Minkowski’s bound D. J. Bernstein 19951102 The signature of a number field K is the pair (r_{1}, r_{2}), where r_{1} is the number of real embeddings of K and r_{2} is the number of conjugate pairs of complex embeddings of K. Let L be a lattice in R^{n}. The determinant of L, written det L, is the volume of any fundamental domain for L. Let S be a convex symmetric region with vol S > 2^{n} det L. Minkowski’s theorem states that S contains a nonzero point of L. Theorem. Let K be a number field of degree n with signature (r_{1}, r_{2}). Let A be the ring of integers of K. Let I be a nonzero ideal of A. Write

λ = \frac{n!}{n^{n}} (\frac{4}{π} )^{r_{2}} \sqrt{|disc A|}.

Let ϵ be a positive real number. Then I contains a nonzero element α with |norm α| ≤ (1 + ϵ)λ Norm I. Proof. Let σ_{1}, …, σ_{r_{1}}, τ_{1}, \overline{τ_{1}}, …, τ_{r_{2}}, \overline{τ_{r_{2}}} be the embeddings of K into C. Define f : K → R^{n} by f(α) = (σ_{1}(α), …, σ_{r_{1}}(α), Re τ_{1}(α), …, Re τ_{r_{2}}(α), Im τ_{1}(α), …, Im τ_{r_{2}}(α)). We have seen that f(I) = \{f(α) : α ∈ I\} is a lattice in R^{n}, with det f(I) = (1/2^{r_{2}})\sqrt{|disc A|} Norm I. For any nonnegative real number t, define S(t) as the set of (x_{1}, …, x_{r_{1}}, y_{1}, …, y_{r_{2}}, z_{1}, …, z_{r_{2}}) ∈ R^{n} such that |x_{1}| + … + |x_{r_{1}}| + 2\sqrt{y_{1}^{2} + z_{1}^{2}} + … + 2\sqrt{y_{r_{2}}^{2} + z_{r_{2}}^{2}} ≤ t. Three facts about S(t): first, S(t) is convex; second, vol S(t) = (1/n!)2^{r_{1}}(π/2)^{r_{2}}t^{n}; third, if f(α) ∈ S(t) then |norm α| ≤ t^{n}/n^{n}. Now set t = ^{n} \sqrt{(1 + ϵ)n^{n}λ Norm I}. Combining various formulas, we find that vol S(t) > 2^{n} det f(I). By Minkowski’s theorem, S(t) contains a nonzero point of f(I); in other words, f(α) ∈ S(t) for some nonzero α ∈ I. Then |norm α| ≤ t^{n}/n^{n} = (1 + ϵ)λ Norm I as claimed. Theorem. Let K be a number field of degree n with signature (r_{1}, r_{2}). Let A be the ring of integers of K. Let I be a nonzero ideal of A. Write

λ = \frac{n!}{n^{n}} (\frac{4}{π} )^{r_{2}} \sqrt{|disc A|}.

Then I contains a nonzero element α with |norm α| ≤ λ Norm I. Proof. This is a compactness argument: norms are integers, so the set of norms of elements of I is discrete. Write m = min {|norm α| : α ∈ I, α ≠ 0}. Suppose m > λ Norm I. Select ϵ so that m > (1 + ϵ)λ Norm I. By the previous theorem, I contains a nonzero element α with |norm α| ≤ (1 + ϵ)λ Norm I. Contradiction.