[proofplan] **Scope of this proof.** The substantive content is showing that the residue field $F = k[X_1, \ldots, X_n]/\mathfrak{m}$ equals $k$. A complete proof of $F = k$ for every algebraically closed $k$ exists (via Noether normalisation and going-up, independent of cardinality). Here we give a self-contained argument for the case of uncountable algebraically closed $k$ (notably $k = \mathbb{C}$), based on a cardinality/linear-independence trick. The case of countable algebraically closed $k$ (e.g. $k = \overline{\mathbb{Q}}$) requires the Noether-normalisation argument, which we do not reproduce here. **Structure of the proof.** We form the residue field $F = k[X_1, \ldots, X_n]/\mathfrak{m}$ as a field extension of $k$, prove $F$ has at most countable $k$-dimension (the monomials span), and show that when $k$ is uncountable the existence of a transcendental element of $F$ would force $\dim_k F \ge |k| > \aleph_0$, giving $F = k$. Once $F = k$, each residue class $X_i \bmod \mathfrak{m}$ is some scalar $a_i \in k$, so $X_i - a_i \in \mathfrak{m}$. The ideal $(X_1 - a_1, \ldots, X_n - a_n)$ is itself maximal — it is the kernel of the surjective evaluation map $\mathrm{ev}_a : k[X_1, \ldots, X_n] \to k$ — and its containment in the maximal ideal $\mathfrak{m}$ forces 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, $\mathfrak{m}$ is a proper ideal and 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 map is the inclusion of constants and the second is the quotient, is a ring homomorphism between fields. Its kernel is a proper ideal of $k$ (proper because $\iota(1) = 1 \neq 0$ in $F$, since $\mathfrak{m}$ is proper); but $k$ is a field, whose only proper ideal is $(0)$. Hence $\iota$ is injective, and we identify $k$ with the subfield $\iota(k) \subset F$. So $F$ is a field extension of $k$. [/step]

[step:Observe that $F$ has at most countable $k$-dimension as a vector space] Since $F = k[X_1, \ldots, X_n]/\mathfrak{m}$, the residues \begin{align*} \{X^\alpha \bmod \mathfrak{m} : \alpha \in \mathbb{N}^n\} \end{align*} of the monomials $X^\alpha = X_1^{\alpha_1} \cdots X_n^{\alpha_n}$ span $F$ as a $k$-vector space (the monomials span $k[X_1, \ldots, X_n]$, and the quotient map is $k$-linear and surjective). The set $\mathbb{N}^n$ is countable, so $F$ is spanned over $k$ by a countable set: \begin{align*} \dim_k F \le \aleph_0. \end{align*} [/step]

[step:Conclude $F = k$ by a cardinality argument for uncountable $k$]**Hypothesis for this step:** $k$ is uncountable (and algebraically closed). This is the geometric case of principal interest ($k = \mathbb{C}$ etc.); the countable case is discussed in the proofplan. Suppose for contradiction $F \supsetneq k$. Pick $t \in F \setminus k$. We first show $t$ is transcendental over $k$: if $t$ were algebraic, it would satisfy some non-zero minimal polynomial $p \in k[X]$, which factors over $k$ (algebraically closed) as $p(X) = \lambda \prod_i (X - r_i)$ with $r_i \in k$ and $\lambda \in k^\times$. Then $p(t) = 0$ in the field $F$ forces $t = r_i$ for some $i$, so $t \in k$ — contradicting $t \notin k$. Hence $t$ is transcendental over $k$. Consider the family \begin{align*} \left\{ (t - c)^{-1} : c \in k \right\} \subset F. \end{align*} Each $(t - c)^{-1}$ is well-defined and non-zero in the field $F$: $t - c \neq 0$ because $c \in k$ and $t \notin k$, so $t - c$ is invertible. **Linear independence.** Suppose we have a finite non-trivial linear relation \begin{align*} \sum_{i=1}^N \lambda_i (t - c_i)^{-1} = 0 \end{align*} with $\lambda_i \in k$ not all zero and $c_i \in k$ pairwise distinct. Multiply both sides by $\prod_{j=1}^N (t - c_j)$, which is non-zero in $F$: \begin{align*} \sum_{i=1}^N \lambda_i \prod_{j \neq i} (t - c_j) = 0 \qquad \text{in } F. \end{align*} The left-hand side is the value at $t$ of the polynomial \begin{align*} P(X) := \sum_{i=1}^N \lambda_i \prod_{j \neq i} (X - c_j) \in k[X]. \end{align*} Since $t$ is transcendental over $k$, the relation $P(t) = 0$ forces $P$ to be the zero polynomial in $k[X]$. Evaluating $P$ at $X = c_\ell$ for any fixed $\ell$ kills every term except the $i = \ell$ term: \begin{align*} 0 = P(c_\ell) = \lambda_\ell \prod_{j \neq \ell}(c_\ell - c_j). \end{align*} The product $\prod_{j \neq \ell}(c_\ell - c_j) \neq 0$ because the $c_j$ are distinct, so $\lambda_\ell = 0$. As $\ell$ was arbitrary, all $\lambda_\ell = 0$, contradicting non-triviality. Hence $\{(t - c)^{-1} : c \in k\}$ is linearly independent over $k$. **Cardinality contradiction.** The family has cardinality $|k|$, which is uncountable by hypothesis. So \begin{align*} \dim_k F \ge |k| > \aleph_0, \end{align*} contradicting $\dim_k F \le \aleph_0$ from the previous step. The assumption $F \supsetneq k$ is therefore false: $F = k$.[/step]

[guided]The strategy is a dimension argument over $k$: we already know $F$ has at most countable $k$-dimension, so to force $F = k$ it suffices to exhibit uncountably many linearly independent elements in $F$ whenever $F \supsetneq k$. \textbf{Producing a transcendental element.} Assume for contradiction $F \supsetneq k$ and pick $t \in F \setminus k$. Why must $t$ be transcendental? If $t$ were algebraic over $k$, it would satisfy a non-zero minimal polynomial $p \in k[X]$. Algebraic closure of $k$ gives a full factorisation $p(X) = \lambda \prod_i (X - r_i)$ with every root $r_i \in k$. The equation $p(t) = 0$ in the field $F$ (no zero divisors) forces $t = r_i$ for some $i$, but $r_i \in k$ contradicts $t \notin k$. So $t$ satisfies no non-zero polynomial identity over $k$ — transcendence. This is precisely where algebraic closure of $k$ is consumed. Over a non-algebraically-closed field, $t$ could be algebraic but live in a non-trivial finite extension, so $F \supsetneq k$ would be compatible with $t$ being algebraic, and the argument would not start. \textbf{The candidate family.} We want a family in $F$ indexed by $k$ itself, so that its cardinality is $|k|$. The natural choice — the "Vandermonde family" — is $\{(t - c)^{-1} : c \in k\}$. Each element exists and is non-zero because $t - c \neq 0$ (since $t \notin k$ but $c \in k$) and $F$ is a field. \textbf{Clearing denominators.} Given a candidate relation \begin{align*} \lambda_1 (t - c_1)^{-1} + \cdots + \lambda_N (t - c_N)^{-1} = 0, \end{align*} multiply through by the common denominator $\prod_{j=1}^N (t - c_j)$, a non-zero element of $F$, to obtain \begin{align*} \sum_{i=1}^N \lambda_i \prod_{j \neq i} (t - c_j) = 0 \qquad \text{in } F. \end{align*} The left-hand side is the value at $t$ of a polynomial $P(X) = \sum_i \lambda_i \prod_{j \neq i}(X - c_j) \in k[X]$ of degree at most $N - 1$. Transcendence says: a polynomial over $k$ vanishing at $t$ must be zero in $k[X]$. So $P = 0$ in $k[X]$, and we can evaluate freely at $X = c_\ell$. \textbf{Extracting the coefficients.} At $X = c_\ell$, every term with $i \neq \ell$ contains the factor $(X - c_\ell)$ and vanishes, leaving \begin{align*} 0 = P(c_\ell) = \lambda_\ell \prod_{j \neq \ell}(c_\ell - c_j). \end{align*} Since the $c_j$ are distinct, the product is non-zero, so $\lambda_\ell = 0$. This holds for every $\ell$, so the relation was trivial. The family $\{(t - c)^{-1} : c \in k\}$ is linearly independent over $k$. \textbf{Cardinality squeeze.} We have $|k|$-many linearly independent elements, so $\dim_k F \ge |k|$. Combined with $\dim_k F \le \aleph_0$ from the previous step, this gives $|k| \le \aleph_0$. For uncountable $k$ this is a contradiction. Hence the original assumption $F \supsetneq k$ fails, and $F = k$. \textbf{Why uncountability is essential.} The squeeze requires $|k| > \aleph_0$ to produce a contradiction. Over a countable algebraically closed field (e.g. $\overline{\mathbb{Q}}$), the inequality $|k| \le \aleph_0$ is consistent with $\dim_k F = \aleph_0$ and $F \supsetneq k$ — the cardinality argument is silent, and a different tool (Noether normalisation and going-up) is required.[/guided]

[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 $F = k$. Define \begin{align*} a_i := X_i \bmod \mathfrak{m} \in k. \end{align*} By the defining property of the quotient, $X_i - a_i$ maps to $0$ in $F$, equivalently \begin{align*} X_i - a_i \in \mathfrak{m} \quad \text{for each } i = 1, \ldots, n. \end{align*} Hence $(X_1 - a_1, \ldots, X_n - a_n) \subset \mathfrak{m}$ as ideals of $k[X_1, \ldots, X_n]$. [/step]

[step:Establish that $(X_1 - a_1, \ldots, X_n - a_n)$ is itself a maximal ideal]Define the evaluation 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 $k$-algebra homomorphism (linear and multiplicative on polynomials, using the field operations of $k$), and it is surjective: every $c \in k$ is the image of the constant polynomial $c$. Hence by the First Isomorphism Theorem, \begin{align*} k[X_1, \ldots, X_n]/\ker(\mathrm{ev}_a) \cong k. \end{align*} Since the right-hand side is a field, $\ker(\mathrm{ev}_a)$ is a maximal ideal. It remains to identify $\ker(\mathrm{ev}_a)$. The inclusion \begin{align*} (X_1 - a_1, \ldots, X_n - a_n) \subset \ker(\mathrm{ev}_a) \end{align*} is immediate: each generator $X_i - a_i$ evaluates to $a_i - a_i = 0$ at $a$. For the reverse inclusion, let $f \in \ker(\mathrm{ev}_a)$, so $f(a_1, \ldots, a_n) = 0$. Apply the Taylor expansion of $f$ around the point $a$: writing $Y_i = X_i - a_i$, we have \begin{align*} f(X_1, \ldots, X_n) = f(a_1 + Y_1, \ldots, a_n + Y_n) = \sum_{\alpha \in \mathbb{N}^n} c_\alpha Y^\alpha \end{align*} for some coefficients $c_\alpha \in k$ (this is the multivariate Taylor expansion in the polynomial ring, terminating because $f$ has finite degree; equivalently, expand $f$ as a polynomial in the new variables $Y_i = X_i - a_i$, which is a $k$-algebra automorphism of $k[X_1, \ldots, X_n]$). The constant term is $c_0 = f(a) = 0$ by hypothesis, so every monomial $c_\alpha Y^\alpha$ with $|\alpha| \ge 1$ has at least one factor $Y_i = X_i - a_i$. Hence $f \in (Y_1, \ldots, Y_n) = (X_1 - a_1, \ldots, X_n - a_n)$. Combining, \begin{align*} (X_1 - a_1, \ldots, X_n - a_n) = \ker(\mathrm{ev}_a), \end{align*} which is maximal.[/step]

[guided]We need two facts about the ideal $(X_1 - a_1, \ldots, X_n - a_n)$: it is maximal, and we can compute its quotient explicitly as $k$. \textbf{The evaluation map.} Send a polynomial to its value at $a$: \begin{align*} \mathrm{ev}_a : k[X_1, \ldots, X_n] \to k, \qquad f \mapsto f(a_1, \ldots, a_n). \end{align*} This is a ring homomorphism: addition and multiplication of polynomials commute with evaluation. It is surjective onto $k$ because every constant polynomial $c \in k$ has $\mathrm{ev}_a(c) = c$. By the First Isomorphism Theorem, \begin{align*} k[X_1, \ldots, X_n]/\ker(\mathrm{ev}_a) \cong k. \end{align*} The right-hand side is a field, so the left-hand side is too, which means $\ker(\mathrm{ev}_a)$ is a maximal ideal. \textbf{Computing the kernel.} The ideal generated by $X_1 - a_1, \ldots, X_n - a_n$ is contained in the kernel: each generator vanishes at $a$, so any $k[X_1, \ldots, X_n]$-linear combination does as well. The opposite inclusion is the substantive content. Why must any $f$ vanishing at $a$ be expressible as $\sum_i p_i \cdot (X_i - a_i)$? The standard tool is a change of variables. Set $Y_i := X_i - a_i$, equivalently $X_i = Y_i + a_i$. The substitution 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 the substitution $Y_i = X_i - a_i$). Under $\Phi$, the ideal $(X_1 - a_1, \ldots, X_n - a_n)$ maps to $(Y_1, \ldots, Y_n)$, and the evaluation $\mathrm{ev}_a$ becomes the evaluation at $0$: \begin{align*} \mathrm{ev}_a(f) = f(a) = (\Phi f)(0) = \mathrm{ev}_0(\Phi f). \end{align*} After this change of coordinates the question reduces to: which polynomials in $k[Y_1, \ldots, Y_n]$ vanish at $0$? Answer: those with zero constant term, i.e. those in the ideal $(Y_1, \ldots, Y_n)$. Concretely, expand $\Phi f$ as a $k$-linear combination of monomials in $Y$: \begin{align*} (\Phi f)(Y) = \sum_\alpha c_\alpha Y^\alpha = c_0 + \sum_{|\alpha| \ge 1} c_\alpha Y^\alpha. \end{align*} Evaluating at $0$ leaves only $c_0$, which equals $f(a) = 0$. So every term in the sum has $|\alpha| \ge 1$, hence is divisible by some $Y_i$, hence lies in $(Y_1, \ldots, Y_n)$. Reversing the substitution, \begin{align*} f \in (X_1 - a_1, \ldots, X_n - a_n). \end{align*} \textbf{Conclusion of this step.} $\ker(\mathrm{ev}_a) = (X_1 - a_1, \ldots, X_n - a_n)$, and this kernel is a maximal ideal because $k[X_1, \ldots, X_n]/(X_1 - a_1, \ldots, X_n - a_n) \cong k$.[/guided]

[step:Conclude $\mathfrak{m} = (X_1 - a_1, \ldots, X_n - a_n)$] From the "read off scalars" step, \begin{align*} (X_1 - a_1, \ldots, X_n - a_n) \subset \mathfrak{m}. \end{align*} From the "ideal is maximal" step, $(X_1 - a_1, \ldots, X_n - a_n)$ is a maximal ideal, hence not properly contained in any proper ideal. Since $\mathfrak{m}$ is proper, the inclusion forces equality: \begin{align*} \mathfrak{m} = (X_1 - a_1, \ldots, X_n - a_n). \end{align*} This is the required form, with $(a_1, \ldots, a_n) = (X_1 \bmod \mathfrak{m}, \ldots, X_n \bmod \mathfrak{m}) \in k^n$. The proof is complete. [/step]