[proofplan] The point is that an ideal in a [polynomial ring](/page/Polynomial%20Ring) has finitely many generators, so an arbitrary family of equations defining $V(I)$ can be replaced by the finite family of generator equations. After choosing generators using [Hilbert's Basis Theorem](/theorems/2904), we prove directly from the definition of $V(I)$ that a point annihilates every element of $I$ exactly when it annihilates the chosen generators. The only minor convention is the zero ideal: if a generating list is empty, we replace it by the single generator $0$. [/proofplan]

[step:Choose a finite generating set for the ideal] Let \begin{align*} R:=k[x_1,\dots,x_n]. \end{align*} By [Hilbert's Basis Theorem](/theorems/2907) (citing a result not yet in the wiki: Hilbert's Basis Theorem), the polynomial ring $R$ is Noetherian because $k$ is a field. Hence the ideal $I\trianglelefteq R$ is finitely generated. If $I=(0)$, set $r:=1$ and $f_1:=0\in R$. Then $I=(f_1)$. If $I\ne (0)$, choose generators $f_1,\dots,f_r\in R$ with $r\in\mathbb N$ such that \begin{align*} I=(f_1,\dots,f_r). \end{align*} Thus in every case we have polynomials $f_1,\dots,f_r\in R$ with $r\in\mathbb N$ and \begin{align*} I=(f_1,\dots,f_r). \end{align*} [/step]

[step:Show that points vanishing on the ideal vanish on the generators] Let $a\in V(I)$. By definition of $V(I)$, every polynomial in $I$ vanishes at $a$. Since $f_i\in I$ for every $i\in\{1,\dots,r\}$, we have \begin{align*} f_i(a)=0 \end{align*} for every $i\in\{1,\dots,r\}$. Therefore $a\in V(f_1,\dots,f_r)$. This proves \begin{align*} V(I)\subset V(f_1,\dots,f_r). \end{align*} [/step]

[step:Show that points vanishing on the generators vanish on the whole ideal]Let $a\in V(f_1,\dots,f_r)$. Then \begin{align*} f_i(a)=0 \end{align*} for every $i\in\{1,\dots,r\}$. Because $I=(f_1,\dots,f_r)$, every polynomial $g\in I$ admits coefficients $h_1,\dots,h_r\in R$ such that \begin{align*} g=\sum_{i=1}^r h_i f_i. \end{align*} Evaluating this identity at $a$ gives \begin{align*} g(a)=\sum_{i=1}^r h_i(a)f_i(a). \end{align*} Since each $f_i(a)=0$, the right-hand side is $0$, so $g(a)=0$. As $g\in I$ was arbitrary, $a\in V(I)$. Hence \begin{align*} V(f_1,\dots,f_r)\subset V(I). \end{align*}[/step]

[guided]We must prove that the finitely many equations $f_1(a)=0,\dots,f_r(a)=0$ force every equation from the ideal $I$ to vanish at $a$. The reason is precisely what it means for $f_1,\dots,f_r$ to generate $I$: every element of $I$ is a polynomial linear combination of these generators. Let $a\in V(f_1,\dots,f_r)$. By definition of $V(f_1,\dots,f_r)$, this means \begin{align*} f_i(a)=0 \end{align*} for every $i\in\{1,\dots,r\}$. Now let $g\in I$ be arbitrary. Since \begin{align*} I=(f_1,\dots,f_r), \end{align*} there exist polynomials $h_1,\dots,h_r\in k[x_1,\dots,x_n]$ such that \begin{align*} g=\sum_{i=1}^r h_i f_i. \end{align*} Evaluation at the point $a\in\mathbb A_k^n$ is compatible with addition and multiplication of polynomials, so evaluating the displayed identity at $a$ gives \begin{align*} g(a)=\sum_{i=1}^r h_i(a)f_i(a). \end{align*} Each factor $f_i(a)$ is $0$, so every summand $h_i(a)f_i(a)$ is $0$. Therefore \begin{align*} g(a)=0. \end{align*} Because $g\in I$ was arbitrary, every polynomial in $I$ vanishes at $a$. By the definition of $V(I)$, this says $a\in V(I)$. Hence \begin{align*} V(f_1,\dots,f_r)\subset V(I). \end{align*}[/guided]

[step:Conclude equality of the two vanishing sets] The two inclusions proved above give \begin{align*} V(I)=V(f_1,\dots,f_r). \end{align*} Thus the affine algebraic set defined by the possibly infinite system of equations indexed by $I$ is defined by the finite system $f_1=\cdots=f_r=0$. [/step]