[proofplan] We argue by contradiction. Assume $f$ vanishes on $S = S_1 \times \cdots \times S_n$. By [Membership in the Vanishing Ideal](/theorems/2616), $f$ decomposes as $f = \sum_{j=1}^n h_j g_j$ where $g_j = \prod_{s \in S_j}(X_j - s)$ has degree $d_j + 1$ and $\deg h_j \leq d - (d_j + 1) = \sum_{i \neq j} d_i - 1$. We then show that no term $h_j g_j$ can contribute the monomial $X_1^{d_1} \cdots X_n^{d_n}$ because the degree bound on $h_j$ is too tight, contradicting the assumption that this monomial has a nonzero coefficient in $f$. [/proofplan] [step:Assume $f$ vanishes on $S$ and decompose via the Vanishing Ideal theorem] Suppose for contradiction that $f(x_1, \ldots, x_n) = 0$ for every $(x_1, \ldots, x_n) \in S = S_1 \times \cdots \times S_n$. For each $j = 1, \ldots, n$, define the vanishing polynomial \begin{align*} g_j(X_j) = \prod_{s \in S_j}(X_j - s) \in \mathbb{F}[X_j], \end{align*} which is monic of degree $|S_j| = d_j + 1$. Since $f$ vanishes on $S$, [Membership in the Vanishing Ideal](/theorems/2616) provides polynomials $h_1, \ldots, h_n \in \mathbb{F}[X_1, \ldots, X_n]$ such that \begin{align*} f = \sum_{j=1}^{n} h_j g_j, \end{align*} with the degree bound $\deg h_j \leq \deg f - |S_j| = d - (d_j + 1)$ for each $j$, where $d = \sum_{i=1}^n d_i$. [guided] The [Membership in the Vanishing Ideal](/theorems/2616) theorem requires: (i) $\mathbb{F}$ is a field, (ii) $S_1, \ldots, S_n$ are non-empty finite subsets of $\mathbb{F}$, (iii) $f$ vanishes on $S_1 \times \cdots \times S_n$. All three conditions hold by hypothesis and our assumption. The theorem produces the decomposition $f = \sum h_j g_j$ with $\deg h_j \leq \deg f - |S_j|$. Let us compute the degree bound explicitly. We have $\deg f = d = \sum_{i=1}^n d_i$ and $|S_j| = d_j + 1$, so \begin{align*} \deg h_j \leq d - (d_j + 1) = \sum_{i=1}^n d_i - d_j - 1 = \sum_{i \neq j} d_i - 1. \end{align*} This bound will be the source of the contradiction: it is too restrictive for any $h_j g_j$ to contain the monomial $X_1^{d_1} \cdots X_n^{d_n}$. [/guided] [/step] [step:Show no summand $h_j g_j$ can produce the monomial $X_1^{d_1} \cdots X_n^{d_n}$] Since the coefficient of $X_1^{d_1} \cdots X_n^{d_n}$ in $f$ is nonzero by hypothesis, and $f = \sum_{j=1}^n h_j g_j$, there must exist some index $j$ for which $h_j g_j$ contributes a nonzero coefficient to $X_1^{d_1} \cdots X_n^{d_n}$. Fix such a $j$. Since $g_j(X_j) = \prod_{s \in S_j}(X_j - s)$ is a polynomial in $X_j$ alone of degree $d_j + 1$, the product $h_j g_j$ can produce the monomial $X_1^{d_1} \cdots X_n^{d_n}$ only if $h_j$ contains a monomial of the form \begin{align*} X_1^{d_1} \cdots X_j^{e_j} \cdots X_n^{d_n} \end{align*} for some $e_j \geq 0$ with $e_j + (\text{a contribution of at least } d_j + 1 - e_j' \text{ from } g_j) = d_j$. More precisely, since $g_j$ has degree $d_j + 1$ in $X_j$ and the monomial we need has exponent $d_j$ in $X_j$, the factor $h_j$ must contribute exponent $e_j \leq d_j$ in $X_j$. But $g_j$ contributes at least degree $1$ in $X_j$ to any monomial in $h_j g_j$ that involves $g_j$ (since the lowest-degree term of $g_j$ in $X_j$ is the constant term $\prod_{s \in S_j}(-s)$, and all other terms have degree $\geq 1$ in $X_j$). However, what matters is the total degree constraint. The total degree of such a monomial in $h_j$ is at least $\sum_{i \neq j} d_i$ (from the exponents $d_i$ in all variables $X_i$ with $i \neq j$, since $g_j$ contributes to the $X_j$-exponent but not to any other variable). Therefore \begin{align*} \deg h_j \geq \sum_{i \neq j} d_i. \end{align*} But the degree bound from the decomposition gives $\deg h_j \leq \sum_{i \neq j} d_i - 1$. This is a contradiction, since $\sum_{i \neq j} d_i > \sum_{i \neq j} d_i - 1$. [guided] The heart of the argument is a degree-counting obstruction. We need to show that the monomial $X_1^{d_1} \cdots X_n^{d_n}$ cannot appear in any $h_j g_j$. Consider a fixed index $j$. The polynomial $g_j$ involves only the variable $X_j$, so in any monomial of the product $h_j g_j$, the exponents in the variables $X_i$ with $i \neq j$ come entirely from $h_j$. For the product $h_j g_j$ to contain the monomial $X_1^{d_1} \cdots X_n^{d_n}$, the factor $h_j$ must supply at least exponent $d_i$ in each variable $X_i$ with $i \neq j$. This means $h_j$ must contain a monomial whose exponents in $X_i$ (for $i \neq j$) are at least $d_i$. The total degree of such a monomial in $h_j$ is at least \begin{align*} \sum_{i \neq j} d_i + e_j \geq \sum_{i \neq j} d_i, \end{align*} where $e_j \geq 0$ is the exponent in $X_j$. Therefore $\deg h_j \geq \sum_{i \neq j} d_i$. On the other hand, the decomposition from [Membership in the Vanishing Ideal](/theorems/2616) gives $\deg h_j \leq \sum_{i \neq j} d_i - 1$. These two inequalities are incompatible: \begin{align*} \sum_{i \neq j} d_i \leq \deg h_j \leq \sum_{i \neq j} d_i - 1, \end{align*} which is impossible. So no $h_j g_j$ can contribute the monomial $X_1^{d_1} \cdots X_n^{d_n}$. Since $f = \sum_{j=1}^n h_j g_j$ and no summand can produce this monomial, the coefficient of $X_1^{d_1} \cdots X_n^{d_n}$ in $f$ must be zero. But this contradicts the hypothesis that the coefficient is nonzero. [/guided] [/step] [step:Conclude that $f$ does not vanish identically on $S$] The assumption that $f$ vanishes on $S$ leads to a contradiction: the decomposition $f = \sum h_j g_j$ cannot account for the monomial $X_1^{d_1} \cdots X_n^{d_n}$ with its nonzero coefficient. Therefore $f$ is not identically zero on $S = S_1 \times \cdots \times S_n$; there exists at least one point $(x_1, \ldots, x_n) \in S$ with $f(x_1, \ldots, x_n) \neq 0$. [/step]