330 VIII The Geometry of Numbers
Choosing somet′> 2 /δ,wefindanewsetofintegersq 1 ′,...,qm′,p′ 1 ,...,p′nsat-
isfying the same requirements withtreplaced byt′, and hence withδ′≤ 1 /t′ <
δ/2. Proceeding in this way, we obtain a sequence of(m+n)-tuples of integers
q( 1 v),...,qm(v),p( 1 v),...,p(nv)for whichδ(v)→0 and hence‖q(v)‖→∞,sincewe
cannot haveq(v)=qfor infinitely manyv.
The hypothesis of the corollary is certainly satisfied if 1,αj 1 ,...,αjmare linearly
independent over the fieldQof rational numbers for somej∈{ 1 ,...,n}.
Minkowski also used his lattice point theorem to give the first proof that the dis-
criminant of any algebraic number field, other thanQ, has absolute value greater
than 1. The proof is given in most books on algebraic number theory.
2 Lattices....................................................
In the previous section we defined the set of lattice points to beZn. However, this de-
finition is tied to a particular coordinate system inRn. It is useful to consider lattices
from a more intrinsic point of view. The key property is ‘discreteness’.
With vector addition as the group operation,Rnis an abelian group. A subgroupΛ
is said to bediscreteif there exists a ball with centreOwhich contains no other point
ofΛ. (More generally, a subgroupHof a topological groupGis said to be discrete if
there exists an open setU⊆Gsuch thatH∩U={e},whereeis the identity element
ofG.)
IfΛis a discrete subgroup ofRn, then any bounded subset ofRncontains at most
finitely many points ofΛsince, if there were infinitely many, they would have an
accumulation point and their differences would accumulate atO. In particular,Λis a
closed subset ofRn.
Proposition 6If x 1 ,...,xmare linearly independent vectors inRn, then the set
Λ={ζ 1 x 1 +···+ζmxm:ζ 1 ,...,ζm∈Z}is a discrete subgroup ofRn.
Proof It is clear thatΛis a subgroup ofRn,sincex,y∈Λimpliesx−y∈Λ.IfΛ
is not discrete, then there existy(v)∈Λwith|y(^1 )|>|y(^2 )|>···and|y(v)|→0as
v→∞.LetVbe the vector subspace ofRnwith basisx 1 ,...,xmand for any vector
x=α 1 x 1 +···+αmxm,whereαk∈R( 1 ≤k≤m), put
‖x‖=max(|α 1 |,...,|αm|).This defines a norm onV.Wehave
y(v)=ζ 1 (v)x 1 +···+ζm(v)xm,whereζk(v)∈Z( 1 ≤k≤m). Since any two norms on a finite-dimensional vector
space are equivalent (Lemma VI.7), it follows thatζk(v)→0asv→∞( 1 ≤k≤m).
Sinceζk(v)is an integer, this is only possible ify(v)=Ofor all largev,whichisa
contradiction.