This is about a (seemingly) basic lemma about rational polyhedral cones that is sometimes used when working with toric varieties and is usually "left to the reader". Unfortunately, I could neither prove it myself nor find a complete proof in the literature. So, I am looking for either a proof or a proper reference.



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> **Lemma.** Let $V$ be an $\mathbb{R}$-vector space of finite dimension, and let $N$ be a $\mathbb{Z}$-structure on $V$ (i.e., a free abelian group with $N\otimes\_{\mathbb{Z}}\mathbb{R}=V$). Let $\sigma$ and $\tau$ be $N$-rational polyhedral cones in $V$, and suppose that $\tau$ is a *subset* of $\sigma$. Then, the following statements are equivalent:
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> (i) $\tau$ is a face of $\sigma$;
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> (ii) If $x,y\in\sigma$ with $x+y\in\tau$, then $x,y\in\tau$;
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> (iii) If $x,y\in\sigma\cap N$ with $x+y\in\tau$, then $x,y\in\tau$.
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Showing that (i) and (ii) are equivalent and imply (iii) is clear. My problem is to show that (iii) implies (i) or (ii), i.e., that it suffices to consider only rational points.


**Note 1.** Trying to show (iii)$\Rightarrow$(i) analogously to (ii)$\Rightarrow$(i) leads to the question whether $(\sigma-\tau)\cap N=(\sigma\cap N)-(\tau\cap N)$, which has a negative answer in general.


**Note 2.** Condition (iii) is sometimes expressed by saying that the monoid $\tau\cap N$ is a face of the monoid $\sigma\cap N$, and similarly for (ii).



A characterisation of faces of rational polyhedral cones
To prove that (iii) implies (i), assume w.l.o.g. that $\tau\neq\sigma$. We first need to show that $\tau\subset \partial \sigma$. If this is not the case then either there exists a hyperplane $H$ containing $\tau$ such that $H$ is not a supporting hyperplane for $\sigma$, or $\tau$ is full-dimensional. If $H$ exists then there are $x,y\in\sigma$ on different sides of $H$, which may be chosen in $N$ as $\sigma$ is rational, so that $x+y\in\tau$, contadicting (iii).


If $H$ does not exist, then $\dim \tau=\dim\sigma$, and there exists $x\in\sigma\setminus\tau$ ($x$ may be chosen to be a generator for an extreme ray of $\sigma$) and $y\in\tau\setminus\partial\tau$ (again, it's possible to choose $x,y\in N$) so that $x+y\in\tau$, again contradicting (iii).


Thus $\tau\subset \partial \sigma$. Let $\sigma'$ be the minimal face of $\sigma$ containing $\tau$. Note that $\sigma'$ is a rational polyhedral cone, and we are basically in the situation as above, with $\sigma$ replaced by $\sigma'$, except that we don't have any more dimensions to spare, and so either $\tau=\sigma'$, as required, or we contradict (iii) as above.


