# Bernstein-Kushnirenko and Bézout’s theorems (weak version)

$\DeclareMathOperator{\conv}{conv} \newcommand{\dprime}{^{\prime\prime}} \DeclareMathOperator{\interior}{interior} \newcommand{\kk}{\mathbb{K}} \newcommand{\kstar}{\kk^*} \newcommand{\kstarn}{(\kk^*)^n} \newcommand{\kstarnn}[1]{(\kk^*)^{#1}} \DeclareMathOperator{\mv}{MV} \newcommand{\pp}{\mathbb{P}} \newcommand{\qq}{\mathbb{Q}} \newcommand{\rnonnegs}{\mathbb{r}^s_{\geq 0}} \newcommand{\rr}{\mathbb{R}} \newcommand{\scrA}{\mathcal{A}} \newcommand{\scrL}{\mathcal{L}} \newcommand{\scrP}{\mathcal{P}} \DeclareMathOperator{\supp}{supp} \DeclareMathOperator{\vol}{vol} \newcommand{\znonneg}{\mathbb{Z}_{\geq 0}} \newcommand{\znonnegs}{\mathbb{Z}^s_{\geq 0}} \newcommand{\zz}{\mathbb{Z}}$
In earlier posts we defined the degree of a projective variety (defined over an algebraically closed field $\kk$) and showed that it can be determined from its Hilbert polynomial. In this post we use these notions to

1. obtain Kushnirenko’s formula for the number of solutions of $n$ generic polynomials on $(\kk \setminus \{0\})^n$,
2. sketch a derivation of the formula in Bernstein’s theorem using Kushnirenko’s formula, and
3. deduce the formula in Bézout’s theorem from Bernstein’s formula.

For the proof of Kushnirenko’s formula we use a beautiful argument of Khovanskii (A. G. Khovanskii, Newton Polyhedron, Hilbert Polynomial, and Sums of Finite Sets, Functional Analysis and Its Applications, vol 26, 1992). Bernstein’s formula then follows from multi-additivity properties of intersection numbers and mixed volumes. The formula from Bézout’s theorem is in turn a special case of Bernstein’s formula. Note that the statements we prove in this post are “weak” forms of these formulae in the sense that we only consider the case that the intersection is finite. In the “strong” form (as e.g. in Sections VII.3 and VIII.3 of How Many Zeroes?) these provide an upper bound for the number of isolated solutions, even if the solution set contains non-isolated points.

Following the “toric” tradition, we write $\kstar$ to denote the “torus” $\kk \setminus \{0\}$. Consider the $n$-dimensional torus $\kstarn$ with coordinates $(x_1, \ldots, x_n)$. Every $\alpha = (\alpha_1, \ldots, \alpha_n) \in \zz^n$ corresponds to a monomial $x_1^{\alpha_1} \cdots x_n^{\alpha_n}$ which we denote by $x^\alpha$. A regular function on $\kstarn$, i.e. a Laurent polynomial in $(x_1, \ldots, x_n)$, is a polynomial in $x_1, x_1^{-1}, \ldots, x_n, x_n^{-1}$, and can be expressed as a sum
$f = \sum_\alpha c_\alpha x^\alpha$
such that $c_\alpha \neq 0$ for at most finitely many $\alpha \in \zz^n$; the support of $f$, denoted $\supp(f)$, is the finite set of all $\alpha$ such that $c_\alpha \neq 0$. We say that $f$ is supported at $\scrA$ if $\supp(f) \subseteq \scrA$.

Theorem 1 (Kushnirenko). Given a finite subset $\scrA$ of $\zz^n$, the number of solutions on $\kstarn$ of $n$ generic Laurent polynomials supported at $\scrA$, counted with appropriate multiplicity, is $n!$ times the volume of the convex hull of $\scrA$.

## Proof of Theorem 1: Step 1 – reduction to count of $|s\scrA|$

Let $\alpha_0, \ldots, \alpha_N$ be the elements of $\scrA$. Consider the map
$\phi_\scrA: \kstarn \to \pp^N$
which maps
$x \mapsto [x^{\alpha_0}: \cdots : x^{\alpha_N}]$
Let $X_\scrA$ be the closure of $\phi_\scrA(\kstarn)$ in $\pp^N$. A linear polynomial $\sum_i c_iz_i$, where $[z_0: \cdots :z_N]$ are homogeneous coordinates on $\pp^N$, restricts on the image of $\phi_\scrA$ to a Laurent polynomial $\sum_i c_ix^{\alpha_i}$ supported at $\scrA$. Since $\dim(X_\scrA) = \dim(\phi_\scrA(\kstarn)) \leq n$, a generic codimension $n$ linear subspace of $\pp^N$ either does not intersect $X_\scrA$ (when $\dim(X_\scrA) < n$), or, when $\dim(X_\scrA) = n$, intersects $X_\scrA$ at $\deg(X_\scrA)$ generic points on the image of $\phi_\scrA$, each with multiplicity $1$ (Theorem 4 of Degree of a projective variety). Since in the latter case, for a generic point $z$ on the image of $\phi_\scrA$ there are precisely $\deg(\phi_\scrA)$ points (counted with appropriate multiplicity) in $\phi_\scrA^{-1}(z)$, we have the following:

Observation 2. The number (counted with appropriate multiplicity) of solutions on $\kstarn$ of $n$ generic Laurent polynomials supported at $\scrA$ is either zero (when $\dim(\phi_\scrA(\kstarn)) < n$), or the degree of $X_\scrA$ times the degree of the map $\phi_\scrA$ (when $\dim(\phi_\scrA(\kstarn)) = n$).

Claim 3. Let $G$ be the subgroup of $\zz^n$ generated by $A – \alpha_0 = \{\alpha_i – \alpha_0: i = 0, \ldots, N\}$. Then

1. there is $r \geq 0$, called the rank of $G$, such that $G \cong \zz^r$,
2. $\dim(\phi(\scrA)) = r$,
3. If $r = n$, then $\deg(\phi_\scrA)$ is the index (as a subgroup) of $G$ in $\zz^n$.

Proof. In the coordinates $(z_1/z_0, \ldots, z_N/z_0)$ on $U_0 := \pp^N \setminus V(z_0)$, the map $\phi_\scrA$ reduces to a map from $\kstarn \to \kstarnn{N}$ given by
$x \mapsto (x^{\alpha_1 – \alpha_0}, \ldots, x^{\alpha_N – \alpha_0})$
Now Claim 3 follows from Proposition VI.1 of How Many Zeroes? (the main ingredient in the proof is the observation that there is a basis $\beta_1, \ldots, \beta_n$ of $\zz^n$ such that $m_1\beta_1, \ldots, m_r\beta_r$ is a basis of $G$ for some $r \geq 0$).

In the case that $r < n$, it is clear that the volume of the convex hull $\conv(\scrA)$ of $\scrA$ is zero, and therefore Theorem 1 follows from Observation 2 and Claim 3. So from now on, assume $r = n$. Due to Observation 2 and Claim 3, in order to prove Theorem 1 it remains to compute $\deg(X_\scrA)$. Let $R = \kk[z_0, \ldots, z_N]/I(X_\scrA)$ be the homogeneous coordinate ring of $X_\scrA$. The theory of Hilbert polynomials (Theorems 1 and 2 of the preceding post) imply the following:

Observation 4. For $s \gg 1$, the dimension (as a vector space over $\kk$) of the degree $s$ graded component $R_s$ of $S$ is given by a polynomial $P_\scrA(s)$ of the form
$P_\scrA(s) = \deg(X_\scrA)\frac{s^n}{n!} + \text{terms of degree} < n$
In particular,
$\deg(X_\scrA) = n! \lim_{s \to \infty} \frac{\dim_\kk(R_s)}{s^n}$

Claim 5. For each $s \geq 0$,
$\dim_\kk(R_s) = |s\scrA|$
where $s\scrA$ is the set of all possible $s$-fold sums $\alpha_{i_1} + \cdots + \alpha_{i_s}$ with $\alpha_{i_j} \in \scrA$.

Proof. Indeed, for each $s \geq 0$, the substitutions $z_i = x^{\alpha_i}$, $i = 0, \ldots, N$, induce a $\kk$-linear map
$\sigma:\kk[z_0, \ldots, z_N]_s \to \scrL_{s\scrA}$
where $\kk[z_0, \ldots, z_N]_s$ is the set of homogeneous polynomials of degree $s$ in $(z_0, \ldots, z_N)$ and $\scrL_{s\scrA}$ is the set of Laurent polynomials supported at $s\scrA$. It is straightforward to check that $\sigma$ is surjective and $\ker(\sigma)$ is the set $I(X_\scrA)_s$ of homogeneous polynomials of degree $s$ in $I(X_\scrA)$. Consequently, as a vector space over $\kk$, $R_s$ is isomorphic to $\scrL_{s\scrA}$. In particular,
$\dim_\kk(R_s) = \dim_\kk(\scrL_{s\scrA}) =|s\scrA|$
which completes the proof of Claim 5.

## Proof of Theorem 1: Step 2 – $|s\scrA|$ in terms of $\vol(\conv(\scrA))$

The following lemma is a special case of Proposition 3.7 from Discriminants, Resultants, and Multidimensional Determinants by Gelfand, Kapranov and Zelevinsky.

Lemma 6. Let $\scrP$ be the closure of a bounded open subset of $\rr^n$ such that $\scrP$ contains the origin and the boundary of $\scrP$ is piecewise linear. Then
$\lim_{s \to \infty} \frac{|s\scrP \cap \zz^n|}{s^n} = \vol(\scrP)$
where by $s\scrP$ we denote the “homothetic” set $\{s\alpha : \alpha \in \scrP\}$.

Proof. We proceed by induction on $n$. If $n = 1$, $\scrP$ is a closed interval, say of length $l$, and the lemma follows from the observation that
$sl \leq |s\scrP \cap \zz| \leq sl + 1$
In the general case we associate to each point $\alpha$ in $\zz^n$ a “lattice cube” which is the translation $\alpha + I^n$ of the standard cube
$I^n := \{(x_1, \ldots, x_n) \in \rr^n: 0 \leq x_i \leq 1$
Let $N(s)$ be the number of lattice cubes contained in $s\scrP$. Since the diameter of $I^n$ is $\sqrt{n}$, it follows that
$0 \leq |s\scrP \cap \zz^n| – N(s) \leq a(s)$
where $a(s)$ is the number of elements in $\zz^n \cap s\scrP$ with distance $\leq \sqrt{n}$ from the boundary of $s\scrP$. Similarly, since the volume of $I^n$ is 1,
$0 \leq \vol(s\scrP) – N(s) \leq b(s)$
where $b(s)$ is the volume of the set of all points in $s\scrP$ with distance $\leq \sqrt{n}$ from the boundary of $s\scrP$. It follows from the induction hypothesis and basic properties of volume that both $a(s)$ and $b(s)$ grow as constant times $s^{n-1}$ as $s \to \infty$. Consequently,
$\lim_{s \to \infty} \frac{|s\scrP \cap \zz^n|}{s^n} =\lim_{s \to \infty} \frac{\vol(s\scrP)}{s^n} =\lim_{s \to \infty} \frac{s^n\vol(\scrP)}{s^n} = \vol(\scrP)$
which completes the proof of the lemma.

We formally state a corollary of the observation that $a(s)$ in the proof of Lemma 6 grows as a constant times $s^{n-1}$ as $s \to \infty$.

Proposition 7. Let $\scrP$ and $s\scrP$ be as in Lemma 6. For a fixed $r \in \rr$, let $\scrP’_{s, r}$ be the subset of $s\scrP$ consisting of all points with distance $\leq r$ from the boundary of $s\scrP$. Then
$\lim_{s \to \infty} \frac{|\scrP’_{s, r} \cap \zz^n|}{s^n} = 0$

The main idea of the proof of the following lemma is the same as that of the standard proof of Gordan’s lemma.

Lemma 8. Pick $\alpha_1, \ldots, \alpha_q \in \zz^n$ such that the subgroup generated by the $\alpha_i$ coincides with $\zz^n$. Then there is a constant $C$ with the following property: for every collection of real numbers $\lambda_1, \ldots, \lambda_q$ such that $\sum_i \lambda_i \alpha_i \in \zz^n$, there are integers $l_1, \ldots, l_q$ such that
\begin{align*} &\sum_i l_i\alpha_i = \sum_i \lambda_i \alpha_i,\ \text{and} \\ &\sum_i |\lambda_i – l_i| < C \end{align*}

Proof. Let
$\scrP := \{\sum_i \lambda_i\alpha_i: 0 \leq \lambda_i \leq 1\} \subseteq \rr^n$
Since $\scrP$ is compact and $\zz^n$ is discrete, the intersection $\scrP \cap \zz^n$ is finite. Since the $\alpha_i$ generate $\zz^n$, for each $\alpha \in \scrP \cap \zz^n$, we can fix a representation $\alpha = \sum_i l_i(\alpha)\alpha_i$ with $l_i(\alpha) \in \zz$. Now given $\beta \in \zz^n$ such that $\beta = \sum_i \lambda_i \alpha_i$ with $\lambda_i \in \rr$,
$\alpha := \sum_i (\lambda_i – \lfloor \lambda_i \rfloor) \alpha_i \in \scrP \cap \zz^n$
where $\lfloor \lambda_i \rfloor$ is the greatest integer $\leq \lambda_i$. Consequently,
$\beta = \sum_{i=1}^q \lfloor \lambda_i \rfloor \alpha_i + \alpha = \sum_{i=1}^q (\lfloor \lambda_i \rfloor + l_i(\alpha)) \alpha_i$
with
$\sum_{i=1}^q |\lambda_i – \lfloor \lambda_i \rfloor – l_i(\alpha)| \leq \sum_{i=1}^q |\lambda_i – \lfloor \lambda_i \rfloor| + \sum_{i=1}^q|l_i(\alpha)| \leq q + Q$
where $Q$ is the maximum of $\sum_i|l_i(\alpha)|$ over all $\alpha \in \scrP \cap \zz^n$. Therefore the lemma holds with $C := q + Q$.

Theorem 9. Let $\scrA = \{\alpha_1, \ldots, \alpha_q\}$ be a finite subset of $\zz^n$ and $\scrP := \conv(\scrA) \subseteq \rr^n$. Assume the subgroup of $\zz^n$ generated by all the pairwise differences $\alpha_i – \alpha_j$ coincides with $\zz^n$. Then there is $r \in \rr$ with the following property: for each positive integer $s$, every element of $s\scrP \cap \zz^n$ whose distance from the boundary of $s\scrP$ is greater than or equal to $r$ belongs to the set $s\scrA$ of all $s$-fold sums $\alpha_{i_1} + \cdots + \alpha_{i_s}$ of elements from $\scrA$.

Proof. Fix a nonnegative integer $s$. We want to understand which points of $s\scrP \cap \zz^n$ can be represented as elements from $s\scrA$. Replacing each $\alpha_i$ by $\alpha_i – \alpha_1$ if necessary, we may assume that one of the $\alpha_i$ is the origin. Then
$s\scrA = \{\sum_i l_i\alpha_i:\ l_i \in \znonneg,\ \sum_i l_i \leq s\}$
Moreover, the $\alpha_i$ satisfy the hypothesis of Lemma 8. Let $C$ be the constant prescribed by Lemma 8. If $\alpha \in s\scrP \cap \zz^n$ then
$\alpha = \sum_i \lambda_i \alpha_i$
with nonnegative real numbers $\lambda_i$ such that $\sum_i \lambda_i \leq s$. By Lemma 8, there are $l_i \in \zz$ such that
$\alpha = \sum_i l_i \alpha_i,\ \text{and}\ \sum_i|\lambda_i – l_i| < C$
Consequently, if the $\lambda_i$ further satisfies the following inequalities:
$\lambda_i \geq C,\ \text{and}\ \sum_i \lambda_i \leq s – C$
then the condition $\sum_i|\lambda_i – l_i| < C$ implies that
$l_i > 0,\ \text{and}\ \sum_i l_i < s$
i.e. $\alpha \in s\scrA$. In short, we proved the following:

Observation 10. Let $\scrP_{s,C}$ be the subset of $s\scrP$ consisting of all $\beta = \sum_i \lambda_i \alpha_i$ such that each $\lambda_i \geq C$ and $\sum_i \lambda_i \leq s – C$. Then
$\scrP_{s,C} \cap \zz^n \subseteq s\scrA$

Now we go back to the proof of Theorem 9. Consider the map $\pi: \rr^q \to \rr^n$ that maps the $i$-th standard unit element in $\rr^q$ to $\alpha_i$ for each $i$. Let
$\Delta := \{(\lambda_1, \ldots, \lambda_q): \lambda_i \geq 0\ \text{for each}\ i,\ \sum_i \lambda_i \leq 1\} \subseteq \rr^q$
be the standard simplex in $\rr^q$. Let $\Delta_{s,C}$ be the sub-simplex of $s\Delta$ defined by the inequalities $\lambda_i \geq C$ for each $i$, and $\sum_i \lambda_i \leq s – C$. Then
$\pi(s\Delta) = s\scrP,\ \text{and}\ \pi(\Delta_{s, C}) = \scrP_{s,C}$
Note that every point in $s\Delta$ whose distance from the boundary of $s\Delta$ is $\geq C\sqrt{q}$ belongs to $\Delta_{s,C}$. Khovanskii then argues (in the proof of Theorem 3 of Newton Polyhedron, Hilbert Polynomial, and Sums of Finite Sets) that this implies that “every point in $s\scrP$ whose distance from the boundary of $s\scrP$ is $\geq C\sqrt{q}||\pi||$ is in $\scrP_{s, C}$”, where
$||\pi|| := \sup_{v \neq 0} \frac{||\pi(v)||}{||v||}$
is the “norm” of $\pi$, which also equals the norm of the largest eigenvalue of $\pi$. However, it is not clear how this conclusion follows from the preceding statement: indeed, it is possible to have $\alpha’$ close to the boundary of $s\Delta$, but $\pi(\alpha’)$ to have large distance from the boundary of $s\scrP$, e.g. when one of the $\alpha_i$ is in the interior of $\scrP$, then $se_i$ is in the boundary of $s\Delta$, but $\pi(se_i) = s\alpha’$ is far away from the boundary of $s\scrP$ when $s$ is large. We are going to close this gap by a more precise modification of the argument.

Indeed, consider the collection $S$ of all sets $\scrA’ \subseteq \scrA$ such that $\conv(\scrA’)$ intersects interior of $\scrP$. For each such $\scrA’$, we fix one element
$\alpha_{\scrA’} \in \conv(\scrA’) \cap \interior(\scrP)$
and two representations of $\alpha_{\scrA’}$ as convex combinations of $\scrA’$ and $\scrA$:
$\alpha_{\scrA’} = \sum_{\alpha_i \in \scrA’} \epsilon’_{\scrA’, i} \alpha_i = \sum_{\alpha_i \in \scrA} \epsilon_{\scrA’, i} \alpha_i$
where
$\sum_{\alpha_i \in \scrA} \epsilon_{\scrA’, i} = \sum_{\alpha_i \in \scrA’} \epsilon’_{\scrA’, i} = 1$
and in addition, all $\epsilon_{\scrA’, i}$ and $\epsilon’_{\scrA’, j}$ are positive. For each $\scrA’$, we also fix a large constant $C_{\scrA’}$ such that which we precisely specify a bit further down. Consider a nonnegative $\rr$-linear combination
$\alpha= \sum_i \lambda_i \alpha_i \in s\scrP$
such that
$\sum_i \lambda_i \leq s$
Assume $\alpha \in \scrP_{s, C}$, but the $\lambda_i$ does not satisfy the defining conditions of $\scrP_{s,C}$ from Observation 10, i.e. either some $\lambda_i < C$ or $\sum_i \lambda_i > s – C$. We want to find some conditions under which it is possible to find a different representation of $\alpha$ which does satisfy the defining conditions of $\scrP_{s,C}$.

Claim 11. Pick $\scrA’ \in S$. If $C_{\scrA’}$ is chosen sufficiently large (independently of $s$), then the following holds: if $\lambda_i > \epsilon’_{\scrA’,i}C_{\scrA’}$ for each $\alpha_i \in \scrA’$, then there is a representation
$\alpha = \sum_i \lambda’_i \alpha_i$
with $\lambda’_i \geq C$ for each $i$, and $\sum_i \lambda’_i \leq s – C$.

Proof. Assume $\lambda_i > \epsilon’_{\scrA’,i}C_{\scrA’}$ for each $\alpha_i \in \scrA’$. Then
\begin{align*} \alpha &= \sum_{\alpha_i \in \scrA’} (\lambda_i – \epsilon’_{\scrA’,i}C_{\scrA’}) \alpha_i + C_{\scrA’} \sum_{\alpha_i \in \scrA’} \epsilon’_{\scrA’, i} \alpha_i + \sum_{\alpha_i \not\in \scrA’}\lambda_i \alpha_i \\ &= \sum_{\alpha_i \in \scrA’} (\lambda_i – \epsilon’_{\scrA’,i}C_{\scrA’}) \alpha_i + C_{\scrA’}\sum_{\alpha_i \in \scrA} \epsilon_{\scrA’, i} \alpha_i + \sum_{\alpha_i \not\in \scrA’}\lambda_i \alpha_i \\ &= \sum_{\alpha_i \in \scrA’} ((\lambda_i – \epsilon’_{\scrA’,i}C_{\scrA’}) + \epsilon_{\scrA’, i}C_{\scrA’}) \alpha_i + \sum_{\alpha_i \not\in \scrA’}(\lambda_i + \epsilon_{\scrA’,i}C_{\scrA’}) \alpha_i \\ \end{align*}
If $C_{\scrA’}$ is chosen such that
$\epsilon_{\scrA’, i}C_{\scrA’} \geq 2C\ \text{for each}\ \alpha_i \in \scrA’$
then in the final representation of $\alpha$ above, each coefficient is $\geq 2C \geq C$, so to ensure that $\alpha \in \scrP_{s, C}$ it only remains to bound the sum of the coefficients by $s – C$. However, that is easy: first note that during the above change of representation for $\alpha$, the sum of the coefficients is unchanged, since we replace a convex combination (where the sum of coefficients is $1$) by another convex combination. Denote the coefficients of the final representation of $\alpha$ above by $\tilde \lambda_i$. Since we originally had $\sum_i \lambda_i \leq s$, after the change of representation we have
$\sum_i \tilde \lambda_i \leq s,\ \text{and}\ \tilde \lambda_i \geq 2C\ \text{for each}\ i$
Now recall that there is $i_0$ such that $\alpha_{i_0}$ is the origin. Consequently we can simply reduce $\tilde \lambda_{i_0}$ to $\tilde \lambda_{i_0} – C$ to satisfy all defining conditions of $\scrP_{s, C}$. This completes the proof of Claim 11.

Now we are ready to prove Theorem 9. Replacing $\scrA$ by $\scrA – \alpha_i$ for some $i$ if necessary, we may assume that the origin is a vertex of $\scrP$. Due to Observation 10, it suffices to show that there is $r$ independent of $s$ such that all elements of $s\scrP \setminus \scrP_{s, C}$ has distance smaller than $r$ from the boundary of $s\scrP$. Pick $\alpha \in s\scrP \setminus \scrP_{s, C}$ and a representation
$\alpha = \sum_i \lambda_i \alpha_i$
such that each $\lambda_i$ is nonnegative and $\sum_i \lambda_i \leq s$. Claim 11 implies that for each subset $\scrA’$ of $\scrA$ such that $\conv(\scrA’)$ intersects the interior of $\scrP$, and each $\alpha_i \in \scrA’$,
$\lambda_i \leq \epsilon’_{\scrA’, i}C_{\scrA’}$
Consequently, if
$C’ := \max_{\scrA’, i} \epsilon’_{\scrA’, i} C_{\scrA’}$
then there are two possible cases:

Case 1: $\lambda_i < C’$ for each $i$. Since by our assumption the origin is a vertex of $s\scrP$, it then follows that the the distance of $\alpha$ from the boundary is bounded by
$r_1 := \max\{||\sum_i \epsilon_i\alpha_i || : 0 \leq \epsilon_i \leq C’\}$

Case 2: there is $i$ such that $\lambda_i \geq C’$. In that case Claim 11 implies that all $\alpha_i$ such that $\lambda_i \geq C’$ are contained in a proper face $\scrP’$ of $\scrP$. We consider two subcases:

Case 2.1: $\scrP’$ contains the origin. In this case $s’\scrP’ \subseteq s\scrP’$ whenever $0 \leq s’ \leq s$, and consequently,
$\alpha’ := \sum_{\alpha_i \in \scrP’} \lambda_i \alpha_i \in s\scrP’$
In particular, $\alpha’$ is on the boundary of $s\scrP$ and
$||\alpha – \alpha’|| = || \sum_{\alpha_i \not\in \scrP’}\lambda_i\alpha_i|| \leq r_1$

Case 2.2: $\scrP’$ does not contain the origin. Let $\scrA^* \subseteq \scrP’$ be the set of all $\alpha_i$ such that $\lambda_i \geq C’$. We are going to show that
$\sum_{\alpha_i \in \scrA^*} \lambda_i \geq s – qC’$
(recall that $q := |\scrA|$). Indeed, otherwise we would have
$\sum_i \lambda_i = \sum_{\alpha_i \in \scrA^*} \lambda_i + \sum_{\alpha_i \not\in \scrA^*} \lambda_i < s – qC’ + (q-1)C’ = s – C’$
Consequently, if $\alpha_{i_0}$ is the origin, then we have another representation for $\alpha$ for which the sum of the coefficients is $\leq s$:
$\alpha = \sum_{i \neq i_0} \lambda_i \alpha_i + (\lambda_{i_0} + C’)\alpha_{i_0}$
However, since $\alpha_{i_0} \not\in \scrP’$, it follows that $\scrA’ := \scrA^* \cup \{\alpha_{i_0}\}$ is one of the sets considered in Claim 11, and therefore Claim 11 would imply that $\alpha \in \scrP_{s, C}$ which is a contradiction to the choice of $\alpha$. This proves that
$\sum_{\alpha_i \in \scrA^*} \lambda_i \geq s – qC’$
Pick $\alpha_{i’} \in \scrP’$ and let
$\alpha’ := \sum_{\alpha_i \in \scrP’} \lambda_i \alpha_i + (s – \sum_{\alpha_i \in \scrP’} \lambda_i) \alpha_{i’} \in s\scrP’$
Then $\alpha’$ is in the boundary of $s\scrP$ and
\begin{align*} ||\alpha – \alpha’|| &= || \sum_{\alpha_i \not\in \scrP’}\lambda_i\alpha_i – (s – \sum_{\alpha_i \in \scrP’} \lambda_i) \alpha_{i’} || \leq r \end{align*}
where
$r := \max\{||\sum_i \epsilon_i\alpha_i || : -qC’ \leq \epsilon_i \leq C’\}$

Since $r \geq r_1$, we showed that in all case the distance of $\alpha$ from the boundary of $s\scrP$ is $\leq r$, as required to complete the proof of Theorem 9.

The following result describes the connection between $|s\scrA|$ and the volume of the convex hull of $\scrA$ that we were aiming for in this section:

Corollary 12. Let $\scrA = \{\alpha_1, \ldots, \alpha_q\}$ be a finite subset of $\zz^n$. Assume the subgroup $G$ of $\zz^n$ generated by all the pairwise differences $\alpha_i – \alpha_j$ is a subgroup of finite index $m$ in $\zz^n$. Then
$\lim_{s \infty} \frac{|s\scrA|}{s^n} = \frac{1}{m}\vol(\conv(\scrA))$
where $s\scrA$ of all $s$-fold sums $\alpha_{i_1} + \cdots + \alpha_{i_s}$ of elements from $\scrA$.

Proof. In the case that $m = 1$, the result follows immediately from combining Lemma 6, Proposition 7 and Theorem 9. In the general case, there is a basis $\beta_1, \ldots, \beta_n$ such that $m_1\beta_1, \ldots, m_n\beta_n$ is a basis of $G$ with
$m = \prod_i m_i$
Note that $\beta_1, \ldots, \beta_n$ is also a basis of $\rr^n$, so the map $\psi:\rr^n \to \rr^n$ that sends
$\beta_i \mapsto \frac{1}{m_i}\beta_i$
is well defined. Note that $\psi$ is an isomorphism over $\rr$ and maps $G$ onto $\zz^n$. The $m = 1$ case of the corollary then implies that
$\lim_{s \infty} \frac{|s\scrA|}{s^n} = \vol(\conv(\psi(\scrA)))$
Now note that the change of coordinates between the standard basis of $\rr^n$ and $\beta_1, \ldots, \beta_n$ is given by a matrix with determinant one (since it is a matrix with integer entries and its inverse is also a matrix with integer entries), and therefore it preserves volumes. However, after this change of coordinates $\psi$ corresponds to a diagonal matrix with diagonal $(1/m_1, \ldots, 1/m_n)$ and determinant $1/\prod_i m_i$. Consequently,
$\vol(\conv(\psi(\scrA))) = \vol(\conv(\scrA))/\prod_i m_i = \vol(\conv(\scrA))/m$
which completes the proof of Corollary 12.

## Proof of Theorem 1: Step 3 – conclusion

Given a finite subset $\scrA$ of $\zz^n$, let $M$ be the number of solutions on $\kstarn$ of $n$ generic Laurent polynomials supported at $\scrA$, counted with appropriate multiplicity. Let $G$ be the subgroup of $\zz^n$ generated by all the pairwise differences of elements of $\scrA$. As explained in the paragraph preceding Observation 4, if the rank of $G$ is smaller than $n$, then both $M$ and the volume of the convex hull of $\scrA$ are zero, so that Theorem 1 is true. On the other hand, if rank equals $n$, then the index of $G$ in $\zz^n$ is finite; call it $m$. Then Observation 2, Claim 3, Observation 4 and Claim 5 imply that
$M = n!m\lim_{s \to \infty} \frac{|s\scrA|}{s^n}$
Corollary 12 then implies that
$M = n! \vol(\conv(\scrA))$
which proves Theorem 1.

## Bernstein and Bézout’s formulae

Given bounded convex sets $\scrP_1, \ldots, \scrP_s$ in $\rr^n$, the function $\rnonnegs \to \rr$ which maps
$(\lambda_1, \ldots, \lambda_s) \mapsto \vol(\lambda_1 \scrP_1 + \cdots + \lambda_s \scrP_s)$
is given by a (necessarily unique) homogeneous polynomial of degree $n$ in the $\lambda_i$ (see e.g. Theorem V.39 of How Many Zeroes? for the case that each $\scrP_i$ is a polytope). For each $\alpha = (\alpha_1, \ldots, \alpha_s) \in \znonnegs$ such that $\sum_i \alpha_i = n$, we write $\nu_\alpha(\scrP_1, \ldots, \scrP_s)$ for the coefficient of $\lambda^\alpha$ in the expression of $\vol(\lambda_1 \scrP_1 + \cdots + \lambda_s \scrP_s)$ as a polynomial in $\lambda_1, \ldots, \lambda_s$. It then follows from elementary properties of commutative semigroups that the map that sends
$(\scrP_1, \ldots, \scrP_n) \mapsto \nu_{(1, \ldots, 1)}(\scrP_1, \ldots, \scrP_n)$
is a symmetric multiadditive function on the set of $n$-tuples of bounded convex bodies in $\rr^n$ (see e.g. How Many Zeroes?, Lemma B.59); it is called the mixed volume of $\scrP_1, \ldots, \scrP_n$ and we denote it by $\mv(\scrP_1, \ldots, \scrP_n)$.

Lemma 13. Any symmetric multiadditive function $\rho: G^n \to \rr$ from a commutative semigroup $G$ is uniquely determined by its “diagonal part”, i.e. the map $G \to \rr$ that maps
$g \in G \mapsto \rho(g, \ldots, g)$
More precisely,
$\rho(g_1, \ldots, g_n) = \frac{1}{n!}\sum_{I \subseteq \{1, \ldots, n\}} (-1)^{n-|I|} \rho(\sum_{i \in I}g_i, \ldots, \sum_{i \in I}g_i)$
In particular, $\mv$ is the unique symmetric multiadditive function on the set of $n$-tuples of bounded convex bodies (or the set of $n$-tuples of convex polytopes) such that $\mv(\scrP, \ldots, \scrP) = n! \vol(\scrP)$.

For a proof of Lemma 13, see e.g. this MathOverflow post or How Many Zeroes?, Corollary B.62.

The following is the (weak form of) the theorem of David Bernstein that we were after.

Theorem 14 (Bernstein). Given finite subsets $\scrA_1, \ldots, \scrA_n$ of $\zz^n$, the number (counted with appropriate multiplicity) of solutions on $\kstarn$ of $f_1, \ldots, f_n$, where each $f_i$ is a generic Laurent polynomial supported at $\scrA_i$, is the mixed volume of the convex hulls of $\scrA_i$.

Proof (sketch). Consider the map $\rho$ that sends $(\scrA_1, \ldots, \scrA_n)$ to the number (counted with appropriate multiplicity) of solutions on $\kstarn$ of generic Laurent polynomials supported at $\scrA_i$. Due to Theorem 1 and Lemma 13, it suffices to show that $\rho$ is symmetric and multiadditive. That can be proved easily once one proves the correct “non-degeneracy” condition that identifies when $f_1, \ldots, f_n$ are “sufficiently generic” (see How Many Zeroes?, Claim VII.28).

As a corollary of Theorem 14 we almost immediately obtain a weak form of Bézout’s theorem.

Theorem 15 (Bézout). The number of solutions (counted with appropriate multiplicity) on $\kk^n$ of generic polynomials $f_1, \ldots, f_n$ of degree respectively $d_1, \ldots, d_n$ is $\prod_i d_i$.

Proof (sketch). Every polynomial $f_i$ of degree $d_i$ in $n$ variables is supported at the simplex $\scrP_i \subseteq \rr^n$ with vertices at the origin and $d_ie_j$, $j = 1, \ldots, n$ (where the $e_j$ are the standard unit elements along the axes of $\rr^n$). If $f_i$ are generic then all solutions of $f_1, \ldots, f_n$ on $\kk^n$ actually belong to $\kstarn$. Consequently, Theorem 15 follows from Theorem 14 if we can show that the mixed volume of $\scrP_1, \ldots, \scrP_n$ is $\prod_i d_i$. However, that follows from an easy computation (see How Many Zeroes?, Claim VIII.2.1).