Number Theory: An Introduction to Mathematics

2 Lattices 331

The converse of Proposition 6 is also valid. In fact we will prove a sharper result:

Proposition 7IfΛis a discrete subgroup ofRn, then there exist linearly independent
vectors x 1 ,...,xminRnsuch that

Λ={ζ 1 x 1 +···+ζmxm:ζ 1 ,...,ζm∈Z}.

Furthermore, if y 1 ,...,ymis any maximal set of linearly independent vectors inΛ,
we can choose x 1 ,...,xmso that

Λ∩〈y 1 ,...,yk〉={ζ 1 x 1 +···+ζkxk:ζ 1 ,...,ζk∈Z} ( 1 ≤k≤m),

where〈Y〉denotes the vector subspace generated by the set Y.

Proof LetS 1 denote the set of allα 1 >0 such thatα 1 y 1 ∈Λand letμ 1 be the infi-
mum of allα 1 ∈S 1. We are going to show thatμ 1 ∈S 1. If this is not the case there
existα 1 (v)∈S 1 withα 1 (^1 )>α( 12 )>···andα 1 (v)→μ 1 asv→∞. Since the ball
|x|≤( 1 +μ 1 )|y 1 |contains only finitely many points ofΛ, this is a contradiction.
Anyα 1 ∈S 1 can be written in the formα 1 =pμ 1 +θ,wherepis a positive integer
and 0≤θ<μ 1 .Sinceθ>0wouldimplyθ∈S 1 , contrary to the definition ofμ 1 ,
we must haveθ=0. Hence if we putx 1 =μ 1 y 1 ,then

Λ∩〈y 1 〉={ζ 1 x 1 :ζ 1 ∈Z}.

Assume that, for some positive integerk( 1 ≤ k < m), we have found vectors
x 1 ,...,xk∈Λsuch that

Λ∩〈y 1 ,...,yk〉={ζ 1 x 1 +···+ζkxk:ζ 1 ,...,ζk∈Z}.

We will prove the proposition by showing that this assumption continues to hold when
kis replaced byk+1.
Anyx∈Λ∩〈y 1 ,...,yk+ 1 〉has the form

x=α 1 x 1 +···+αkxk+αk+ 1 yk+ 1 ,

whereα 1 ,...,αk+ 1 ∈R.LetSk+ 1 denote the set of allαk+ 1 >0 which arise in such
representations and letμk+ 1 be the infimum of allαk+ 1 ∈Sk+ 1. We are going to show
thatμk+ 1 ∈Sk+ 1 .Ifμk+ 1 ∈/Sk+ 1 ,thereexistα(kv+) 1 ∈Sk+ 1 withαk(^1 +) 1 >αk(^2 +) 1 >···

andα(kv+) 1 →μk+ 1 asv→∞.ThenΛcontains a point

x(v)=α 1 (v)x 1 +···+αk(v)xk+αk(v+) 1 yk+ 1 ,

whereα(jv)∈R( 1 ≤j≤k). In fact, by subtracting an integral linear combination of

x 1 ,...,xkwe may assume that 0≤α(jv)< 1 ( 1 ≤j≤k). Since only finitely many
points ofΛare contained in the ball|x|≤|x 1 |+···+|xk|+( 1 +μk+ 1 )|yk+ 1 |,this
is a contradiction.
Henceμk+ 1 >0andΛcontains a vector

xk+ 1 =α 1 x 1 +···+αkxk+μk+ 1 yk+ 1.