[proofplan] We prove directly from the quotient description of the [coordinate ring](/page/Coordinate%20Ring). Let $R := k[x_1,\ldots,x_n]$ and suppose a residue class in $R/I(X)$ is nilpotent. Evaluating the corresponding power at every point of $X$ shows that the original polynomial vanishes at every point of $X$, because fields have no nonzero nilpotent elements. Hence the residue class was already zero, so the quotient has no nonzero nilpotents. [/proofplan] [step:Translate nilpotence in the coordinate ring into vanishing on $X$] Let $R := k[x_1,\ldots,x_n]$, and let $\pi: R \to R/I(X)$ be the quotient homomorphism. We must prove that every nilpotent element of $R/I(X)$ is zero. Let $\overline{f} \in R/I(X)$ be a nilpotent element, where $f \in R$ is a polynomial and $\overline{f} := \pi(f)$ denotes its residue class modulo $I(X)$. By nilpotence, there exists $m \in \mathbb{N}$ such that \begin{align*} \overline{f}^{m} = 0 \end{align*} in $R/I(X)$. Since multiplication in the quotient is induced by multiplication in $R$, this means \begin{align*} \overline{f^{m}} = 0. \end{align*} By the definition of the zero residue class, this is equivalent to \begin{align*} f^{m} \in I(X). \end{align*} Using the definition of $I(X)$, we obtain, for every point $a \in X$, \begin{align*} f^{m}(a)=0. \end{align*} Since evaluation at $a$ is compatible with multiplication of polynomials, this gives \begin{align*} f(a)^{m}=0 \end{align*} in the field $k$. [/step] [step:Use that a field has no nonzero nilpotent elements] Fix $a \in X$. From the previous step, $f(a)^m=0$ in $k$. We claim that $f(a)=0$. Indeed, if $f(a) \neq 0$, then $f(a)$ is invertible in the field $k$. Multiplying the equality $f(a)^m=0$ by $f(a)^{-1}$ repeatedly gives \begin{align*} 1 = 0, \end{align*} which contradicts the field axioms. Therefore $f(a)=0$. Since $a \in X$ was arbitrary, $f$ vanishes at every point of $X$. Hence, by the definition of the vanishing ideal, \begin{align*} f \in I(X). \end{align*} [guided] We now use the only algebraic property of the base field needed in the argument: fields have no nonzero nilpotent elements. Fix a point $a \in X$. The previous step gave \begin{align*} f(a)^m = 0 \end{align*} as an equality in $k$. Why does this force $f(a)=0$? Suppose instead that $f(a) \neq 0$. Since $k$ is a field, every nonzero element of $k$ is invertible, so $f(a)^{-1}$ exists. Multiplying the equality $f(a)^m=0$ by $f(a)^{-1}$ gives \begin{align*} f(a)^{m-1}=0. \end{align*} Repeating this cancellation $m$ times gives \begin{align*} 1=0, \end{align*} which is impossible in a field. Therefore the supposition $f(a)\neq 0$ is false, and we must have \begin{align*} f(a)=0. \end{align*} Because the point $a \in X$ was arbitrary, the polynomial $f$ vanishes at every point of $X$. By the definition \begin{align*} I(X) := \{g \in R : g(b)=0 \text{ for every } b \in X\}, \end{align*} this proves \begin{align*} f \in I(X). \end{align*} [/guided] [/step] [step:Conclude that the coordinate ring is reduced] Since $f \in I(X)$, its residue class modulo $I(X)$ is zero: \begin{align*} \overline{f}=0 \end{align*} in $R/I(X)$. Thus every nilpotent element of $R/I(X)$ is zero. Recalling that $k[X] := R/I(X)$, this proves that $k[X]$ has no nonzero nilpotent elements. Therefore $k[X]$ is reduced. [/step]