+5b^2>0$ for all non-zero $z$. This is because $$4(a^2+ab+5b^2)\geq 0\\(2a+b)^2+19b^2\geq 0$$ Also $N(z)$ is a non-negative integer.

$z$ is a unit $\iff N(z)=1$

Proof:

• If $z$ is a unit, then there is a $w\in\Bbb Z[\theta]$ such that $zw=1$. Then $N(z)N(w)=1$. So $N(z)=1$

• If $N(z)=1$, then $z\bar z=1$, so $z$ is a unit.

The units in $\Bbb Z[\theta]$ satisfy $N(z)=1$. To find the units notice that $$a^2+ab+5b^2=1\\(2a+b)^2+19b^2=4$$ gives $b=0$, and so $a=\pm 1$. Therefore the units of $\Bbb Z[\theta]$ are $\pm 1$.

Let $z$ be a non-unit, and a non-associate divisor of $2$. Then $zw=2$, and so $N(z)N(w)=4$.

• If $N(z)=1$, then $z$ is a unit, which is impossible.
• If $N(z)=4$, then $N(w)=1$ so $w$ is a unit, and then $z$ is associate to $2$.

Therefore $N(z)=2$.

Answer 2

1. Let $z \in \mathbb{Z}[\theta]$ be a unit, with inverse $u \in \mathbb{Z}[\theta]$. You can check by hand that $\zeta=z\overline{z},\upsilon=u \overline{u} \in \mathbb{Z}$, and since $zu=1$, you can deduce that $\zeta\upsilon=1$. Taking coordinates, you can see that $\zeta \geq 0$ so that $\zeta=1$, QED.

2. The proof is actually very similar to 1: let $z$ be a non-unit, non associate divisor of $2$, then write $2=uz$ with $u$ a non-unit. Then $(z\overline{z})(u\overline{u})=(zu)\overline{zu}=2\cdot 2=4$, and $\zeta=z\overline{z},\upsilon=u\overline{u}$ are positive integers. Moreover, if $\zeta=1$, this means that $z$ is a unit (since then $\overline{z}=z^{-1} \in \mathbb{Z}[\theta]$, which is assumed false. Similarly, if $\zeta=4$, then $\upsilon=1$ and $u$ is thus a unit so that $z$ and $2$ are associated, which we assumed was also false. Hence $\zeta=2$.

Answer 3

From a Galois-theory perspective, $\mathbb{Q}(\sqrt{-19})$ is an extension of degree $2$ of $\mathbb{Q}$, thus there are only $2$ linearly independent immersions of $\mathbb{Q}(\sqrt{-19})$ into the algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$ in $\mathbb{C}$: the identity $\sigma_0$ and the one which changes the sign of $\sqrt{-19}$: $\sigma_1$. This happens because the minimal polynomial of $\sqrt{-19}$ has $\pm\sqrt{-19}$ as roots. Now you can define a norm: $$N:\mathbb{Q}(\sqrt{-19})\longrightarrow \mathbb{Q}$$ by $N(\alpha)=\sigma_0(\alpha)\sigma_1(\alpha)$ which is basically the multiplication of $\alpha$ by its conjugate: $\bar{\alpha}$ (remember $\sigma_1$ changes the sign of $\sqrt{-19}$). As you can see $N$ is multiplicative and it can be restricted to the ring of integers of $\mathbb{Q}(\sqrt{-19})$: $\mathbb{Z}[\theta]$ as $-19\equiv1$ $mod\hspace{1mm} 4$. Notice that on such ring $N$ is integer valued: its image is in $\mathbb{N}$ and clearly $N(1)=1$. Thus for any unit $u$ it holds: $$1=N(1)=N(uu^{-1})=N(u)N(u^{-1})$$ Hence $N(u)=1$. Since $N$ multiplies its arguments by their conjugate, $u\bar{u}=1$.

Answer 4

Here is an approach to finding the units that does not use complex conjugates or norms.

Suppose $z=a+b\theta$ where $a$ and $b$ are integers and $\theta$ is your generator $(1+\sqrt{-19})/2$. By definition, if $z$ is a unit then there must be an inverse $z^{-1}=c+d\theta$ where $c$ and $d$ are also integers and $zz^{-1}=1$.

We may determine the required values of $c$ and $d$ in terms of $a$ and $b$. To wit,

$zz^{-1}=ac+(bc+ad)\theta+bd\theta^2=1$

For $\theta=(1+\sqrt{-19})/2$ we have $\theta^2-\theta+5=0$, thus $\theta^2=\theta-5$. Plugging that into the above yields

$(ac-5bd)+[bc+(a+b)d]\theta=1$

and thus a linear system of equations for $c$ and $d$:

$ac-5bd=1$

$bc+(a+b)d=0$

This may be solved for $c$ and $d$ by the usual methods for solving linear systems yielding the following:

$c=\dfrac{a+b}{a^2+ab+5b^2}=\dfrac{a+b}{(a+\frac12b)^2+\frac{19}4b^2}$

$d=-\dfrac{b}{a^2+ab+5b^2}=-\dfrac{b}{(a+\frac12b)^2+\frac{19}4b^2}$

where we complete the square in the $a^2+ab$ terms and thus render the common denominator as a sum of two nonnegative quantities.

In the field $\mathbb Q[\theta]$ this gives a well-defined inverse $z^{-1}=c+d\theta$ except for the case $(a,b)=(0,0)$ which corresponds to $z=0$. In the more limited domain $\mathbb Z[\theta]$ we require $c$ and $d$ to be integers. But then if $b$ is nonzero (thus making $d$ also nonzero) we need

$b^2\ge(a+\frac12b)^2+\frac{19}4b^2\ge\frac{19}4b^2$

which is a contradiction for all nonzero $b$. Thus $b$ must be $0$, forcing $a\not=0$ and then $c=1/a$. The latter is then an integer only for $a=\pm1$ making those values the only units.

The contradiction that excludes all nonzero values of $b$ arose from the relatively large coefficient of $b^2$ in the denominator that emerged from the linear system, which in turn came from the relatively large absolute value $\sqrt{19}/2$ we chose for the imaginary part of $\theta$. What if we tried an absolutely smaller imaginary part? If we had chosen $\theta=(1+\sqrt{-3})/2$ instead of $\theta=(1+\sqrt{-19})/2$, then we would have found that $d$ could be an integer for $b=\pm1$ as well as $b=0$, thus giving $\mathbb Z[(1+\sqrt{-3})/2]$ its additional "primitive" or "nontrivial" units besides $\pm 1$.