## 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 $$m < 0$$ 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 $$(d \alpha^3)^2 = 2 d^2 \sqrt{2} \ne \sqrt{3}$$.

Ben Lynn blynn@cs.stanford.edu 💡