proof of Minkowski’s bound


The proof of Minkowski’s bound will rely on Minkowski’s lattice point theorem (http://planetmath.org/MinkowskisTheorem), but we first need to establish some lemmas.

Lemma 1.

Let M be a real number and suppose that for every non-zero ideal a of the ring of integers OK there exists a non-zero xa with norm N(x)MN(a).

Then, every ideal classMathworldPlanetmath of OK has a representative a satisfying N(a)M.

Proof.

Let [𝔟] be an ideal class represented by the ideal 𝔟. Choosing a non-zero x𝔟 then x𝔟-1 is an ideal of 𝒪K and, by the condition of the lemma, contains a non-zero y satisfying N(y)MN(x𝔟-1). Then, 𝔞x-1y𝔟 is an ideal representing [𝔟] and N(𝔞)=N(y)/N(x𝔟-1)M. ∎

If the real embeddings of K are denoted by σk:K (k=1,,r1) and the complex embeddings are τk:K together with their complex conjugates τ¯k (k=1,,r2), then we define

j:Kr1×r2,
j(x)=(σ1(x),,σr1(x),τ1(x),,τr2(x)).

Also note that r1×r2 is isomorphicPlanetmathPlanetmathPlanetmathPlanetmath as a real vector space to r1+2r2=n given by the isomorphismPlanetmathPlanetmath

f:r1×r2n,
f(x1,,xr1,y1,,yr2)=(x1,,xr1,(y1),,(yr2),(y1),,(yr2)).

As f and j are linear maps (with respect to the field of rationals ), the combinationMathworldPlanetmath fj gives a -linear map from K to n. The image will be a lattice, and we can compute its volume.

Lemma 2.

If a is a non-zero ideal of OK, then Γ=fj(a) is a lattice in Rn (http://planetmath.org/LatticeInMathbbRn). Its fundamental mesh has volume

vol(Γ)=2-r2|DK|N(𝔞).
Proof.

The proof of this is to be added. ∎

Lemma 3.

For any L>0, let S be the set in Rr1×Cr2 consisting of points (x1,,xr1,y1,yr2) satisfying

k=1r1|xk|+2k=1r2|yk|L.

Then, f(S) has volume (2r1-r2πr2/n!)Ln.

Proof.

The proof of this is to be added. ∎

Proof of Minkowski’s bound

For an ideal 𝔞 and any constant b>1, let L>0 be given by

2r1-r2πr2n!Ln=2nb2-r2|DK|N(𝔞).

Letting S be the set given in Lemma 3 and Γ=fj(𝔞), Lemmas 2 and 3 give vol(S)>2nvol(Γ). As S is convex and symmetric about the origin, Minkowski’s theorem tells us that there is a non-zero x𝔞 with fj(x)S.

As the geometric mean is always bounded above by the arithmetic mean, we get the inequality

N(x)=k=1r1|σk(x)|k=1r2|τk(x)|2n-n(k=1r1|σk(x)|+2k=1r2|τk(x)|)nn-nLn=bMK|DK|N(𝔞)

where MK=(n!/nn)(4/π)r2. If we choose b such that bMK|DK|N(𝔞) is less than the smallest integer greater than MK|DK|N(𝔞), then this gives N(x)MK|DK|N(𝔞) and Minkowski’s bound follows from Lemma 1.

Title proof of Minkowski’s bound
Canonical name ProofOfMinkowskisBound
Date of creation 2013-03-22 18:33:41
Last modified on 2013-03-22 18:33:41
Owner gel (22282)
Last modified by gel (22282)
Numerical id 5
Author gel (22282)
Entry type Proof
Classification msc 11R29
Classification msc 11H06
Related topic MinkowskisTheorem
Related topic MinkowskisConstant
Related topic IdealClass