[proofplan]
We prove directly from the definition of the vanishing ideal. First we show that the zero polynomial belongs to $I(X)$. Then we verify closure under addition and additive inverses by evaluating at an arbitrary point of $X$. Finally we verify absorption under multiplication by an arbitrary polynomial in $k[x_1,\dots,x_n]$, which is exactly the remaining condition for being an ideal.
[/proofplan]
[step:Fix the polynomial ring and the vanishing condition]
Let
\begin{align*}
R:=k[x_1,\dots,x_n]
\end{align*}
be the [polynomial ring](/page/Polynomial%20Ring) over $k$. By definition,
\begin{align*}
I(X)=\{f\in R: f(a)=0 \text{ for every } a\in X\}.
\end{align*}
If $X=\varnothing$, this condition is vacuous, so $I(X)=R$, which is an ideal of $R$. Hence assume for the rest of the proof that $X$ is arbitrary, with the understanding that the same pointwise arguments below are also vacuous when $X=\varnothing$.
[/step]
[step:Show that the vanishing set is closed under the additive group operations]
The zero polynomial $0_R\in R$ satisfies $0_R(a)=0$ for every $a\in X$, so $0_R\in I(X)$.
Let $f,g\in I(X)$. For every $a\in X$, the defining property of $I(X)$ gives
\begin{align*}
f(a)=0 \quad \text{and} \quad g(a)=0.
\end{align*}
Therefore
\begin{align*}
(f+g)(a)=f(a)+g(a)=0+0=0.
\end{align*}
Since this holds for every $a\in X$, we have $f+g\in I(X)$.
Let $f\in I(X)$. For every $a\in X$,
\begin{align*}
(-f)(a)=-f(a)=-0=0.
\end{align*}
Thus $-f\in I(X)$. Hence $I(X)$ is an additive subgroup of $R$.
[guided]
The first part of the ideal criterion is that the subset must be an additive subgroup of the ambient ring $R=k[x_1,\dots,x_n]$. We therefore check the three additive facts: the zero element is present, sums stay inside, and additive inverses stay inside.
The zero polynomial $0_R\in R$ evaluates to zero at every point $a\in X$:
\begin{align*}
0_R(a)=0.
\end{align*}
So $0_R$ satisfies the defining condition for membership in $I(X)$, and therefore $0_R\in I(X)$.
Now take $f,g\in I(X)$. This means that both polynomials vanish at every point of $X$. Thus, for an arbitrary point $a\in X$, we have
\begin{align*}
f(a)=0 \quad \text{and} \quad g(a)=0.
\end{align*}
Polynomial evaluation is compatible with addition, so
\begin{align*}
(f+g)(a)=f(a)+g(a)=0+0=0.
\end{align*}
Because the point $a\in X$ was arbitrary, $f+g$ vanishes on all of $X$. Hence $f+g\in I(X)$.
Finally take $f\in I(X)$. For every $a\in X$, we have $f(a)=0$, and evaluation is compatible with additive inverses. Therefore
\begin{align*}
(-f)(a)=-f(a)=-0=0.
\end{align*}
So $-f$ also vanishes on all of $X$, and $-f\in I(X)$. These three checks prove that $I(X)$ is an additive subgroup of $R$.
[/guided]
[/step]
[step:Show that multiplication by any polynomial preserves vanishing on $X$]
Let $h\in R$ and let $f\in I(X)$. For every $a\in X$, the defining property of $I(X)$ gives $f(a)=0$. Since polynomial evaluation is compatible with multiplication,
\begin{align*}
(hf)(a)=h(a)f(a)=h(a)\cdot 0=0.
\end{align*}
Thus $hf$ vanishes at every point of $X$, so $hf\in I(X)$.
[/step]
[step:Conclude that the vanishing set is an ideal]
We have shown that $I(X)$ is an additive subgroup of $R$ and that for every $h\in R$ and every $f\in I(X)$, the product $hf$ lies in $I(X)$. These are precisely the defining closure conditions for an ideal of the commutative ring $R=k[x_1,\dots,x_n]$. Therefore $I(X)$ is an ideal of $k[x_1,\dots,x_n]$.
[/step]
