[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 $R$.[/step]

[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: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]