Geometry of Numbers

2 Finiteness of class group

Consider the \(r\) real embeddings of a number field \(K\), and the \(s\) complex conjugate pair embeddings. The product of these is an embedding \(K \to \RR ^r\times \CC ^s\). \(\cO _K\) is a free \(\ZZ \)-module of rank \(r+2s = n\).

Lemma 2.1. \(\cO _K\) is a lattice in \(\RR ^r\times \CC ^s\) of volume \(2^{-s}\sqrt {|\disc (\cO _K)|}\).

Proof. Given a basis of \(\cO _K\), the discriminant is the determinant of the matrix of the conjugates of the basis. Note that \(\real (z) = \frac {z+\bar {z}}2\) and similarly \(\imag (z) = \frac {z-\bar {z}}2\), so that we get that the square root of the absolute value of the discriminant of \(\cO _K\) is \(2^{s}A(O_K)\). □

Lemma 2.2. The region in \((\sigma ^\RR _i,\sigma ^\CC _j)=\RR ^r \times \CC ^s\) where \(\sum |\sigma ^\RR _i|+2\sum |\sigma ^\CC _j| \leq x\) has measure \((\frac {\pi }{2})^s2^r\frac {x^{n}}{n!}\)

Proof. By scaling it is clear that the area should be \(cx^{r+2s}\), where \(c\) is some constant. Adding one more real component gives area \(\int _0^xcy^{r+2s}2dy = \frac {2c}{r+2s+1}x^{(r+1)+2s}\). Adding a complex component gives area \(\int _0^{\frac {x}{2}}c(x-2y)^{r+2s}2\pi (\frac x 2 -y)dy\). Integrating by parts yields \(\int _0^{\frac x 2} c(x-2y)^{r+2s} =\frac {\pi }{2} \frac {cx^{r+2s+2}}{(r+2s+1)(r+2s+2)}\). By induction, we see that the measure is \((\frac {\pi }{2})^s2^{r}\frac {x^{r+2s}}{(r+2s)!}\). □

Corollary 2.3. Given a non-zero fractional ideal \(I\) in \(\cO _K\), there is a non-zero element of norm at most \(2^{r+s}\sqrt {|\disc K|}N(I)\).

Proof. Consider the convex symmetric region \(\sum _1^r|\sigma ^\RR _i| + \sum _1^s2|\sigma ^\CC _i| \leq x\), where \(x^n = \frac {2^{n-r}n!}{\pi ^s}\sqrt {|\disc K|}N(I)\). It is compact and has area \(2^n\) times the area of \(N(I)\), so there is a nonzero lattice point \(x\). Then by AM-GM, \(N(x) \leq (\frac {\sum _1^r|\sigma ^\RR _i| + \sum _1^s2|\sigma ^\CC _i|}{n})^n = (\frac {4}{\pi })^s\frac {n!}{n^n}\sqrt {|\disc K|} N(I)\). □

Theorem 2.4. Every ideal class contains an ideal of norm \(\leq (\frac {4}{\pi })^s\frac {n!}{n^n}\sqrt {|\disc K|}\).

Proof. Choose an element \(x\) satisfying the norm bound in the previous result. Then \((x)I^{-1}\) is a proper ideal in the inverse class with the desired norm bound. □

Corollary 2.5. The class group is finite.

Proof. There are finitely many ideals of a given norm. □