Number Fields

Let $K$ be a number field of degree $n$ over $\mathbb{Q}$. Let ${\sigma }_1,...,{\sigma }_n$ be the embeddings of $K$ in $ℂ$. We write $[a_{ij}]$ for the matrix with $a_{ij}$ in the $i$th row and $j$ column, and $\mid a_{ij}\mid$ for the determinant of this matrix. Define the discriminant of ${\alpha }_1,...,{\alpha }_n\in K$ by

$$\mathrm{disc}({\alpha }_1,...,{\alpha }_n)=\mid {\sigma }_i({\alpha }_j){\mid }^2$$

As with the trace and norm, the discriminant may be generalized by replacing $\mathbb{Q}$ with any number field.

Theorem: $\mathrm{disc}({\alpha }_1,...,{\alpha }_n)=\mid T({\alpha }_i{\alpha }_j)\mid$

Proof: Recall some basic facts about determinants:

$$\mid a_{ij}\mid =\mid a_{ji}\mid ,\mid AB\mid =\mid A\mid \mid B\mid$$

Then the theorem follows from

$$[{\sigma }_j({\alpha }_i)][{\sigma }_i({\alpha }_j)]=[{\sigma }_1({\alpha }_i{\alpha }_j)+...+{\sigma }_n({\alpha }_i{\alpha }_j)]=[T({\alpha }_i{\alpha }_j)]$$

Corollary: $\mathrm{disc}({\alpha }_1,...,{\alpha }_n)\in \mathbb{Q}$, and if the ${\alpha }_i$ are integral, then $\mathrm{disc}({\alpha }_1,...,{\alpha }_n)\in \mathbb{Z}$.

Theorem: $\mathrm{disc}({\alpha }_1,...,{\alpha }_n)=0$ if and only if ${\alpha }_1,...,{\alpha }_n$ are linearly dependent over $\mathbb{Q}$.

Proof: If the ${\alpha }_j$ are linearly dependent over $\mathbb{Q}$ then so are the columns of the matrix $[{\sigma }_i({\alpha }_j)]$ and its determinant will be zero. Conversely if $\mathrm{disc}({\alpha }_1,...,{\alpha }_n)=0$, then the rows $R_i$ of the matrix $[T({\alpha }_i{\alpha }_j)]$ are linearly dependent. Suppose the ${\alpha }_i$ are linearly independent. Then let $a_1,..,a_n$ be rationals, not all zero, such that $a_1{R}_1+...+a_n{R}_n=0$. Then consider $\alpha =a_1{\alpha }_1+...+a_n{\alpha }_n\ne 0$. Note we have $T(\alpha {\alpha }_j)=0$ for all $j$. The ${\alpha }_j$ form a basis for $K$ over $\mathbb{Q}$, and since $\alpha \ne 0$, so do the $\alpha {\alpha }_j$. But this implies $T(\beta )=0$ for all $\beta \in K$, a contradiction since there exist elements of nonzero trace (for example $T(1)=n$).

Theorem: Let $K=\mathbb{Q}[\alpha ]$ and let ${\alpha }_1,...,{\alpha }_n$ be the conjugates of $\alpha$ over $\mathbb{Q}$. Let $f$ be the minimal polynomial of $\alpha$. Then

$$\mathrm{disc}(1,\alpha ,...,{\alpha }^{n-1})=\prod _{1\le r<s\le n}({\alpha }_r-{\alpha }_s{)}^2=\pm N^K(f\prime (\alpha ))$$

where the sign is positive if and only if $n=\mathrm{0,1}(\mathrm{mod}4)$.

Proof: First observe $\mid {\sigma }_i({\alpha }^{j-1})\mid =\mid ({\sigma }_i(\alpha ){)}^{j-1}\mid =\mid {\alpha }_i^{j-1}\mid$ is a Vandermonde determinant, yielding the first equality.

Next we have

$$\prod _{r<s}({\alpha }_r-{\alpha }_s{)}^2=\pm \prod _{r\ne s}({\alpha }_r-{\alpha }_s)$$

where the plus sign holds if and only if $n=\mathrm{0,1}(\mathrm{mod}4)$. Since $f\prime$ has rational coefficients, we have

$$N^K(f\prime (\alpha ))=\prod _{r=1}^n{\sigma }_r(f\prime (\alpha ))=\prod _{r=1}^{n}f\prime ({\sigma }_r(\alpha ))=\prod _{r=1}^{n}f\prime ({\alpha }_r)$$

Then the second equality follows from the fact that for all $r$

$$f\prime ({\alpha }_r)=\prod _{s\ne r}({\alpha }_r-{\alpha }_s)$$

(To see this, let $f(x)=(x-{\alpha }_r)g(x)$. Then by the product rule, $f\prime ({\alpha }_r)=g({\alpha }_r)$, and the roots of $g$ are all the roots of $f$ except for ${\alpha }_r$.)

Let us now apply this theorem to compute $\mathrm{disc}(1,\omega ,...,{\omega }^{p-2})$ for $\omega =e^{2\pi i/p}$ where $p$ is an odd prime. We have $f(x)=1+x+...+x^{p-1}$. Since $x^p-1=(x-1)f(x)$, we have $p{x}^{p-1}=f(x)+(x-1)f\prime (x)$ showing that

$$f\prime (\omega )=\frac{p}{\omega (\omega -1)}$$

Taking norms gives

$$N(f\prime (\omega ))=\frac{N(p)}{N(\omega )N(\omega -1)}=\frac{p^{p-1}}{N(\omega -1)}$$

Since $(1-\omega )...(1-{\omega }^{p-1})=p$, we have $N(\omega -1)=N(1-\omega )=p$, hence $N(f\prime (\omega ))=p^{p-2}$.

Write $\mathrm{disc}(\alpha )$ for $\mathrm{disc}(1,\alpha ,...,{\alpha }_n)$ where $\alpha$ is a degree $n$ algebraic integer over $\mathbb{Q}$ and $K$ is taken to be $\mathbb{Q}[\alpha ]$.

We now know that $\mathrm{disc}(\omega )=\pm p^{p-2}$ for prime $p$. For $\omega =e^{2\pi i/m}$ for general $m$, the expression is more complicated, but it can be readily shown that $\mathrm{disc}(\omega )\mid m^{\varphi (m)}$ as follows. Let $f$ be the minimal polynomial of $\omega$. Then Since $x^m-1=f(x)g(x)$ for some $g\in \mathbb{Z}[x]$ we have $m=\omega f\prime (\omega )g(\omega )$. Taking norms gives

$$m^{\varphi (m)}=\pm \mathrm{disc}(\omega )N(\omega g(\omega ))$$

and we are done since $N(\omega g(\omega ))\in \mathbb{Z}$.