## Trace and Norm: Applications

Let $R$ be the number ring corresponding the the number field $K$. Then it can be easily shown that every unit $u$ satisfies $N(u) = \pm 1$. Also, if $a \in R$ satisfies $N(a) = 1$, then $1 = n(a)^{n/d} = a(a^{n/d - 1}(\sigma_2(a)...\sigma_n(a))^{n/d})$ thus $a$ is indeed factor of unity in $R$ since all conjugates of $a$ must also be algebraic integers.

Hence an element $a \in R$ is a unit if and only if $N(a) = {\pm 1}$. For example, for squarefree $m \lt 0$ we have that the only units in the number ring of $\mathbb{Q}[\sqrt{m}]$ are ${\pm 1}$ if $m \ne {-1},{-3}$. If $m = {-1}$ we also have $i, -i$, and if $m = {-3}$, we have the primitive sixth roots of unity. On the other hand $\mathbb{Z}[\sqrt{2}]$ has infinitely many units generated by the powers of $(1+\sqrt{2})$ (which correspond to the solutions of $a^2 - 2b^2 = {\pm 1}$ over the integers).

For $a \in R$, if $N(a)$ is prime (in $\mathbb{Z}$) then clearly $a$ is irreducible. For example, $9 + \sqrt{10}$ is irreducible in $\mathbb{Z}[\sqrt{10}]$.

Note $x^2 + 5 y^2$ cannot be 2 or 3 if $x,y$ are integers so all elements of $\mathbb{Z}[\sqrt{-5}]$ of norm 6 are irreducible. Hence in $\mathbb{Z}[\sqrt{-5}]$ we have $2\cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})$ as an example of nonunique factorization.

The trace can be used to show that certain elements are not contained in certain fields. For example, consider the field $\mathbb{Q}[\alpha]$ where $\alpha = \root{4}{2}$. We wish to determine if $\sqrt{3}$ is contained in this field. If it is, we have the equation $\sqrt{3} = a + b \alpha + c \alpha^2 + d \alpha^3$. Now $T(\sqrt{3}) = T(\alpha) = T(\alpha^2) = T(\alpha^3) = 0$, and $T(a) = 4a$, thus we must have $a=0$. Dividing both sides of the equation by $\alpha$ gives $\root{4}{9/2} = b + c \alpha + d \alpha + a(\alpha^3)/2$, and taking traces now shows $b = 0$. Similarly, dividing by $\alpha$ again shows $c = 0$. We can divide again to show $d = 0$ and thus obtain a contradiction, or simply say that $(d \alpha^3)^2 = 2 d^2 \sqrt{2} \ne \sqrt{3}$.