# Showing a quotient $\mathbb{Z}$ module is free

In Fulton's "Introduction to Toric Varieties" he repeatedly uses the following fact.

Let $\sigma$ be a strongly convex rational polyhedral cone in a lattice $N$ and let $N_{\sigma}$ be the subgroup generated by the elements in $\sigma\cap N,$ so $N_{\sigma} = (\sigma \cap N) + (-\sigma \cap N).$ Then $N/N_{\sigma}$ is a lattice (finite rank free $\mathbb{Z}$-module).

I suspect one shows that $N_{\sigma}$ is a saturated subgroup of $N,$ which means the following: If $u\in N$ and $m\in \mathbb{Z}$ are such that $mu\in N_{\sigma},$ then $u\in N_{\sigma}.$ Showing this implies $N/N_{\sigma}$ is torsion free and hence free. But I still can't prove the result.

Here's a failed attempt:

If $np\in N_{\sigma}$ then $np=s_1 - s_2$ for some $s_i\in \sigma\cap N,$ and hence $p = m^{-1}s_1 - m^{-1} s_2.$ Sure, $p$ is in $\sigma$ and $N,$ $m^{-1}s_i$ are in $\sigma$ and $s_i\in N,$ but we also need to ensure $m^{-1}s_i$ are in $N$ if we want to deduce the required result this way. In fact it does not even appear to be true unless the $s_i$ are primitive (the components have gcd 1).

I'm not seeing a solution to this. Please help me.

Let us be completely clear about what you need. You want to show that $N/N_{\sigma}$ is torsion free. The elements of $N/N_{\sigma}$ are the cosets $u+N_{\sigma}$ where $u$ is in $N.$ The action of $n\in \mathbb{Z}$ on $u+N_{\sigma}$ sends it to $nu + N_{\sigma}.$ A torsion element is an element of $N/N_{\sigma}$ such that there is a non-zero $n\in \mathbb{Z}$ such that $n(u+N_{\sigma}) = nu + N_{\sigma}$ is the zero element of $N/N_{\sigma}.$ That is true if and only if $nu \in N_{\sigma}.$ So you want to show that $nu \in N_{\sigma}$ implies $nu\in N_{\sigma}.$ You've shown that you know why $u$ is in $\sigma$ or $-\sigma.$ Now, from the start, we know $u\in N.$ So $u\in \sigma \cap N$ or $u\in -\sigma \cap N.$ Hence $u\in (\sigma \cap N) + (-\sigma \cap N)= N_{\sigma},$ which is what you want.
I don't think we speak of saturated cones $\sigma$, but saturated semi-groups. The statement is that $N \cap \sigma$ is saturated, which is clear by definition of $\sigma$ ($\sigma$ is rational cone, and so if $np \in \sigma \cap N$, for some $p \in N$, then $np=\sum c_i\rho_i$, where $c_i \geq 0$ and $\rho_i$ are the cone generators. But then $p=\sum (\frac{c_i}{n}) \rho_i$, so $p \in \sigma$). Now, this implies that $N_\sigma$ is saturated as a subgroup as well.
So let $\bar{q}$ represent any torsion element in $N/N_\sigma$. Then $nq \in N_\sigma$ for some $n$. But $N_\sigma$ is saturated, so $q \in N_\sigma$. I.e. $\bar q = 0$ in $N/N_\sigma$. So $N/N_\sigma$ is torsion free.
• I don't see why $N_{\sigma}$ is saturated. You've shown that if $np\in \sigma\cap N$ then $p\in \sigma,$ but is it true that $p \in N ?$ Oct 24, 2013 at 4:25
• @Katie: Yes, because we suppose $p \in N$ from the start. The definition of a saturated semigroup $S$ is: if $n \cdot x \in S$ for $x \in S$, then $x \in S$. Oct 24, 2013 at 7:08