[proofplan] The proof first identifies the Zariski closure of $Y$ with the common zero locus $V(I(Y))$. This follows directly from the definition of the Zariski topology and the order-reversing relation between containment of subsets and containment of vanishing ideals. Then density becomes the assertion that $V(I(Y))=\mathbb A_k^n$. Finally, because $k$ is infinite, the only polynomial in $k[x_1,\dots,x_n]$ vanishing on all of $k^n$ is the zero polynomial, so this equality is equivalent to $I(Y)=(0)$. [/proofplan] [step:Identify the closure of $Y$ as the zero locus of its vanishing ideal] Let $R:=k[x_1,\dots,x_n]$. For an ideal $J\trianglelefteq R$, define \begin{align*} V(J):=\{a\in k^n: f(a)=0 \text{ for every } f\in J\}. \end{align*} By definition of $I(Y)$, every $f\in I(Y)$ vanishes at every point of $Y$, so $Y\subset V(I(Y))$. Hence $V(I(Y))$ is a Zariski closed subset of $\mathbb A_k^n$ containing $Y$. Now let $C\subset \mathbb A_k^n$ be any Zariski closed subset with $Y\subset C$. Since closed subsets are common zero loci of families of polynomials, there is an ideal $J\trianglelefteq R$ such that $C=V(J)$. If $f\in J$ and $y\in Y$, then $y\in C=V(J)$, so $f(y)=0$. Therefore $J\subset I(Y)$. It follows that every point annihilating all polynomials in $I(Y)$ also annihilates all polynomials in $J$, hence \begin{align*} V(I(Y))\subset V(J)=C. \end{align*} Thus $V(I(Y))$ is contained in every Zariski closed subset containing $Y$, and since it is itself closed and contains $Y$, it is the Zariski closure $\overline{Y}$. [guided] Let $R:=k[x_1,\dots,x_n]$. For an ideal $J\trianglelefteq R$, define its zero locus by \begin{align*} V(J):=\{a\in k^n: f(a)=0 \text{ for every } f\in J\}. \end{align*} We want to identify the smallest Zariski [closed set](/page/Closed%20Set) containing $Y$. The natural candidate is $V(I(Y))$, because $I(Y)$ was defined to consist exactly of the polynomials that vanish on every point of $Y$. First, $Y\subset V(I(Y))$. Indeed, if $y\in Y$ and $f\in I(Y)$, then the defining property of $I(Y)$ gives $f(y)=0$. Therefore $y$ satisfies all equations from $I(Y)$, and hence $y\in V(I(Y))$. Since $V(I(Y))$ is a common zero locus of polynomials, it is Zariski closed. Now we prove minimality. Let $C\subset \mathbb A_k^n$ be any Zariski closed subset such that $Y\subset C$. By the definition of the [Zariski topology on affine space](/theorems/9408), there is an ideal $J\trianglelefteq R$ with $C=V(J)$. Take any $f\in J$. Since every point $y\in Y$ lies in $C=V(J)$, the polynomial $f$ vanishes at every $y\in Y$. Hence $f\in I(Y)$. This proves \begin{align*} J\subset I(Y). \end{align*} Zero loci reverse inclusions of ideals: if $J\subset I(Y)$, then a point satisfying every equation in $I(Y)$ certainly satisfies every equation in $J$. Therefore \begin{align*} V(I(Y))\subset V(J)=C. \end{align*} So $V(I(Y))$ is a closed set containing $Y$ and is contained in every closed set containing $Y$. This is precisely the Zariski closure of $Y$: \begin{align*} \overline{Y}=V(I(Y)). \end{align*} [/guided] [/step] [step:Reduce density to polynomials vanishing on all of affine space] By the previous step, \begin{align*} \overline{Y}=V(I(Y)). \end{align*} Therefore $Y$ is Zariski dense in $\mathbb A_k^n$ if and only if $\overline{Y}=\mathbb A_k^n$, which is equivalent to \begin{align*} V(I(Y))=k^n. \end{align*} Thus it remains to characterize when every polynomial in $I(Y)$ vanishes on all of $k^n$. [/step] [step:Use infinitude of $k$ to rule out nonzero polynomials vanishing everywhere] We use the following elementary polynomial identity lemma. [claim:Polynomial identity lemma over an infinite field] Let $m\in \mathbb N$. If $F\in k[x_1,\dots,x_m]$ satisfies $F(a)=0$ for every $a\in k^m$, then $F=0$. [/claim] [proof] If $m=0$, then $k[x_1,\dots,x_m]=k$. A constant $F\in k$ vanishes at the unique point of $k^0$ precisely when $F=0$. Assume $m\ge 1$. We argue by induction on $m$. For $m=1$, a nonzero polynomial in $k[x_1]$ has degree $d\ge 0$ and has at most $d$ roots in the field $k$. Since $k$ is infinite, it cannot vanish at every element of $k$. Hence a polynomial in $k[x_1]$ vanishing on all of $k$ must be zero. Now assume the result holds for $m-1$, where $m\ge 2$. Write $F$ as a polynomial in $x_m$ with coefficients in $k[x_1,\dots,x_{m-1}]$: \begin{align*} F=\sum_{j=0}^d F_j x_m^j \end{align*} for some integer $d\ge 0$ and coefficient polynomials $F_j\in k[x_1,\dots,x_{m-1}]$. Fix $b=(b_1,\dots,b_{m-1})\in k^{m-1}$. Define the one-variable polynomial $F_b\in k[x_m]$ by \begin{align*} F_b(x_m):=F(b_1,\dots,b_{m-1},x_m). \end{align*} By hypothesis, $F_b(t)=0$ for every $t\in k$. The one-variable case gives $F_b=0$, so each coefficient satisfies $F_j(b)=0$. Since this holds for every $b\in k^{m-1}$, the induction hypothesis gives $F_j=0$ for every $j$. Therefore $F=0$. [/proof] Suppose first that $I(Y)=(0)$. Then $V(I(Y))=V((0))=k^n$, because the zero polynomial vanishes at every point of $k^n$. Hence $Y$ is dense by the previous step. Conversely, suppose $Y$ is dense. Then $V(I(Y))=k^n$. Let $f\in I(Y)$. Since $V(I(Y))=k^n$, the polynomial $f$ vanishes at every point of $k^n$. By the polynomial identity lemma over the infinite field $k$, applied with $m=n$ and $F=f$, we get $f=0$. Hence every element of $I(Y)$ is zero, so $I(Y)=(0)$. [/step] [step:Conclude the equivalence] We have shown that \begin{align*} Y \text{ is Zariski dense in } \mathbb A_k^n \iff V(I(Y))=k^n \iff I(Y)=(0). \end{align*} This proves the desired density criterion. [/step]