[proofplan] We argue by contradiction. Assuming the common zero at the origin is unique, we construct a polynomial $f$ that vanishes on all of $\mathbb{F}_q^n$ by combining the indicator function for the common zero set (built from the Fermat identity $1 - g^{q-1}$) with a correction term that forces vanishing at the origin. The degree condition $\sum \deg f_i < n$ ensures that the indicator product has total degree strictly less than $n(q-1)$, so the monomial $X_1^{q-1} \cdots X_n^{q-1}$ of degree $n(q-1)$ receives its coefficient entirely from the correction term. Since this coefficient is nonzero, [Alon's Combinatorial Nullstellensatz](/theorems/2617) provides the contradiction. [/proofplan]

[step:Translate so that the assumed unique common zero is at the origin]Suppose for contradiction that $f_1, \ldots, f_m$ have exactly one common zero in $\mathbb{F}_q^n$. After replacing each $f_i(X_1, \ldots, X_n)$ by $f_i(X_1 - z_1, \ldots, X_n - z_n)$ where $(z_1, \ldots, z_n)$ is that common zero, we may assume the unique common zero is $(0, \ldots, 0)$. This translation preserves the degree of each polynomial, so $\sum_{i=1}^m \deg f_i < n$ still holds.[/step]

[guided]The goal is to derive a contradiction from the assumption that $f_1, \ldots, f_m$ share exactly one common zero. Translation by a fixed vector is an affine automorphism of $\mathbb{F}_q^n$ that preserves the degree of each polynomial: if $g(X) = f(X - z)$, then each monomial of $f$ maps to a polynomial of the same total degree in the shifted variables. So we lose nothing by placing the unique zero at the origin, which simplifies the construction in the next step.[/guided]

[step:Construct a polynomial that vanishes on all of $\mathbb{F}_q^n$]Write $P(X) = \prod_{i=1}^m (1 - f_i(X)^{q-1})$ and $Q(X) = \prod_{i=1}^n \prod_{s \in \mathbb{F}_q^\times}(X_i - s)$. By Fermat's little theorem in $\mathbb{F}_q$, every $a \in \mathbb{F}_q^\times$ satisfies $a^{q-1} = 1$, and $0^{q-1} = 0$. Therefore $1 - f_i(x)^{q-1}$ equals $1$ when $f_i(x) = 0$ and $0$ when $f_i(x) \neq 0$, so $P(x) = 1$ if $x$ is a common zero of $f_1, \ldots, f_m$ and $P(x) = 0$ otherwise. For $Q$: the factor $\prod_{s \in \mathbb{F}_q^\times}(X_i - s)$ vanishes at $x_i$ if and only if $x_i \in \mathbb{F}_q^\times$, i.e., $x_i \neq 0$. So the full product $Q(x) = 0$ if and only if some $x_i \neq 0$, i.e., $x \neq (0, \ldots, 0)$. In particular, $Q(0, \ldots, 0) = \prod_{i=1}^n \prod_{s \in \mathbb{F}_q^\times}(-s) \neq 0$. Set $\gamma = 1 / Q(0, \ldots, 0) \in \mathbb{F}_q^\times$ and define \begin{align*} f(X) = P(X) - \gamma\, Q(X). \end{align*} We verify $f$ vanishes on every $x \in \mathbb{F}_q^n$: - **$x = (0, \ldots, 0)$:** Here $P(0) = 1$ (the origin is a common zero) and $\gamma \, Q(0) = 1$ by the choice of $\gamma$, so $f(0) = 0$. - **$x \neq (0, \ldots, 0)$:** Since the origin is the unique common zero, $x$ is not a common zero of $f_1, \ldots, f_m$, so $P(x) = 0$. Since $x \neq 0$, some coordinate $x_i \neq 0$, so $Q(x) = 0$. Therefore $f(x) = 0 - 0 = 0$.[/step]

[guided]The construction of $f$ combines two ingredients. The indicator polynomial $P$ detects the common zero set $Z = \{x \in \mathbb{F}_q^n : f_1(x) = \cdots = f_m(x) = 0\}$: it equals $1$ on $Z$ and $0$ elsewhere. The correction polynomial $Q$ detects the origin: it equals $0$ at every $x \neq 0$ (because some coordinate is nonzero, making the corresponding factor vanish) and is nonzero at $x = 0$ (where each factor evaluates to $\prod_{s \in \mathbb{F}_q^\times}(-s) \neq 0$). By assumption, $Z = \{0\}$. So $P$ and $Q$ have complementary supports: $P(x) = 0$ for $x \neq 0$ (since $x \notin Z$), and $Q(x) = 0$ for $x \neq 0$ (since some $x_i \neq 0$). The only point where either is nonzero is $x = 0$, and there $P(0) = 1$ while $Q(0) \neq 0$. Choosing $\gamma = 1/Q(0)$ makes $f(0) = 1 - 1 = 0$, so $f$ vanishes everywhere on $\mathbb{F}_q^n$.[/guided]

[step:Compute the coefficient of $X_1^{q-1} \cdots X_n^{q-1}$ in $f$]Set $S_i = \mathbb{F}_q$ for each $i$, so $|S_i| = q$ and $d_i = q - 1$. The target monomial is $X_1^{q-1} \cdots X_n^{q-1}$, of total degree $n(q-1)$. **Contribution from $P$.** Each factor $1 - f_i(X)^{q-1}$ has degree $(q-1) \deg f_i$, so $\deg P \leq (q-1) \sum_{i=1}^m \deg f_i$. By hypothesis $\sum_{i=1}^m \deg f_i < n$, so \begin{align*} \deg P \leq (q-1) \sum_{i=1}^m \deg f_i < n(q-1). \end{align*} Since the monomial $X_1^{q-1} \cdots X_n^{q-1}$ has total degree $n(q-1) > \deg P$, the polynomial $P$ contributes zero to the coefficient of this monomial. **Contribution from $Q$.** Each inner product $\prod_{s \in \mathbb{F}_q^\times}(X_i - s)$ is monic of degree $q - 1$ in $X_i$, so $Q$ has total degree $n(q-1)$ and its unique monomial of maximal degree is $X_1^{q-1} \cdots X_n^{q-1}$ with coefficient $1$. Therefore the coefficient of $X_1^{q-1} \cdots X_n^{q-1}$ in $f = P - \gamma Q$ is $0 - \gamma \cdot 1 = -\gamma \neq 0$.[/step]

[guided]Why does the degree bound $\sum \deg f_i < n$ matter here? The indicator product $P$ has degree at most $(q-1)\sum \deg f_i$, and the target monomial has degree $n(q-1)$. The strict inequality $\sum \deg f_i < n$ multiplies by $(q-1)$ to give $(q-1)\sum \deg f_i < n(q-1)$, which means $P$ cannot produce the monomial $X_1^{q-1} \cdots X_n^{q-1}$ -- it simply does not have enough total degree. The correction term $-\gamma Q$ has degree exactly $n(q-1)$, and its top monomial is $X_1^{q-1} \cdots X_n^{q-1}$ with coefficient $-\gamma$. Since no cancellation with $P$ is possible (the degree gap prevents it), the coefficient of $X_1^{q-1} \cdots X_n^{q-1}$ in $f$ is exactly $-\gamma \neq 0$. This is the crux: the degree condition on the $f_i$ creates a gap between $\deg P$ and $n(q-1)$ that protects the leading monomial of $Q$ from cancellation.[/guided]

[step:Apply the Nullstellensatz to reach a contradiction] The polynomial $f \in \mathbb{F}_q[X_1, \ldots, X_n]$ vanishes on $S_1 \times \cdots \times S_n = \mathbb{F}_q^n$ with $|S_i| = q = d_i + 1$ where $d_i = q - 1$. Its total degree is $\max(\deg P, \deg Q) = n(q-1) = \sum_{i=1}^n d_i$, and the coefficient of $X_1^{d_1} \cdots X_n^{d_n}$ is $-\gamma \neq 0$. By [Alon's Combinatorial Nullstellensatz](/theorems/2617), $f$ cannot vanish identically on $S_1 \times \cdots \times S_n$. This contradicts the vanishing established above, so the assumption that $f_1, \ldots, f_m$ have exactly one common zero is false. [/step]