[proofplan] We prove the identity algebraically by showing that the ideal generated by $F_1,\dots,F_m$ is the whole [polynomial ring](/page/Polynomial%20Ring). If that ideal were proper, a maximal ideal containing it would give a finite type [field extension](/page/Field%20Extension) of $\mathbb{C}$; [Zariski's lemma](/theorems/2872) and algebraic closedness identify that residue field with $\mathbb{C}$, producing a common zero of the $F_j$. This contradiction gives $1\in (F_1,\dots,F_m)$, and writing this ideal membership explicitly produces the desired polynomials $Q_j$. [/proofplan] [step:Assume the generated ideal is proper and pass to a maximal residue field] Let $R := \mathbb{C}[z_1,\dots,z_n]$ denote the [polynomial ring](/page/Polynomial%20Ring) in the coordinate functions $z_1,\dots,z_n$, and define the ideal \begin{align*} I := (F_1,\dots,F_m) \subset R. \end{align*} Suppose, for contradiction, that $I \neq R$. Since $R$ is a unital commutative ring and $I$ is a proper ideal, the standard maximal ideal existence theorem gives a maximal ideal $\mathfrak{m} \subset R$ such that $I \subset \mathfrak{m}$. Define the quotient map \begin{align*} \pi: R &\to K := R/\mathfrak{m} \\ G &\mapsto G + \mathfrak{m}. \end{align*} Because $\mathfrak{m}$ is maximal, $K$ is a field. The canonical structure map \begin{align*} \iota: \mathbb{C} &\to K \\ c &\mapsto c + \mathfrak{m} \end{align*} is injective: if $c \in \mathbb{C}\setminus\{0\}$, then $c$ is a unit in $R$, so $c \notin \mathfrak{m}$ because $\mathfrak{m}$ is proper. Moreover $K$ is generated as a $\mathbb{C}$-algebra by the elements $\pi(z_1),\dots,\pi(z_n)$, since every element of $R$ is a polynomial in $z_1,\dots,z_n$ with coefficients in $\mathbb{C}$, where $\mathbb{C}$ acts on $K$ through $\iota$. [guided] The contradiction begins by translating the failure of the desired identity into ideal language. Let \begin{align*} R := \mathbb{C}[z_1,\dots,z_n] \end{align*} denote the [polynomial ring](/page/Polynomial%20Ring), and define \begin{align*} I := (F_1,\dots,F_m) \subset R. \end{align*} If the theorem were false at the ideal level, then $1 \notin I$, equivalently $I \neq R$. In a unital commutative ring, every proper ideal is contained in a maximal ideal, so there exists a maximal ideal $\mathfrak{m} \subset R$ with $I \subset \mathfrak{m}$. We then form the residue field at this maximal ideal. Define \begin{align*} \pi: R &\to K := R/\mathfrak{m} \\ G &\mapsto G + \mathfrak{m}. \end{align*} The quotient $K$ is a field precisely because $\mathfrak{m}$ is maximal. The scalar field embeds into this quotient through the map \begin{align*} \iota: \mathbb{C} &\to K \\ c &\mapsto c + \mathfrak{m}. \end{align*} This map is injective: if $c \neq 0$, then $c$ is a unit in $R$, and a proper ideal cannot contain a unit, so $c \notin \mathfrak{m}$. Thus we may regard $\mathbb{C}$ as a subfield of $K$ through $\iota$. The elements $\pi(z_1),\dots,\pi(z_n)$ generate $K$ as a $\mathbb{C}$-algebra: every element of $K$ is represented by some polynomial $G \in R$, and $G$ is obtained by evaluating a polynomial expression in the coordinate functions $z_1,\dots,z_n$. [/guided] [/step] [step:Use Zariski's lemma and algebraic closedness to identify the residue field with $\mathbb{C}$] By [Zariski's lemma](/theorems/2872), every field that is finitely generated as an algebra over a field is algebraic over that base field. Applying this to the [field extension](/page/Field%20Extension) $\iota(\mathbb{C}) \subset K$, the field $K$ is algebraic over $\iota(\mathbb{C})$. Since $\mathbb{C}$ is algebraically closed and $\iota: \mathbb{C}\to K$ is an isomorphism from $\mathbb{C}$ onto $\iota(\mathbb{C})$, the field $\iota(\mathbb{C})$ is algebraically closed. Hence every element of $K$ lies in $\iota(\mathbb{C})$. Therefore there are complex numbers $a_1,\dots,a_n \in \mathbb{C}$ such that \begin{align*} \pi(z_i) = a_i \quad \text{for each } i \in \{1,\dots,n\}. \end{align*} Let $a := (a_1,\dots,a_n) \in \mathbb{C}^n$. [guided] The quotient field $K$ is not an arbitrary [field extension](/page/Field%20Extension) of $\mathbb{C}$; it is generated by finitely many elements $\pi(z_1),\dots,\pi(z_n)$ as a $\mathbb{C}$-algebra through the injective structure map $\iota: \mathbb{C}\to K$. [Zariski's lemma](/theorems/2872) says that if a field is finitely generated as an algebra over a base field, then it is algebraic over that base field. Its hypotheses are satisfied here because $K$ is a field and is generated as an algebra over $\iota(\mathbb{C})$ by the finite set $\{\pi(z_1),\dots,\pi(z_n)\}$. Hence $K$ is algebraic over $\iota(\mathbb{C})$. Now we use the specific base field in the theorem. Since $\mathbb{C}$ is algebraically closed and $\iota$ identifies $\mathbb{C}$ isomorphically with $\iota(\mathbb{C})$, the subfield $\iota(\mathbb{C})$ is algebraically closed. If $\alpha \in K$, then $\alpha$ is algebraic over $\iota(\mathbb{C})$, so its [minimal polynomial](/page/Minimal%20Polynomial) over $\iota(\mathbb{C})$ must have degree $1$. Thus $\alpha \in \iota(\mathbb{C})$. In particular, for each coordinate generator there exists $a_i \in \mathbb{C}$ such that \begin{align*} \pi(z_i) = a_i, \qquad i \in \{1,\dots,n\}. \end{align*} We collect these scalars into the point \begin{align*} a := (a_1,\dots,a_n) \in \mathbb{C}^n. \end{align*} This is the point that will contradict the hypothesis that the polynomials have no common zero. [/guided] [/step] [step:Evaluate the generators at the residue point to obtain a common zero] For every polynomial $G \in R$, the homomorphism property of $\pi$ and the identities $\pi(z_i)=a_i$ give \begin{align*} \pi(G) = G(a) \in \mathbb{C} \subset K. \end{align*} Since $I \subset \mathfrak{m}$ and $F_j \in I$ for each $j \in \{1,\dots,m\}$, we have $\pi(F_j)=0$ in $K$. Hence \begin{align*} F_j(a)=0 \quad \text{for each } j \in \{1,\dots,m\}. \end{align*} This contradicts the hypothesis that $F_1,\dots,F_m$ have no common zero in $\mathbb{C}^n$. Therefore $I=R$. [guided] We now interpret the quotient map as evaluation at the point $a$. Let $G \in R$ be any polynomial. Since $G$ is built from the variables $z_1,\dots,z_n$ using addition, multiplication, and complex scalar multiplication, and since $\pi$ is a $\mathbb{C}$-algebra homomorphism, substituting $\pi(z_i)=a_i$ gives \begin{align*} \pi(G)=G(a) \in \mathbb{C} \subset K. \end{align*} Apply this to the generators $F_j$. Because $I=(F_1,\dots,F_m)$ is contained in $\mathfrak{m}$, each $F_j$ belongs to $\mathfrak{m}$. Therefore its image in the quotient field is zero: \begin{align*} \pi(F_j)=0. \end{align*} Using the evaluation identity for $G=F_j$, this becomes \begin{align*} F_j(a)=0 \quad \text{for each } j \in \{1,\dots,m\}. \end{align*} Thus $a \in \mathbb{C}^n$ is a common zero of $F_1,\dots,F_m$, contradicting the theorem's hypothesis. The assumption that $I$ was proper is therefore false, so $I=R$. [/guided] [/step] [step:Write the unit ideal membership as the required Bezout identity] Since $I=R$, the element $1 \in R$ belongs to the ideal generated by $F_1,\dots,F_m$. By the definition of a finitely generated ideal, there exist polynomials $Q_1,\dots,Q_m \in R$ such that \begin{align*} 1 = \sum_{j=1}^m Q_jF_j. \end{align*} This is exactly the required identity in $\mathbb{C}[z_1,\dots,z_n]$. [/step]