Let $P \subset \mathbb{R}^d$ be a $d$-dimensional polytope, where all vertices lie on integral coordinates, and let $L(P,n)$ denote the number of integral lattice points contained in the scaled polytope $n \cdot P$, i.e. $L(P,n) := \# ((n \cdot P) \cap \mathbb{Z}^d)$. Then we know by a theorem of Ehrhart:
- The generating function $E(P,t) := \sum_{n=0}^\infty L(P,n) \cdot t^n$ is a rational function of the form $E(P,t) = \frac{h(t)}{(1-t)^{d+1}}$, where $h$ is a polynomial with $h(1) \neq 0$.
- $L(P,n)$ is a polynomial in $n$ for all positive integers.
I wonder if these results can also be obtained by the following algebraic approach: Let $k$ be an arbitrary field and $M := \text{Cone}(P \times \{1\}) \cap \mathbb{Z}^{d+1}$ considered as submonoid of $\mathbb{R}^{d+1}$. Then the Noetherian monoid algebra $k[M]$ is $\mathbb{Z}$-graded with respect to the $(d+1)$-th coordinate, and the corresponding Hilbert series of $k[M]$ coincides with $E(P,t)$. By the Hilbert-Serre theorem this series is a rational function of the form $E(P,t) = \frac{f(t)}{\prod_i 1-t^{e_i}}$. It is also clear that if we can show that all $e_i$ can be chosen to be $1$, it follows that $L(P,n)$ is a polynomial function for all sufficiently large $n$.
Is there an elegant way to proceed with this approach to get the same results as above?
The answer to my question is "Yes". Both points can be shown in a purely algebraic setting. Surprisingly, it is quite easy to show that $L(P,n)$ is a polynomial function for all sufficiently large $n$. The hard part is to show that it is actually a polynomial function for all positive integers, which needs much deeper results. All this can be found in [Bruns, Herzog - Cohen-Macaulay-Rings].
The easy part: In contrast to my explanations before the question, we need two different monoids here. Let $M = \mathbb{R}_+(P \times \{1\}) \cap \mathbb{Z}^{n+1}$ and let $N$ be the submonoid of $M$ generated by $\{ (p,1) : p \in P \cap \mathbb{Z}^n \}$. It is easy to see that $M$ is integral over $N$, i.e. for each $x \in M$ there is an $n \in \mathbb{Z}_+$ with $n \cdot x \in N$ (Actually, $M$ is the integral closure of $N$ in $\mathbb{Z}^{n+1}$.) Therefore $k[N] \leq k[M]$ is an integral ring extension, and hence we can consider $k[M]$ as finitely generated $k[N]$-module. Both rings are graded with respect to the last coordinate, and $k[N]$ is (by definition) generated by homogeneous elements of degree $1$ as a $k$-algebra. Now by the Hilbert-Serre-Theorem the Hilbert series of the $k[N]$-module $k[M]$ (= the Ehrhart series of $P$) has the form $E(P,t) = \frac{h(t)}{(1-t)^m}$, where $m$ equals the number of those homogeneous generators of $k[N]$.
Let $E(P,t) = \sum_{n=0}^\infty f(n) \cdot X^n$ be the general form of the generating function. Then because of its special form, we just discovered, it follows that $f(n)$ equals $p(n)$ for some polynomial function $p$ and all sufficiently large $n$. Furthermore, the largest integer $n$ for which $f(n) = p(n)$ fails to hold is given by $\deg(E(P,t)) = \deg(h) - m$ (see for example [Stanley - Enumerative Combinatorics] for these well known results). The last quantity is also called the a-invariant of the graded ring and is denoted by $a(k[M])$. So in order to show the assertion, it remains to show $a(k[M]) < 0$. Here begins the hard part.
The hard part: I don't understand all of this in detail yet. The proof of Theorem 6.3.11 in the book of Bruns and Herzog exploits the fact that $k[M]$ is Cohen-Macaulay. For such graded rings the a-invariant can be read off from the so called *canonical module. In this case the *canonical module of $k[M]$ can be specified in terms of the interior of $M$, and from this it is easy to show that $a(k[M])$ is actually negative.
Maybe I add further explanations here if I got a better understanding someday, but this may take a while.