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Let $\Lambda \subseteq \mathbb{Z}^n$ be a full rank sublattice. We find an upper triangular basis matrix $B \in \mathfrak{ut}(\mathbb{Z},n)$ of $\Lambda$. Is $B$ unique up to the right action of $\mathfrak{ut}(\mathbb{Z},n) \cap GL(\mathbb{Z},n)$ (unimodular upper triangular matrices)?

EDIT: my aim was to define a binary operation $$\cap :\mathfrak{ut}(\mathbb{Z},n) \times \mathfrak{ut}(\mathbb{Z},n) \to \mathfrak{ut}(\mathbb{Z},n) $$ corresponding to the intersection of lattices. It should be well-defined if we mod out the above action. By using Bezout, in $2$ dimensions I figured out the following very ugly formula for the representative:

$$A\cap B= \begin{pmatrix} lcm(a_{11},b_{11}) & lcm(a_{11},b_{11})(1-xy) \\ 0 & x \end{pmatrix},$$ where $$x=\frac{a_{22}b_{22}gcd(a_{11},b_{11})}{gcd(gcd(a_{11},b_{11})gcd(a_{22},b_{22}),a_{22}b_{12}-a_{12}b_{22})}\\ y=\frac{a_{12}b_{12}}{lcm(lcm(a_{11},a_{12})a_{22}b_{12},lcm(b_{11},b_{12})b_{22}a_{12})}$$ Do you see anything familiar here? I would like to generalize this, but I have no idea how to start for higher dimensions.

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    $\begingroup$ I believe that this follows from the existence and uniqueness of Hermite normal form. $\endgroup$
    – Richard Stanley
    Commented Sep 8, 2016 at 21:08
  • $\begingroup$ that's right, thx! $\endgroup$
    – Bipolar Minds
    Commented Sep 13, 2016 at 15:23

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Converted from a comment by another user:

I believe that this follows from the existence and uniqueness of Hermite normal form.

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