General proof that a product of nonzero homogeneous polynomials is nonzero (under certain conditions). Background, Notation, Definitions: Given a set $X$, I define the set $M(X)$ of monomials with $X$-indeterminates to be the set of elements of $\omega^X$ having finite support. Given $m_0,m_1\in M(X)$, I define the operation $*$ on $M(X)$ by $$(m_0*m_1)(x):=m_0(x)+_\omega m_1(x).$$ $\langle M(X),*\rangle$ is then a commutative, cancellative monoid, with the zero element of $\omega^X$ as the identity.
Given a ring $R$, it is then natural to define the set $R[X]$ of polynomials with $R$-coefficients and $X$-indeterminates to be the set of elements of $R^{M(X)}$ having finite support. We define the addition and multiplication operations $\oplus$ and $\odot$ on $R[X]$ in terms of the addition and multiplication operations $+$ and $\cdot$ on $R$ as follows: $$(p_0\oplus p_1)(m):=p_0(m)+p_1(m)$$ $$(p_0\odot p_1)(m):=\underset{m_0*m_1=m}{\sum_{m_0,m_1\in M(X)}}p_0(m_0)\cdot p_1(m_1).$$ Then $\langle R[X],\oplus,\odot\rangle$ is a ring. It will be commutative when $R$ is, with unity when $R$ has one.
I define the function $\deg:M(X)\to\omega$ by $$\deg(m):=\sum_{x\in X}m(x),$$  and the function $\sigma:\bigl(R[X]\smallsetminus\{0_{R[X]}\}\bigr)\to\omega$ by $$\sigma(p):=\max\{\deg(m):m\in M(X),p(m)\ne0_R\}.$$ It is readily seen that $\deg(m_0*m_1)=\deg(m_0)+_\omega\deg(m_1)$ and that $\sigma(p_0\odot p_1)\le\sigma(p_0)+_\omega\sigma(p_1)$ whenever $p_0,p_1,p_0\odot p_1\ne 0_{R[X]}$.
I define the set $H(R,X)$ of homogeneous polynomials with $R$-coefficients and $X$-indeterminates to be the set of all $p\in R[X]\smallsetminus\{0_{R[X]}\}$ such that $$\sigma(p)=\min\{\deg(m):m\in M(X),p(m)\ne0_R\}.$$ It is readily seen that $H(R,X)\cup\{0_{R[X]}\}$ is a sub-semigroup of $\langle R[X],\odot\rangle$ (a sub-monoid if $R$ is unital).

The Actual Question: It seems clear to me that $R$ has the zero product property ($a\cdot b=0_R$ implies $a=0_R$ or $b=0_R$) if and only if $\langle H(R,X),\odot\rangle$ is a semigroup. In that case, the restriction of $\sigma$ to $H(R,X)$ should be a semigroup homomorphism--that is, $\sigma(h_0\odot h_1)=\sigma(h_0)+_\omega\sigma(h_1)$.
Unfortunately, I have been banging my head against the wall trying to prove these for some time now. In particular, I'm having trouble showing that whenever $R$ has the zero product property, then $h_0\odot h_1\ne0_{R[X]}$ whenever $h_0,h_1\in H(R,X)$. I've tried to proceed by induction on the cardinalities of the supports of $h_0,h_1$, but I can't figure out how to make the induction step click.
Any suggestions, hints, or nice proofs of this?
 A: Let $f = \sum r_i x^i$ and $g = \sum s_j x^j$ be two polynomials. I claim that in the product $fg$ there is a term whose coefficient has the form $r_i s_j$ for some $i, j$. This suffices to prove the claim under the assumption that $R$ has no zero divisors, and reduces the claim to the following straightforward geometric argument: 
Since $f, g$ both have finitely many terms, we may assume WLOG that $X$ is finite, say $|X| = n$. Let 
$$\text{supp}(f) = \{ i \in \mathbb{Z}^n : r_i \neq 0 \}$$
denote the support. Then $\text{supp}(f)$ and $\text{supp}(g)$ are two finite sets of points in $\mathbb{Z}^n \subset \mathbb{R}^n$. Let $H$ be a hyperplane in $\mathbb{R}^n$ such that none of its translates passes through two or more points of either $\text{supp}(f)$ or $\text{supp}(g)$ (a generic hyperplane will have this property). Let $v \in \mathbb{R}^n$ be a vector orthogonal to $H$, and say that a point in a subset of $\mathbb{R}^n$ is extremal if $\langle v, - \rangle$ attains a maximum there. By construction, $\text{supp}(f)$ and $\text{supp}(g)$ have unique extremal points $i_0, j_0$ (if there is more than one extremal point then some translate of $H$ passes through all of them). 
Now, $\text{supp}(fg)$ is contained in the Minkowski sum $\{ i + j : i \in \text{supp}(f), j \in \text{supp}(g) \}$. Furthermore, $\langle v, i + j \rangle = \langle v, i \rangle + \langle v, j \rangle$, from which it follows that
$$\langle v, i + j \rangle \le \langle v, i_0 + j_0 \rangle$$
with equality iff $i = i_0, j = j_0$. In particular, $i + j \neq i_0 + j_0$ unless $i = i_0, j = j_0$. Hence the coefficient of $x^{i_0 + j_0}$ in $fg$ is $r_{i_0} s_{j_0}$ and the conclusion follows. 
The geometric picture is visualizable when $n = 2$. Here imagine two collections of points in the plane, and take e.g. "leftmost points" (although you may have to tilt the plane slightly if there is more than one leftmost point in each collection).  
