Number Fields

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 $x^2+5y^2$ 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 $\alpha =\sqrt[4]{2}$. 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 $\sqrt[4]{9/2}=b+c\alpha +d\alpha +a({\alpha }^3)/2$, 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 $(d{\alpha }^3{)}^2=2d^2\sqrt{2}\ne \sqrt{3}$.