2

In this question, by a lattice I mean a full-rank subgroup of the group ${\mathbb Z}^2$. What I would like to know is:

Can one give a comprehensible description of those lattices $\Lambda\subset{\mathbb Z}^2$ for which the quotient group ${\mathbb Z}^2/\Lambda$ is cyclic?

It seems that, using a rather awkward argument, I can prove that if $\binom{a}{0}$ and $\binom{b}{c}$ form a reduced basis of $\Lambda$, then ${\mathbb Z}^2/\Lambda$ is cyclic if and only if $\gcd(a,b,c)=1$. Can one give a simple proof or a convincing explanation for this?

In general,

Given two lattices $\Lambda_1<\Lambda_2$, what is the necessary and sufficient condition for $\Lambda_2/\Lambda_1$ to be cyclic?

I would be happy to have the answer in terms of the generating matrices of the lattices involved, or in any other reasonable terms. Thank you!

W-t-P
  • 4,629

1 Answers1

1

By the fundamental theorem for finitely generated modules over a p.i.d., a quotient group of $\mathbf Z^2$ will be cyclic if and only it has only $1$ invariant factor $\neq 1$.

Bernard
  • 175,478
  • Could you give a reference, preferably to a web resource? And (sorry for my ignorance) how do I find practically the invariant factors of ${\mathbb Z}^2/\Lambda$? – W-t-P Apr 09 '15 at 20:15
  • You can look at wikipedia. The effective computation is called Smith normal form. For Z$^2$, it should be fast. To the usual three operations on rows and columns, you have to add an arithmetic operation derived from Bézout' identity. – Bernard Apr 09 '15 at 20:27
  • I know how to find a reduced basis of a lattice, but your answer seems to refer to the invariant factors of the quotient group? In fact, I am afraid I do not quite get the ideology behind your answer: if I "know" the quotient group, then it should not be a problem to decide whether it is cyclic; my problem is to "understand" the quotient group. – W-t-P Apr 10 '15 at 10:01
  • @W-t-P: Your lattice is, I suppose, by a matrix in $M_2(\mathbf Z)$. When this matrix is put in Smith normal form, you have the invariant factors, and you know whether the quotient is cyclic. I'm not sure to grasp what kind of answer you're after. – Bernard Apr 10 '15 at 11:12
  • So, you are looking at the invariant factors of the matrix of the lattice itself, not of its quotient group (as I understood it from your answer). Anyway, I am still a little uncertain. If $\Lambda$ is generated by the two vectors $(2,0)$ and $(0,3)$, then the invariant factors are $2$ and $3$, both distinct from $1$, but the ${\mathbb Z}^2/\Lambda$ is cyclic of order $6$. – W-t-P Apr 10 '15 at 12:43
  • No, they are not the invariant factors, because each invariant factor divides the next one. If I remember well, this lattice is exactly $\mathbf Z^2$. – Bernard Apr 10 '15 at 12:49
  • you are right that $2$ and $3$ are not the invariant factors; I will have to do my homework about the Smith normal form. But the lattice generated by $(2,0)$ and $(0,3)$ is most certainly not the whole ${\mathbb Z}^2$, it has index $6$ in ${\mathbb Z}^2$. – W-t-P Apr 10 '15 at 12:57
  • You're right, I was thinking of $(2,0)$ and $(3,0)$. The invariant factors are $1$ and $6$ (maybe in reverse order, depending on how you describe the algorithm). – Bernard Apr 10 '15 at 13:00