[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*}
[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]
[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]
