# Computing a set of coset representatives for $\mathbb{Z}^n / \Lambda$

Let $\Lambda$ be an $n$ dimensional sublattice of the integer lattice $\mathbb{Z}^n$. The quotient $\mathbb{Z}^n/\Lambda$ has order $\sqrt{\det{\Lambda}}$.

What is the best/standard way to compute a set of coset representatives for this quotient?

Edit: I initially forgot to take the square root of $\det{\Lambda}$, which is likely the reason for KCronrad's initial comment.

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Pedantic point: the quotient has order the absolute value of the determinant. – KConrad May 12 '10 at 3:02
Hi, Robby. If you have a Z-basis for the lattice, can you use a "lower-left" rule on the fundamental parallelopiped spanned by the basis? Meaning the points of Z^n internal, then on lower left faces. I don't know, I just made it up. – Will Jagy May 12 '10 at 3:02
One thing you might try to do is find a basis e_1,...,e_n of Z^n and positive integers a_1,...,a_n such that a_1e_1,...,a_ne_n is a basis of your lattice. Then Z^n/Lambda is represented by sums c_1e_1 + ... + c_ne_n with c_i running from 0 to a_i - 1. A suitable normal form associated to any matrix whose columns are a known basis of the lattice should let you read off what the a_i's (and e_i's?) are. – KConrad May 12 '10 at 3:04
It might help if you explain how you are actually being "given" the lattice: as the solution space to a system of linear equations, as the dual to some other lattice,... – KConrad May 12 '10 at 3:07
While (as Keith suggests) you can use the Smith normal form, you can also use the Hermite normal form. Find (using integer row operations) a generator matrix for $\Lambda$ which is upper triangular. If the diagonal entries are $d_1,\dots,d_n$ then coset reps are the $\sum a_i e_i$ where $0\le a_i < | d_i|$. – Robin Chapman May 12 '10 at 7:01

## 1 Answer

As KConrad suggested (why only in the comments?), Smith's normal formal is your best bet. Its running time is insensitive to $m=|{\mathbb Z}^n/\Lambda |$ (unless you need to use arbitrary long entries in your matrix) and behaves as $n^3$.

You may also try coset enumeration, whose running time is usually unbounded but may be bounded in this case by something like $(mn)^2$.

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Well, I guess you get the brownie points then :) – Robby McKilliam May 12 '10 at 9:18