[proofplan] By the definition of an [affine variety](/page/Affine%20Variety), $X$ is the common zero locus of some possibly infinite set of polynomials. We replace that set by the ideal it generates, which does not change the vanishing set. The [Hilbert Basis Theorem](/theorems/860) then makes this ideal finitely generated, and those finitely many generators give the desired finite list of defining equations. [/proofplan] [step:Write the affine variety as a common zero locus] Let \begin{align*} R := k[x_1, \ldots, x_n]. \end{align*} Since $X \subset \mathbb{A}^n_k$ is an affine variety, there exists a subset $S \subset R$ such that \begin{align*} X = V(S). \end{align*} [guided] Set \begin{align*} R := k[x_1, \ldots, x_n]. \end{align*} The statement assumes that $X$ is an affine variety in $\mathbb{A}^n_k$. By the definition of an affine variety, this means that $X$ is cut out as the common zero locus of some collection of polynomial equations. Thus there is a subset $S \subset R$ such that \begin{align*} X = V(S). \end{align*} The important point is that $S$ is not assumed finite. The rest of the proof is precisely the argument that finitely many equations suffice. [/guided] [/step] [step:Replace the defining set by its generated ideal] Let $I \trianglelefteq R$ be the ideal generated by $S$, so \begin{align*} I := (S). \end{align*} The hypotheses of [citetheorem:9517] are satisfied with the same field $k$, the same integer $n$, and the subset $S \subset k[x_1, \ldots, x_n]$. Therefore \begin{align*} V(S)=V(I). \end{align*} Since $X=V(S)$, it follows that \begin{align*} X=V(I). \end{align*} [guided] The set $S$ may be infinite, so we pass from $S$ to the ideal generated by $S$. Define $I \trianglelefteq R$ by \begin{align*} I := (S), \end{align*} meaning that $I$ is the smallest ideal of $R$ containing every polynomial in $S$. We now apply [citetheorem:9517]. Its hypotheses require a field, a [natural number](/page/Natural%20Number), and a subset of the [polynomial ring](/page/Polynomial%20Ring) over that field. Here the field is the given field $k$, the natural number is the given $n \in \mathbb{N}$, and the subset is exactly $S \subset R = k[x_1, \ldots, x_n]$. The theorem therefore gives \begin{align*} V(S)=V((S)). \end{align*} Since $I=(S)$ by definition, this is \begin{align*} V(S)=V(I). \end{align*} Finally, the previous step gave $X=V(S)$, so substitution gives \begin{align*} X=V(I). \end{align*} The point of this step is that ideals are algebraically better behaved than arbitrary collections of equations, while [citetheorem:9517] guarantees that this replacement does not change the geometric zero locus. [/guided] [/step] [step:Use the Hilbert Basis Theorem to choose finitely many ideal generators] The field $k$ is Noetherian as a ring, since its only ideals are $(0)$ and $k$. We use the Hilbert Basis Theorem in the following form: if $A$ is a Noetherian ring, then $A[t]$ is a Noetherian ring. Applying this theorem iteratively to \begin{align*} k,\quad k[x_1],\quad k[x_1,x_2],\quad \ldots,\quad k[x_1,\ldots,x_n] \end{align*} shows that \begin{align*} R = k[x_1, \ldots, x_n] \end{align*} is Noetherian. Hence every ideal of $R$ is finitely generated, so the ideal $I \trianglelefteq R$ is finitely generated. Choose generators $g_1, \ldots, g_r \in I$ such that \begin{align*} I=(g_1, \ldots, g_r). \end{align*} If $r \geq 1$, set $m:=r$ and $f_i:=g_i$ for $1 \leq i \leq r$. If $r=0$, then $I=(0)$; set $m:=1$ and $f_1:=0 \in R$. In either case, there exist $m \in \mathbb{N}$ and $f_1,\ldots,f_m \in R$ such that \begin{align*} I=(f_1,\ldots,f_m). \end{align*} [guided] We now need finiteness. The external algebraic result used here is the Hilbert Basis Theorem, in the precise form: if $A$ is a Noetherian ring, then the polynomial ring $A[t]$ is Noetherian. First we verify the starting hypothesis. The field $k$ is Noetherian as a ring because its only ideals are $(0)$ and $k$, so every ideal of $k$ is generated by finitely many elements. Therefore the Hilbert Basis Theorem applies to $A=k$ and gives that $k[x_1]$ is Noetherian. Applying the same theorem again to $A=k[x_1]$ gives that $k[x_1,x_2]$ is Noetherian. Repeating this argument $n$ times yields that \begin{align*} R = k[x_1, \ldots, x_n] \end{align*} is Noetherian. By the definition of a Noetherian ring, every ideal in a Noetherian ring is finitely generated. Since $I \trianglelefteq R$ is an ideal and $R$ is Noetherian, there are elements $g_1,\ldots,g_r \in I$ such that \begin{align*} I=(g_1,\ldots,g_r). \end{align*} If $r \geq 1$, define $m:=r$ and define $f_i:=g_i$ for each index $i \in \{1,\ldots,r\}$. Then \begin{align*} I=(f_1,\ldots,f_m). \end{align*} If $r=0$, then the generated ideal is the zero ideal $I=(0)$. Since the theorem statement requires $m \in \mathbb{N}$ and Androma's convention is that natural numbers start at $1$, define $m:=1$ and $f_1:=0 \in R$. Then \begin{align*} I=(f_1)=(0). \end{align*} Thus in all cases there exist $m \in \mathbb{N}$ and polynomials $f_1,\ldots,f_m \in R$ such that \begin{align*} I=(f_1,\ldots,f_m). \end{align*} This is the only place where the Hilbert Basis Theorem is used: it converts the arbitrary ideal $I$ into one generated by finitely many equations. [/guided] [/step] [step:Identify the vanishing set of the finite generating list] Since $I=(f_1,\ldots,f_m)$, applying [citetheorem:9517] to the finite subset $\{f_1,\ldots,f_m\}\subset R$ gives \begin{align*} V(f_1,\ldots,f_m)=V((f_1,\ldots,f_m)). \end{align*} Because $(f_1,\ldots,f_m)=I$, this becomes \begin{align*} V(f_1,\ldots,f_m)=V(I). \end{align*} Combining this with $X=V(I)$ gives \begin{align*} X=V(f_1,\ldots,f_m). \end{align*} This proves that the affine variety $X$ is defined by finitely many polynomial equations. [guided] It remains to translate the finite generation of the ideal back into a finite list of equations. We know from the previous step that \begin{align*} I=(f_1,\ldots,f_m). \end{align*} Apply [citetheorem:9517] to the subset $\{f_1,\ldots,f_m\}\subset R$. The hypotheses are satisfied because the field is $k$, the integer is $n$, and each $f_i$ lies in $R=k[x_1,\ldots,x_n]$. Therefore \begin{align*} V(f_1,\ldots,f_m)=V((f_1,\ldots,f_m)). \end{align*} Since $(f_1,\ldots,f_m)=I$, the right-hand side is $V(I)$, so \begin{align*} V(f_1,\ldots,f_m)=V(I). \end{align*} The generated-ideal step already proved \begin{align*} X=V(I). \end{align*} Substituting this equality gives \begin{align*} X=V(f_1,\ldots,f_m). \end{align*} Thus the same affine variety originally defined by a possibly infinite family of equations is defined by the finite list $f_1,\ldots,f_m$. [/guided] [/step]