[proofplan] The strategy is a cardinality argument on the residue field $F := k[X_1, \ldots, X_n]/\mathfrak{m}$. We show $F = k$ by contradiction: if $F$ properly contains $k$, pick $t \in F \setminus k$. Since $k$ is algebraically closed, $t$ must be transcendental over $k$. The set $\{(t - c)^{-1} : c \in k\}$ is then linearly independent over $k$ (the standard partial-fractions argument), forcing $\dim_k F \geq |k|$. But $F$ is a quotient of the countably-dimensional $k$-vector space $k[X_1, \ldots, X_n]$, so $\dim_k F \leq \aleph_0$. Since $|k|$ is uncountable, this is a contradiction. Hence $F = k$, and the residue classes $a_i := X_i \bmod \mathfrak{m}$ lie in $k$. The point ideal $(X_1 - a_1, \ldots, X_n - a_n)$ is contained in $\mathfrak{m}$ and is itself maximal, forcing equality. [/proofplan] [step:Form the residue field $F = k[X_1, \ldots, X_n]/\mathfrak{m}$ as an extension of $k$] Let $\mathfrak{m} \subset k[X_1, \ldots, X_n]$ be a maximal ideal. By definition of maximality, the quotient \begin{align*} F := k[X_1, \ldots, X_n] / \mathfrak{m} \end{align*} is a field. The composition \begin{align*} \iota : k \hookrightarrow k[X_1, \ldots, X_n] \twoheadrightarrow F, \end{align*} where the first arrow is the inclusion of $k$ as constant polynomials and the second is the quotient projection, is a ring homomorphism between fields. Its kernel is a proper ideal of $k$ (proper because $\iota(1) = 1_F \neq 0_F$, the unit and zero in the field $F$ being distinct since $\mathfrak{m}$ is a proper ideal). The only proper ideal of a field is $(0)$, so $\iota$ is injective. We henceforth identify $k$ with the subfield $\iota(k) \subseteq F$, and view $F$ as a field extension of $k$. [/step] [step:Bound $\dim_k F$ above by $\aleph_0$ via the countable monomial basis] The polynomial ring $k[X_1, \ldots, X_n]$ is, as a $k$-vector space, the free $k$-module on the set of monomials \begin{align*} \mathcal{M} := \{ X^\alpha := X_1^{\alpha_1} \cdots X_n^{\alpha_n} : \alpha = (\alpha_1, \ldots, \alpha_n) \in \mathbb{N}^n \}. \end{align*} The set $\mathcal{M}$ is countable: there is an explicit bijection $\mathcal{M} \to \mathbb{N}$ obtained by lexicographically ordering multi-indices, or equivalently, $\mathcal{M}$ is in bijection with $\mathbb{N}^n$, which is a finite product of countable sets and hence countable. Therefore \begin{align*} \dim_k k[X_1, \ldots, X_n] = |\mathcal{M}| = \aleph_0. \end{align*} The quotient map $k[X_1, \ldots, X_n] \to F$ is a surjective $k$-linear map, so \begin{align*} \dim_k F \leq \dim_k k[X_1, \ldots, X_n] = \aleph_0. \end{align*} In particular, $\dim_k F$ is a finite or countably infinite cardinal: $\dim_k F \in \{1, 2, 3, \ldots\} \cup \{\aleph_0\}$. [/step] [step:Suppose for contradiction $F \supsetneq k$ and pick a transcendental element $t \in F$] Suppose for contradiction that $F \neq k$ (i.e.\ $F$ properly contains $k$). Pick any element \begin{align*} t \in F \setminus k. \end{align*} We claim $t$ is transcendental over $k$. Suppose for the sake of contradiction (a nested contradiction inside the main one) that $t$ is algebraic over $k$. Then $t$ satisfies some non-zero polynomial $P \in k[T]$: \begin{align*} P(t) = 0 \quad \text{in } F. \end{align*} Since $k$ is algebraically closed, every non-zero polynomial $P \in k[T]$ factors completely over $k$: \begin{align*} P(T) = c \prod_{j = 1}^{d} (T - \alpha_j) \qquad (c \in k^\times, \; \alpha_1, \ldots, \alpha_d \in k). \end{align*} Substituting $T = t$ in $F$: \begin{align*} 0 = P(t) = c \prod_{j = 1}^{d} (t - \alpha_j) \quad \text{in } F. \end{align*} Since $F$ is a field (hence an integral domain), at least one factor vanishes: $t - \alpha_j = 0$ for some $j$, i.e.\ $t = \alpha_j \in k$. But this contradicts our choice $t \in F \setminus k$. Hence $t$ is not algebraic over $k$ — that is, $t$ is transcendental over $k$. In particular, the subring $k[t] \subseteq F$ generated by $k$ and $t$ is isomorphic to the polynomial ring $k[T]$ in one indeterminate via $T \mapsto t$, and the subfield $k(t) \subseteq F$ generated by $k$ and $t$ is isomorphic to the field of rational functions $k(T)$ in one indeterminate (the field of fractions of $k[T]$). [guided] The transcendence claim is the content of "$k$ algebraically closed" combined with "$t \notin k$". An algebraically closed field is one for which every non-constant polynomial has a root (in the field itself). Equivalently, every algebraic element of any extension field already lies in the algebraically closed ground field — this is the contrapositive form we use here. \textbf{Why does the algebraic closure of $k$ enter?} Without algebraic closure, $F$ could contain genuinely new algebraic elements (e.g.\ $\sqrt{2}$ in an extension of $\mathbb{Q}$) that are not transcendental but also not in $k$. The cardinality argument that follows would not apply, because the linear-independence statement uses transcendence in an essential way: only for transcendental $t$ is the polynomial ring $k[t]$ free of $k$-relations. \textbf{What we have set up.} A copy of the rational function field $k(T)$ inside $F$, via $T \mapsto t$. This is the structure we now exploit: $k(T)$ has uncountably many "simple poles" $(T - c)^{-1}$ for $c \in k$, when $k$ is uncountable, and these are linearly independent over $k$. [/guided] [/step] [step:Show $\{(t - c)^{-1} : c \in k\}$ is $k$-linearly independent in $F$] We claim that the set \begin{align*} S := \{ (t - c)^{-1} : c \in k \} \subseteq F \end{align*} is linearly independent over $k$. Note first that for each $c \in k$, the element $t - c$ is non-zero in $F$ (since $t$ is transcendental over $k$, so in particular $t \neq c$ as elements of $F$). Hence $(t - c)^{-1}$ is well-defined in the field $F$. To prove linear independence, suppose for contradiction there is a non-trivial $k$-linear relation \begin{align*} \sum_{j = 1}^{m} \lambda_j (t - c_j)^{-1} = 0 \quad \text{in } F, \end{align*} with $\lambda_1, \ldots, \lambda_m \in k$ not all zero, and $c_1, \ldots, c_m \in k$ pairwise distinct. (We may assume the $c_j$ are pairwise distinct: any repetitions can be combined into a single coefficient.) Multiply both sides by the non-zero element $\prod_{j = 1}^{m} (t - c_j) \in F^\times$: \begin{align*} 0 &= \left( \prod_{j=1}^{m} (t - c_j) \right) \cdot \sum_{j = 1}^{m} \lambda_j (t - c_j)^{-1} \\ &= \sum_{j = 1}^{m} \lambda_j \prod_{i \neq j} (t - c_i). \end{align*} Define the polynomial \begin{align*} Q(T) := \sum_{j = 1}^{m} \lambda_j \prod_{i \neq j} (T - c_i) \in k[T]. \end{align*} We have shown $Q(t) = 0$ in $F$. Since $t$ is transcendental over $k$ (Step 3), the only polynomial in $k[T]$ vanishing at $t$ is the zero polynomial. Hence $Q = 0$ as a polynomial in $k[T]$. We now show this forces $\lambda_j = 0$ for every $j$, contradicting the assumption of a non-trivial relation. Fix any index $j_0 \in \{1, \ldots, m\}$ and evaluate the polynomial $Q$ at $T = c_{j_0} \in k$: \begin{align*} Q(c_{j_0}) = \sum_{j = 1}^{m} \lambda_j \prod_{i \neq j} (c_{j_0} - c_i). \end{align*} For $j \neq j_0$, the product $\prod_{i \neq j} (c_{j_0} - c_i)$ contains the factor with $i = j_0$, namely $(c_{j_0} - c_{j_0}) = 0$, so the entire term vanishes. Only the $j = j_0$ term survives: \begin{align*} 0 = Q(c_{j_0}) = \lambda_{j_0} \prod_{i \neq j_0} (c_{j_0} - c_i). \end{align*} The product $\prod_{i \neq j_0} (c_{j_0} - c_i)$ is non-zero, since the $c_i$ are pairwise distinct (in particular $c_{j_0} - c_i \neq 0$ for $i \neq j_0$, and $k$ is a field, so the product of non-zero elements is non-zero). Hence $\lambda_{j_0} = 0$. Since $j_0$ was arbitrary, $\lambda_j = 0$ for all $j$, contradicting the assumed non-triviality of the linear relation. Therefore $S$ is linearly independent over $k$. [guided] This is the standard "partial fractions" linear independence: distinct simple poles of rational functions in a single variable are linearly independent over the base field. The proof packages this in two stages — clear denominators to convert a relation among $(t - c_j)^{-1}$ into a polynomial identity in $t$, then use transcendence of $t$ to lift the polynomial identity to an identity in $k[T]$, and finally evaluate at the $c_j$ to extract the coefficients. \textbf{Why does multiplication by $\prod (t - c_j)$ work?} It is a unit in $F$ — non-zero because $t \neq c_j$ in $F$ for each $j$, and a non-zero element of a field is invertible. So multiplying a relation by it preserves the truth-value (zero stays zero), and it clears all the denominators, leaving a polynomial relation $Q(t) = 0$. \textbf{Where does transcendence of $t$ enter?} Without transcendence, we could have $Q(t) = 0$ for some non-zero polynomial $Q \in k[T]$ — namely, if $t$ is algebraic and $Q$ is its minimal polynomial multiplied by anything. With transcendence, the only polynomial vanishing at $t$ is the zero polynomial, so the relation upgrades from "$Q(t) = 0$ in $F$" to "$Q = 0$ as a polynomial in $k[T]$", which is a much stronger statement. \textbf{Where does evaluation at $c_{j_0}$ extract $\lambda_{j_0}$?} The polynomials $\prod_{i \neq j} (T - c_i)$ are the **Lagrange basis polynomials** for the interpolation nodes $c_1, \ldots, c_m$, up to a normalising scalar. Evaluating the $j$-th basis polynomial at $c_{j_0}$ gives zero unless $j = j_0$. So evaluating $Q$ at $c_{j_0}$ isolates the coefficient $\lambda_{j_0}$ — the same trick used to prove existence and uniqueness of polynomial interpolation. \textbf{What if $k$ is not algebraically closed?} The argument above only used transcendence of $t$, not algebraic closure of $k$. Algebraic closure was used in Step 3 to deduce transcendence from "$t \notin k$". Without algebraic closure, $t$ could be algebraic of degree $\geq 2$, and we would have to weaken or abandon the argument. (For example, $\mathbb{R}[X]/(X^2 + 1) = \mathbb{C}$ has $\dim_{\mathbb{R}} = 2$, finite, and the maximal ideal is not of the form $(X - a)$ for $a \in \mathbb{R}$.) [/guided] [/step] [step:Derive a cardinality contradiction using the uncountability of $k$] By Step 4, the set $S = \{(t - c)^{-1} : c \in k\} \subseteq F$ is linearly independent over $k$. The map $c \mapsto (t - c)^{-1}$ from $k$ to $S$ is injective: if $(t - c_1)^{-1} = (t - c_2)^{-1}$ in $F$, then $t - c_1 = t - c_2$ (taking inverses, both sides non-zero), so $c_1 = c_2$. Hence \begin{align*} |S| = |k|. \end{align*} A linearly independent set in a $k$-vector space has cardinality at most the dimension of the vector space (this is the standard fact that any linearly independent set extends to a basis, so its cardinality bounds the dimension from below). Therefore \begin{align*} \dim_k F \geq |S| = |k|. \end{align*} Since $k$ is uncountable, $|k| > \aleph_0$, and so \begin{align*} \dim_k F > \aleph_0. \end{align*} But by Step 2, $\dim_k F \leq \aleph_0$. This is the contradiction. The contradiction was derived from the assumption $F \neq k$ in Step 3. Hence \begin{align*} F = k. \end{align*} [/step] [step:Read off scalars $a_i \in k$ from the residues of the coordinate functions] For each $i \in \{1, \ldots, n\}$, the residue class $X_i \bmod \mathfrak{m}$ is an element of $F$, which by Step 5 equals $k$. Define \begin{align*} a_i := X_i \bmod \mathfrak{m} \in k \qquad (i = 1, \ldots, n). \end{align*} By the defining property of the quotient $F = k[X_1, \ldots, X_n]/\mathfrak{m}$, the element $X_i - a_i$ maps to $0$ in $F$, equivalently $X_i - a_i \in \mathfrak{m}$. Hence \begin{align*} (X_1 - a_1, \ldots, X_n - a_n) \subseteq \mathfrak{m}. \end{align*} [/step] [step:Show $(X_1 - a_1, \ldots, X_n - a_n)$ is itself maximal, forcing equality] Define the evaluation $k$-algebra homomorphism \begin{align*} \mathrm{ev}_a : k[X_1, \ldots, X_n] &\to k \\ f &\mapsto f(a_1, \ldots, a_n). \end{align*} This is a ring homomorphism (polynomial addition and multiplication commute with evaluation), and surjective onto $k$ (every $c \in k$ is the image of the constant polynomial $c$). By the First Isomorphism Theorem, \begin{align*} k[X_1, \ldots, X_n] / \ker(\mathrm{ev}_a) \cong k. \end{align*} Since $k$ is a field, the right-hand side is a field, hence $\ker(\mathrm{ev}_a) \subset k[X_1, \ldots, X_n]$ is a maximal ideal. We claim $\ker(\mathrm{ev}_a) = (X_1 - a_1, \ldots, X_n - a_n)$. \textbf{Inclusion $(X_1 - a_1, \ldots, X_n - a_n) \subseteq \ker(\mathrm{ev}_a)$.} Each generator $X_i - a_i$ evaluates to $a_i - a_i = 0$ at the point $a = (a_1, \ldots, a_n)$, so $X_i - a_i \in \ker(\mathrm{ev}_a)$ for each $i$. The kernel is an ideal, hence contains the ideal generated by the $X_i - a_i$. \textbf{Inclusion $\ker(\mathrm{ev}_a) \subseteq (X_1 - a_1, \ldots, X_n - a_n)$.} Let $f \in \ker(\mathrm{ev}_a)$, so $f(a_1, \ldots, a_n) = 0$. Substitute $X_i = Y_i + a_i$ (equivalently $Y_i := X_i - a_i$) — this defines a $k$-algebra automorphism \begin{align*} \Phi : k[X_1, \ldots, X_n] &\to k[Y_1, \ldots, Y_n] \\ f(X_1, \ldots, X_n) &\mapsto f(Y_1 + a_1, \ldots, Y_n + a_n) \end{align*} (the inverse is $Y_i \mapsto X_i - a_i$). Expand $\Phi f$ as a $k$-linear combination of monomials in $Y$: \begin{align*} (\Phi f)(Y) = \sum_{\alpha \in \mathbb{N}^n} c_\alpha Y^\alpha = c_0 + \sum_{|\alpha| \geq 1} c_\alpha Y^\alpha. \end{align*} Evaluating both sides at $Y = 0$ gives $(\Phi f)(0) = c_0$ on one hand, and $f(a_1, \ldots, a_n) = 0$ on the other, so $c_0 = 0$. Hence \begin{align*} (\Phi f)(Y) = \sum_{|\alpha| \geq 1} c_\alpha Y^\alpha, \end{align*} and every term has $|\alpha| \geq 1$, so is divisible by some $Y_i$ — i.e.\ lies in the ideal $(Y_1, \ldots, Y_n) \subset k[Y_1, \ldots, Y_n]$. Reversing the substitution: $\Phi f \in (Y_1, \ldots, Y_n)$ corresponds to $f \in \Phi^{-1}((Y_1, \ldots, Y_n)) = (X_1 - a_1, \ldots, X_n - a_n)$. This is the desired inclusion. Combining both inclusions, \begin{align*} \ker(\mathrm{ev}_a) = (X_1 - a_1, \ldots, X_n - a_n). \end{align*} Hence $(X_1 - a_1, \ldots, X_n - a_n)$ is a maximal ideal of $k[X_1, \ldots, X_n]$. [/step] [step:Conclude $\mathfrak{m} = (X_1 - a_1, \ldots, X_n - a_n)$] By Step 6, the inclusion $(X_1 - a_1, \ldots, X_n - a_n) \subseteq \mathfrak{m}$ holds. By Step 7, $(X_1 - a_1, \ldots, X_n - a_n)$ is a maximal ideal — equivalently, it is not properly contained in any proper ideal. Since $\mathfrak{m}$ is itself proper, the chain \begin{align*} (X_1 - a_1, \ldots, X_n - a_n) \subseteq \mathfrak{m} \subsetneq k[X_1, \ldots, X_n] \end{align*} forces equality at the first inclusion. Therefore \begin{align*} \mathfrak{m} = (X_1 - a_1, \ldots, X_n - a_n) \end{align*} with $(a_1, \ldots, a_n) = (X_1 \bmod \mathfrak{m}, \ldots, X_n \bmod \mathfrak{m}) \in k^n$. This is the asserted form, completing the proof. [/step]