ProofComplete

The Class Group - Key Proof

We prove the finiteness of the class group using Minkowski's geometry of numbers, a cornerstone of algebraic number theory.

ProofFiniteness of the Class Group

Let KK be a number field of degree n=r1+2r2n = r_1 + 2r_2 where r1r_1 is the number of real embeddings and r2r_2 is the number of pairs of complex conjugate embeddings.

Step 1: The Minkowski embedding

Define the Minkowski embedding ϕ:KRr1×Cr2\phi: K \to \mathbb{R}^{r_1} \times \mathbb{C}^{r_2} by: ϕ(α)=(σ1(α),,σr1(α),τ1(α),,τr2(α))\phi(\alpha) = (\sigma_1(\alpha), \ldots, \sigma_{r_1}(\alpha), \tau_1(\alpha), \ldots, \tau_{r_2}(\alpha)) where σi\sigma_i are real embeddings and τj\tau_j are representatives of complex conjugate pairs.

Under this embedding, OK\mathcal{O}_K maps to a lattice Λ\Lambda in Rn\mathbb{R}^n (identifying Cr2\mathbb{C}^{r_2} with R2r2\mathbb{R}^{2r_2}).

Step 2: Convex body argument

Consider the region StS_t defined by: St={(x1,,xn)Rr1×Cr2:i=1r1xi+2j=1r2xjt}S_t = \left\{(x_1, \ldots, x_n) \in \mathbb{R}^{r_1} \times \mathbb{C}^{r_2} : \sum_{i=1}^{r_1}|x_i| + 2\sum_{j=1}^{r_2}|x_j| \leq t\right\}

This region is:

  • Convex and symmetric about the origin
  • Has volume V(St)=2r1(2π)r2n!tnV(S_t) = \frac{2^{r_1}(2\pi)^{r_2}}{n!}t^n

Step 3: Minkowski's lattice theorem

Choose tt such that V(St)2nvol(Λ)V(S_t) \geq 2^n \text{vol}(\Lambda), where vol(Λ)=2r2ΔK\text{vol}(\Lambda) = 2^{-r_2}\sqrt{|\Delta_K|}.

By Minkowski's theorem, StS_t contains a nonzero lattice point, i.e., there exists 0αOK0 \neq \alpha \in \mathcal{O}_K with: i=1r1σi(α)j=1r2τj(α)2(n!nn)(4π)r2ΔK\prod_{i=1}^{r_1}|\sigma_i(\alpha)| \cdot \prod_{j=1}^{r_2}|\tau_j(\alpha)|^2 \leq \left(\frac{n!}{n^n}\right)\left(\frac{4}{\pi}\right)^{r_2}\sqrt{|\Delta_K|}

The left side equals NK/Q(α)|N_{K/\mathbb{Q}}(\alpha)|, giving a bound on the norm.

Step 4: Bounding ideal norms

Let a\mathfrak{a} be any nonzero ideal. Choose 0αa0 \neq \alpha \in \mathfrak{a} with norm bounded as above. Then: N(a)N((α))=N(α)MKN(\mathfrak{a}) \leq N((\alpha)) = |N(\alpha)| \leq M_K

where MKM_K is the Minkowski bound.

Step 5: Ideal class representatives

Every ideal class [a][\mathfrak{a}] contains an ideal b\mathfrak{b} with N(b)MKN(\mathfrak{b}) \leq M_K.

For each norm NMKN \leq M_K, there are only finitely many ideals with that norm (they correspond to divisors of (N)(N) in OK\mathcal{O}_K).

Therefore, there are only finitely many ideals of bounded norm, giving finitely many ideal class representatives.

Conclusion: hK=Cl(K)<h_K = |\text{Cl}(K)| < \infty.

ExampleExplicit Bound for Quadratic Fields

For K=Q(d)K = \mathbb{Q}(\sqrt{d}) with dd square-free:

  • n=2n = 2, r1=2r_1 = 2 if d>0d > 0 (real quadratic), r1=0,r2=1r_1 = 0, r_2 = 1 if d<0d < 0 (imaginary quadratic)
  • Discriminant: ΔK=d\Delta_K = d or 4d4d depending on dmod4d \bmod 4

For imaginary quadratic: MK=2πΔKM_K = \frac{2}{\pi}\sqrt{|\Delta_K|}

For real quadratic: MK=12ΔKM_K = \frac{1}{2}\sqrt{|\Delta_K|}

These explicit bounds enable hand computation of class numbers for small discriminants.

Remark

The Minkowski bound is not sharp: often hKh_K is much smaller than the number of ideals below the bound. Improved bounds exist (Bach bound, assuming GRH), useful for computational verification of class number 1.

The proof technique—using geometry of numbers to bound algebraic invariants—appears throughout number theory and is a hallmark of Minkowski's revolutionary approach.