Let $Q \subseteq \mathbb R^n$ be a rational polyhedron and let $Q_I=Convexhull(Q \cap \mathbb Z^n)$. By finite basis theorem, we have $Q=P+C$ for some rational polytope $P$ and finitely generated cone $C$ where $C=\mathbb R_+ a_1+ \cdots+ \mathbb R_+a_s$ and $a_i \in \mathbb Z^n$ for $1 \leq i \leq s$. Now consider a linear map $T:\mathbb R^s \rightarrow \mathbb R^n $ where $T(e_i)=a_i$ and let $B=T([0,1]^s)$ be a polytope in $\mathbb R^n$ whose elements have the form $\lambda_1 a_1+\cdots+\lambda_s a_s$ with $0 \leq \lambda_i \leq 1$.
I was reading the proof to show $Q_I$ is a polyhedron from Monomial Algebras by R.H.Villareal. I could not able to prove $$Q_I= Convexhull((P+B)\cap \mathbb Z^n)+C.$$ Because if the above equality is proved then by finite basis theorem, $Q_I$ is a polyhedron.
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The righthand side is clearly contained in the lefthand side.
We now want to show the lefthand side (i.e., $Q_I$) is contained in the righthand side. Since $Q_I$ is the convex hull of its lattice points, and the righthand side is convex, it's enough to show that the lattice points of $Q_I$ are contained in the righthand side.
Now suppose we have a lattice point $x\in Q_I\cap \mathbb Z^n=Q\cap \mathbb Z^n$. We can write $x=p+c$, with $p\in P$ and $c\in C$. Write $c=c'+c''$, where $c''$ is a $\mathbb Z_+$ combination of the $a_i$, and $c'$ is a rational combination of the $a_i$, with all coefficients between zero and one. But now $c'\in B$, so $p+c'\in (P+B)\cap \mathbb Z^n$, and we are done.