[proofplan] The coordinate ring is a quotient of the polynomial algebra $k[x_1,\dots,x_n]$ by the vanishing ideal $I(X)$. Finite generation follows because the residue classes of the coordinate functions generate the quotient as a $k$-algebra. Reducedness follows directly from the pointwise definition of $I(X)$: if $F^m$ vanishes on every point of $X$, then $F$ vanishes on every point of $X$ because a nonzero element of a field is invertible and hence cannot be nilpotent. [/proofplan]

[step:Generate the coordinate ring by the residue classes of the coordinate functions]Let \begin{align*} A:=k[x_1,\dots,x_n] \end{align*} and let \begin{align*} \pi:A&\to k[X] \end{align*} be the quotient map, so $\pi(f)=\bar f$ denotes the residue class of $f\in A$ modulo $I(X)$. For each $i\in\{1,\dots,n\}$, define \begin{align*} \bar x_i:=\pi(x_i)\in k[X]. \end{align*} We claim that $\bar x_1,\dots,\bar x_n$ generate $k[X]$ as a $k$-algebra. Let $\alpha\in k[X]$. Since $\pi$ is surjective by the definition of the [quotient ring](/page/Quotient%20Ring), there exists $F\in A$ such that $\alpha=\pi(F)$. Because $A=k[x_1,\dots,x_n]$, the polynomial $F$ is obtained from the elements $x_1,\dots,x_n$ and scalars in $k$ by finitely many additions and multiplications. Applying the $k$-algebra homomorphism $\pi$, the element $\alpha=\pi(F)$ is obtained from $\bar x_1,\dots,\bar x_n$ and scalars in $k$ by the same finite algebraic expression. Hence \begin{align*} k[X]=k[\bar x_1,\dots,\bar x_n], \end{align*} so $k[X]$ is a finitely generated $k$-algebra.[/step]

[guided]The quotient map \begin{align*} \pi:A&\to k[X] \end{align*} is the map sending each polynomial to its residue class modulo $I(X)$. Thus, if $f\in A$, then $\pi(f)=\bar f\in k[X]$. In particular, for each coordinate function $x_i\in A$, we define its residue class by \begin{align*} \bar x_i:=\pi(x_i). \end{align*} To prove finite generation, we must show that every element of $k[X]$ can be written as a polynomial expression in finitely many chosen elements with coefficients in $k$. The natural chosen elements are the residue classes $\bar x_1,\dots,\bar x_n$, because the [polynomial ring](/page/Polynomial%20Ring) $A=k[x_1,\dots,x_n]$ is generated by $x_1,\dots,x_n$. Let $\alpha\in k[X]$. Since $k[X]$ is the quotient $A/I(X)$, the quotient map $\pi:A\to k[X]$ is surjective. Therefore there is some polynomial $F\in A$ such that \begin{align*} \alpha=\pi(F). \end{align*} The polynomial $F$ is a finite $k$-linear combination of monomials in $x_1,\dots,x_n$. Applying the $k$-algebra homomorphism $\pi$ replaces each $x_i$ by $\bar x_i$ and leaves each scalar in $k$ unchanged. Therefore $\pi(F)$ is a polynomial expression in $\bar x_1,\dots,\bar x_n$ with coefficients in $k$. Since $\alpha=\pi(F)$ was arbitrary, every element of $k[X]$ lies in the $k$-subalgebra generated by $\bar x_1,\dots,\bar x_n$. Hence \begin{align*} k[X]=k[\bar x_1,\dots,\bar x_n]. \end{align*} This proves that $k[X]$ is finitely generated as a $k$-algebra.[/guided]

[step:Prove directly that the vanishing ideal is radical] We prove that $I(X)$ is radical from its pointwise definition. Let $F\in A$, and suppose there exists $m\in\mathbb N$ with $m\ge 1$ such that \begin{align*} F^m\in I(X). \end{align*} By the definition of $I(X)$, for every point $a\in X$ we have \begin{align*} F(a)^m=0 \end{align*} in the field $k$. Fix $a\in X$. If $F(a)\ne 0$, then $F(a)$ is invertible in $k$, and multiplying the equality $F(a)^m=0$ by $F(a)^{-m}$ gives $1=0$, contradicting the field axioms. Hence $F(a)=0$. Since $a\in X$ was arbitrary, $F(a)=0$ for every $a\in X$. Therefore $F\in I(X)$ by the definition of $I(X)$. Hence \begin{align*} I(X)=\sqrt{I(X)}. \end{align*} [/step]

[step:Show every nilpotent element of the coordinate ring is zero] Let $\alpha\in k[X]$ be nilpotent. By definition, there exists $m\in\mathbb N$ with $m\ge 1$ such that \begin{align*} \alpha^m=0. \end{align*} Since $\pi:A\to k[X]$ is surjective, choose $F\in A$ such that $\alpha=\pi(F)=\bar F$. Then \begin{align*} 0=\alpha^m=\pi(F)^m=\pi(F^m). \end{align*} Thus $F^m\in \ker(\pi)$. Since $\pi$ is the quotient map modulo $I(X)$, its kernel is \begin{align*} \ker(\pi)=I(X). \end{align*} Therefore \begin{align*} F^m\in I(X). \end{align*} By the radicality of $I(X)$ established above, this implies \begin{align*} F\in I(X). \end{align*} Hence \begin{align*} \alpha=\pi(F)=0. \end{align*} So the only nilpotent element of $k[X]$ is zero, which means that $k[X]$ is reduced. [/step]

[step:Combine finite generation and reducedness] The first step proves that $k[X]$ is generated as a $k$-algebra by the finite set $\{\bar x_1,\dots,\bar x_n\}$. The preceding step proves that $k[X]$ has no nonzero nilpotent elements. Therefore $k[X]$ is a reduced finitely generated $k$-algebra, as claimed. [/step]