[proofplan] Write $R=k[x_1,\dots,x_n]$ and identify $k[X]$ with the quotient $R/I(X)$. For a point $a\in X$, evaluation at $a$ gives a maximal ideal of $R$ containing $I(X)$, hence a maximal ideal of the quotient. Conversely, every maximal ideal of $k[X]$ pulls back to a maximal ideal of $R$ containing $I(X)$; the [Weak Nullstellensatz](/theorems/2123) identifies it with a unique point ideal $\mathfrak m_a$, and the containment $I(X)\subset \mathfrak m_a$ forces $a\in X$. Finally, distinct points are separated by the coordinate functions. [/proofplan] [step:Define the point ideal in the quotient by evaluation] Let $R:=k[x_1,\dots,x_n]$, and let $\pi:R\to k[X]$ denote the quotient map, so $\pi(f)=\overline f$ for every $f\in R$. Fix $a=(a_1,\dots,a_n)\in X$. Define the evaluation homomorphism $\operatorname{ev}_a:R\to k$ by $\operatorname{ev}_a(f):=f(a)$ for every $f\in R$. Its kernel is \begin{align*} \mathfrak m_a:=\{f\in R:f(a)=0\}. \end{align*} Since every element of $I(X)$ vanishes at every point of $X$, we have \begin{align*} I(X)\subset \mathfrak m_a. \end{align*} Therefore evaluation descends to a well-defined $k$-algebra homomorphism $\overline{\operatorname{ev}}_a:k[X]\to k$ given by $\overline{\operatorname{ev}}_a(\overline f):=f(a)$. This is well-defined because if $\overline f=\overline g$, then $f-g\in I(X)$, hence $(f-g)(a)=0$, so $f(a)=g(a)$. The kernel of $\overline{\operatorname{ev}}_a$ is exactly \begin{align*} \ker \overline{\operatorname{ev}}_a=\{\overline f\in k[X]:f(a)=0\}=\mathfrak m_{a,X}. \end{align*} The map $\overline{\operatorname{ev}}_a$ is surjective because constants in $R$ map to the corresponding elements of $k$. Hence \begin{align*} k[X]/\mathfrak m_{a,X}\cong k. \end{align*} Since the quotient is a field, $\mathfrak m_{a,X}$ is a maximal ideal of $k[X]$. [/step] [step:Pull a maximal ideal of the coordinate ring back to a point ideal] Let $\mathfrak M\trianglelefteq k[X]$ be a maximal ideal. Define its inverse image under the quotient map by \begin{align*} \mathfrak m:=\pi^{-1}(\mathfrak M)\subset R. \end{align*} Then $I(X)\subset \mathfrak m$. Also, the induced quotient map gives an isomorphism \begin{align*} R/\mathfrak m\cong k[X]/\mathfrak M. \end{align*} Since $\mathfrak M$ is maximal, $k[X]/\mathfrak M$ is a field, and therefore $R/\mathfrak m$ is a field. Hence $\mathfrak m$ is a maximal ideal of $R$. By the Weak Nullstellensatz, because $k$ is algebraically closed, there is a unique point $a=(a_1,\dots,a_n)\in \mathbb A_k^n$ such that \begin{align*} \mathfrak m=(x_1-a_1,\dots,x_n-a_n)=\{f\in R:f(a)=0\}. \end{align*} It remains to check that $a\in X$. For an ideal $L\trianglelefteq R$, write $V(L):=\{p\in\mathbb A_k^n:h(p)=0\text{ for every }h\in L\}$. Since $I(X)\subset \mathfrak m$, every $f\in I(X)$ satisfies $f(a)=0$. Thus \begin{align*} a\in V(I(X)). \end{align*} Because $X$ is an affine algebraic set, $X=V(J)$ for some ideal $J\trianglelefteq R$. Since $J\subset I(X)$, the inclusion-reversing property of zero loci gives \begin{align*} V(I(X))\subset V(J)=X. \end{align*} Therefore $a\in X$. Finally, since $\mathfrak m=\pi^{-1}(\mathfrak M)$ and $\mathfrak m=\mathfrak m_a$, the ideal $\mathfrak M$ is the image of $\mathfrak m_a$ in $R/I(X)$: \begin{align*} \mathfrak M=\mathfrak m_a/I(X)=\{\overline f\in k[X]:f(a)=0\}=\mathfrak m_{a,X}. \end{align*} Thus every maximal ideal of $k[X]$ has the form $\mathfrak m_{a,X}$ for some $a\in X$. [guided] We start with a maximal ideal $\mathfrak M$ of the [quotient ring](/page/Quotient%20Ring) $k[X]=R/I(X)$ and recover a point of $X$ from it. The quotient map is $\pi:R\to k[X]$. Define \begin{align*} \mathfrak m:=\pi^{-1}(\mathfrak M). \end{align*} This ideal contains $I(X)$ because every element of $I(X)$ maps to $0$ in the quotient, and $0\in\mathfrak M$. The reason for passing back to $R$ is that [maximal ideals of the polynomial ring](/theorems/2125) over an [algebraically closed field](/page/Algebraically%20Closed%20Field) are controlled by the Weak Nullstellensatz. First we verify that $\mathfrak m$ is maximal in $R$. The quotient map $\pi$ identifies the quotient of $R$ by $\mathfrak m$ with the quotient of $k[X]$ by $\mathfrak M$: \begin{align*} R/\mathfrak m\cong k[X]/\mathfrak M. \end{align*} Since $\mathfrak M$ is maximal, $k[X]/\mathfrak M$ is a field. Therefore $R/\mathfrak m$ is a field, so $\mathfrak m$ is maximal. Now we apply the Weak Nullstellensatz. Its hypotheses are met because $R=k[x_1,\dots,x_n]$ is a [polynomial ring](/page/Polynomial%20Ring) over the algebraically closed field $k$, and $\mathfrak m$ is a maximal ideal of $R$. The theorem gives a unique point $a=(a_1,\dots,a_n)\in \mathbb A_k^n$ such that \begin{align*} \mathfrak m=(x_1-a_1,\dots,x_n-a_n). \end{align*} Equivalently, \begin{align*} \mathfrak m=\{f\in R:f(a)=0\}. \end{align*} We still need to prove that this point lies on $X$, not merely in the ambient [affine space](/page/Affine%20Space). For an ideal $L\trianglelefteq R$, write $V(L):=\{p\in\mathbb A_k^n:h(p)=0\text{ for every }h\in L\}$. Since $I(X)\subset \mathfrak m$, every polynomial in $I(X)$ vanishes at $a$. Thus \begin{align*} a\in V(I(X)). \end{align*} Because $X$ is affine algebraic, there is an ideal $J\trianglelefteq R$ such that $X=V(J)$. Every polynomial in $J$ vanishes on $X$, so $J\subset I(X)$. Zero loci reverse inclusions of ideals, hence \begin{align*} V(I(X))\subset V(J)=X. \end{align*} Therefore $a\in X$. Finally, because $\mathfrak m=\pi^{-1}(\mathfrak M)$, the ideal $\mathfrak M$ is precisely the image of $\mathfrak m$ in the quotient $R/I(X)$. Since $\mathfrak m=\{f\in R:f(a)=0\}$, this image is \begin{align*} \mathfrak M=\{\overline f\in k[X]:f(a)=0\}. \end{align*} This is exactly $\mathfrak m_{a,X}$. [/guided] [/step] [step:Separate distinct points by coordinate functions] Let $a,b\in X$ and suppose \begin{align*} \mathfrak m_{a,X}=\mathfrak m_{b,X}. \end{align*} For each index $i\in\{1,\dots,n\}$, define the polynomial \begin{align*} g_i:=x_i-a_i\in R. \end{align*} Then $g_i(a)=0$, so \begin{align*} \overline{g_i}\in \mathfrak m_{a,X}. \end{align*} By the assumed equality of ideals, \begin{align*} \overline{g_i}\in \mathfrak m_{b,X}. \end{align*} Thus $g_i(b)=0$, which means \begin{align*} b_i-a_i=0. \end{align*} Therefore $a_i=b_i$ for every $i\in\{1,\dots,n\}$, and hence $a=b$. The assignment $a\mapsto \mathfrak m_{a,X}$ is injective. The previous step proved surjectivity onto the maximal ideals of $k[X]$. Combining injectivity and surjectivity, the map $X\to \operatorname{MaxSpec}(k[X])$ given by $a\mapsto \mathfrak m_{a,X}$ is a bijection. [/step]