Exercise A0.V.2.18

Let $\delta = (1 + \sqrt{-19})/2$ and let $R = \mathbb{Z}[\delta].$ Define norm $N(a + b\sqrt{-19}) = a^2 + 19b^2.$

Lemma L2.2

$N(a + b\delta) = a^2 + ab + 5b^2$

Proof

$$\begin{align*} N(a + b\delta) &= N\left(a + b\frac{1+\sqrt{-19}}{2}\right) \\ &= N\left(\left(a + \frac{b}{2}\right) + \frac{b}{2}\sqrt{-19}\right) \\ &= \left(a + \frac{b}{2}\right)^2 + 19\left(\frac{b}{2}\right)^2 \\ &= a^2 + ab + \frac{b^2}{4} + \frac{19b^2}{4} \\ &= a^2 + ab + 5b^2 \\ \end{align*}$$

$\blacksquare$

Proposition A

The $4$ least values taken by $N(a + b\delta)$ are $0, 1, 4,$ and $5.$ Moreover, $N(a + b\delta) \geq 5$ if $b \neq 0.$

Proof

Let $f(a,b) = a^2 + ab + 5b^2.$ For a fixed $b,$ the extremum of this parabola is at $a = -b/2.$ We use this to find a lower bound for the value of $f(a,b).$

$$\begin{align*} f\left(-\frac{b}{2}, b\right) &= \left(-\frac{b}{2}\right)^2 + \left(-\frac{b}{2}\right)b + 5b^2 \\[1em] &= \frac{b^2}{4} - \frac{2b^2}{4} + \frac{20b^2}{4}\\[1em] &= \frac{19b^2}{4} \end{align*}$$

Now, if $b = \pm 1,$ then $f(a,b)$ is at least $4.75,$ and since $f(a,b)$ is an integer for integer values of $a,$ it must be greater than or equal to $5.$

Observe that the lower bound grows monotonically with $|b|.$ This forces $f(a,b) \geq 5$ whenever $|b| \geq 1$ and $a,b \in \mathbb{Z}.$ Hence, $N(a + b\delta) \geq 5$ whenever $a,b \in \mathbb{Z}$ and $b \neq 0.$

We observe that $f(a, 0) = a^2,$ and the minimal integer values of $a^2$ are $0, 1, 4, 9, \ldots \ .$

Observe also that $f(1,-1) = 5,$ attaining the lower bound for non-zero integer $b.$

Combining our results, we see that indeed, the minimal values attained by $f(a,b)$ with integer arguments (and hence by $N(a + b\delta)$) are $0,1,4,5.$

$\blacksquare$

Proposition B

The units of $R$ are $\pm 1.$

Proof

Analogous to the arguments given in 1.17A3 and 1.17A4.

$\blacksquare$

Proposition C

Suppose there exists a non-zero non-unit element $c \in R$ such that every $a \in R$ can be written as $a = qc + r$ for some $q, r \in R$ with $r = 0$ or $r \in R^{\times}.$ Then $c$ divides $2$ or $3,$ and thus is equal to $\pm 2$ or $\pm 3.$

Proof

Consider $2 = qc + r,$ which can be written as $2 - r = qc.$ If $r = 0,$ then $c \mid 2.$ If $r = -1,$ then $c \mid 3.$ Since $c$ is not a unit, it cannot be the case that $r = 1,$ as that would force $1 = qc.$

Observe that if $4 = N(2) = N(qc) = N(q)N(c),$ then $N(q) = 1$ necessarily.

Observe that if $9 = N(3) = N(qc) = N(q)N(c),$ but $N(x)$ cannot be equal to $3,$ so $N(c) = 9,$ and hence $N(q) = 1.$

This forces $c \in \{\pm 2, \pm 3\}$

$\blacksquare$

Proposition D

There does not exist $q \in R$ such that $\delta = qc + r$ with $c \in \{\pm 2, \pm 3\}$ and $r \in \{0, \pm 1\}.$

Proof

Let $q = a + \delta b.$ Then $N(qc) = c^2 \cdot (a^2 + ab + 5b^2).$ If $c = \pm 2,$ then $N(qc)$ is a multiple of $4.$ If $c = \pm 3,$ then $N(qc)$ is a multiple of $9.$

Observe that $N(\delta) = N(\delta - 1) = 5,$ $N(\delta + 1) = 7.$

So, if $\delta = qc + r$ held, then we would have $\delta - r = qc,$ which requires $N(\delta - r) = N(qc).$ But $N(\delta - r) \in \{5,7\}$ is prime, while $N(qc)$ is composite. This is an impossibility.

$\blacksquare$

Proposition E

$R$ is not a Euclidean domain.

Proof

We know from 2.17 that a Euclidean domain must have an element $c$ such that the remainder of Euclidean division by $c$ is either a unit or $0.$ We have demonstrated in Proposition C and Proposition D that no such element exists in $R.$ Therefore, $R$ is not a Euclidean domain.

$\blacksquare$