[proofplan] **Scope of this proof.** A full proof of the Weak Nullstellensatz exists for every algebraically closed field $k$; the classical route is via Noether normalisation and going-up, reducing $F = k[X_1, \ldots, X_n]/\mathfrak{m}$ to the case $n = 1$ where algebraic closure gives $F = k$ directly. Here we give a self-contained, cardinality-based argument that works when $k$ is uncountable (the case of greatest geometric interest, e.g. $k = \mathbb{C}$). For countable algebraically closed $k$ (e.g. $k = \overline{\mathbb{Q}}$), the same conclusion holds but requires the going-up / Noether-normalisation argument, which we do not reproduce here. **Structure of the uncountable argument.** We reduce the question to maximal ideals and prove that every maximal ideal $\mathfrak{m} \subset k[X_1, \ldots, X_n]$ has the form $(X_1 - a_1, \ldots, X_n - a_n)$ for some $(a_1, \ldots, a_n) \in k^n$, which immediately exhibits a point of $V(\mathfrak{m}) \subset V(I)$. Setting $F = k[X_1, \ldots, X_n]/\mathfrak{m}$ — a field extension of $k$ — the question becomes: must $F = k$? For uncountable $k$, a transcendental element of $F$ would generate uncountably many linearly independent inverses $(t-c)^{-1}$, contradicting the countable spanning set $\{X^{\alpha} \bmod \mathfrak{m}\}$ of $F$. [/proofplan] [step:Reduce to showing every maximal ideal of $k[X_1, \ldots, X_n]$ has the form $(X_1 - a_1, \ldots, X_n - a_n)$] Let $I \subsetneq k[X_1, \ldots, X_n]$ be proper. Since $k[X_1, \ldots, X_n]$ is a commutative ring with unit, the standard application of *Zorn's Lemma* to the poset of proper ideals containing $I$ (ordered by inclusion; any chain of proper ideals has the union as a proper ideal upper bound, proper because $1$ cannot lie in it without already lying in some chain member) produces a maximal element. Thus every proper ideal is contained in a maximal ideal. Let $\mathfrak{m} \supset I$ be such a maximal ideal. Then $V(\mathfrak{m}) \subset V(I)$ (by the order-reversing property of $V$ under inclusion of ideals). It therefore suffices to prove $V(\mathfrak{m}) \neq \varnothing$. We claim: \begin{align*} \mathfrak{m} = (X_1 - a_1, \ldots, X_n - a_n) \quad \text{for some } (a_1, \ldots, a_n) \in k^n. \end{align*} Granting this, the point $a := (a_1, \ldots, a_n)$ satisfies $(X_i - a_i)(a) = 0$ for each $i$, so each generator of $\mathfrak{m}$ vanishes at $a$. Since $\mathfrak{m}$ is generated by these, every element of $\mathfrak{m}$ vanishes at $a$, i.e. $a \in V(\mathfrak{m}) \subset V(I)$. The remaining steps prove the claim. [/step] [step:Form the residue field $F = k[X_1, \ldots, X_n]/\mathfrak{m}$ and reduce the claim to showing $F = k$] Since $\mathfrak{m}$ is a maximal ideal of the commutative ring $k[X_1, \ldots, X_n]$, 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*} is a ring homomorphism between fields, hence injective (its kernel is a proper ideal of $k$, which has only $0$ and $k$ as ideals; the image of $1$ is non-zero, so the kernel is $0$). We identify $k$ with $\iota(k) \subset F$ and view $F$ as a field extension of $k$. \textbf{Suppose for now that $F = k$.} Then for each $i$, the residue class $X_i \bmod \mathfrak{m}$ lies in $k \subset F$; call it $a_i \in k$. By definition of the quotient, $X_i - a_i \in \mathfrak{m}$ for each $i$, so \begin{align*} (X_1 - a_1, \ldots, X_n - a_n) \subset \mathfrak{m}. \end{align*} Conversely, $(X_1 - a_1, \ldots, X_n - a_n)$ is itself a maximal ideal: the quotient \begin{align*} k[X_1, \ldots, X_n]/(X_1 - a_1, \ldots, X_n - a_n) \cong k \end{align*} via the evaluation map $f(X_1, \ldots, X_n) \mapsto f(a_1, \ldots, a_n)$ (well-defined since the kernel is exactly the ideal of polynomials vanishing at $a$, which contains every $X_i - a_i$). Since $(X_1 - a_1, \ldots, X_n - a_n)$ is maximal and contained in the proper ideal $\mathfrak{m}$, equality holds: \begin{align*} \mathfrak{m} = (X_1 - a_1, \ldots, X_n - a_n). \end{align*} The remaining task is therefore to prove $F = k$. [/step] [step:Note that $F$ is countable-dimensional over $k$ 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 (since the monomials $\{X^\alpha\}$ span the polynomial ring as a $k$-vector space, and the quotient map is $k$-linear and surjective). The set $\mathbb{N}^n$ is countable, so $F$ is spanned by a countable set over $k$; equivalently, \begin{align*} \dim_k F \le \aleph_0. \end{align*} [/step] [step:Assume $k$ is uncountable and derive $F = k$ by a cardinality argument] \textbf{Hypothesis for this step:} $k$ is uncountable (in addition to being algebraically closed). The case $k = \mathbb{C}$ is the principal example. Suppose for contradiction that $F \neq k$. Then there exists $t \in F \setminus k$. We claim $t$ is transcendental over $k$. Indeed, if $t$ were algebraic over $k$, it would satisfy some non-zero minimal polynomial $p(X) \in k[X]$. Since $k$ is algebraically closed, $p$ factors into linear factors over $k$, and $p(t) = 0$ forces $t$ to equal a root of $p$ — but every root of $p$ lies in $k$, contradicting $t \notin k$. So $t$ is transcendental over $k$. Consider the family \begin{align*} \left\{ \frac{1}{t - c} : c \in k \right\} \subset F. \end{align*} Since $t$ is transcendental, $t - c \neq 0$ in $F$ for every $c \in k$ (otherwise $t = c \in k$), so each element $(t - c)^{-1}$ is a well-defined non-zero element of the field $F$. We claim this family is linearly independent over $k$. \textbf{Linear independence.} Suppose for contradiction there is a non-trivial finite linear relation \begin{align*} \sum_{i=1}^N \lambda_i (t - c_i)^{-1} = 0 \end{align*} with the $c_i \in k$ pairwise distinct, $\lambda_i \in k$, and not all $\lambda_i$ zero. Multiplying through by $\prod_{j=1}^N (t - c_j)$ (a non-zero element of the field $F$) yields \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 a polynomial expression in $t$ of degree $\le N - 1$ with coefficients in $k$. Since $t$ is transcendental, this polynomial expression must be the zero polynomial in $k[X]$ — every coefficient vanishes. Evaluating at $X = c_k$ for any fixed $k$ kills every term except the $i = k$ term: \begin{align*} 0 = \sum_{i=1}^N \lambda_i \prod_{j \neq i} (c_k - c_j) = \lambda_k \prod_{j \neq k} (c_k - c_j). \end{align*} The product $\prod_{j \neq k}(c_k - c_j)$ is non-zero because the $c_i$ are pairwise distinct, so $\lambda_k = 0$. As $k$ was arbitrary, all $\lambda_k = 0$, contradicting non-triviality. Hence the family $\{(t-c)^{-1} : c \in k\}$ is linearly independent over $k$. \textbf{Cardinality contradiction.} The family $\{(t-c)^{-1} : c \in k\}$ has cardinality $|k|$, which is uncountable. So \begin{align*} \dim_k F \ge |k| > \aleph_0. \end{align*} This contradicts Step 3, which gave $\dim_k F \le \aleph_0$. Therefore the assumption $F \neq k$ is false, and $F = k$. [guided] The strategy is a dimension argument: $F$ has at most countable dimension over $k$ (we've already established this), so we will derive a contradiction by exhibiting $|k|$-many linearly independent elements in $F$ when $|k|$ is uncountable. \textbf{Step A — Find a transcendental element if $F \neq k$.} Suppose for contradiction $F \supsetneq k$. Pick any $t \in F \setminus k$. Why must $t$ be transcendental over $k$? If $t$ were algebraic, it would be a root of some non-zero polynomial $p \in k[X]$. Because $k$ is algebraically closed, $p$ factors as $p(X) = \alpha \prod_i (X - r_i)$ with $r_i \in k$ and $\alpha \in k^\times$. The equation $p(t) = 0$ holds in the field $F$ (which has no zero divisors), so $t = r_i$ for some $i$. But every $r_i \in k$, contradicting $t \notin k$. So $t$ is transcendental — meaning it satisfies no non-zero polynomial relation over $k$. Notice: this is exactly where algebraic closure of $k$ is used. Without it, $t$ might be algebraic over $k$ but live in a non-trivial finite extension, so $F$ would not equal $k$ but the dimension argument would not immediately give a contradiction. \textbf{Step B — Construct the candidate independent family.} Consider the elements $(t - c)^{-1}$ as $c$ ranges over $k$. Why these? They are a "Vandermonde-style" family parametrized by $k$ itself, so the family has cardinality $|k|$. Each $(t - c)^{-1}$ exists and is non-zero in $F$: $t - c \neq 0$ because $c \in k$ but $t \notin k$ (transcendental), and $F$ is a field so we can invert. \textbf{Step C — Prove linear independence by clearing denominators.} Suppose \begin{align*} \lambda_1 (t - c_1)^{-1} + \cdots + \lambda_N (t - c_N)^{-1} = 0 \end{align*} with the $c_i \in k$ pairwise distinct. We want to show all $\lambda_i = 0$. The standard trick is to clear denominators: multiply through by $\prod_{j=1}^N (t - c_j)$, a non-zero element of $F$. The result is \begin{align*} \sum_{i=1}^N \lambda_i \prod_{j \neq i} (t - c_j) = 0 \quad \text{in } F. \end{align*} The left-hand side is a polynomial expression in $t$ with coefficients in $k$, of degree at most $N-1$. Now we use the transcendence of $t$: a polynomial over $k$ that vanishes when evaluated at the transcendental element $t$ must be the zero polynomial in $k[X]$ (this is the definition of transcendence). Therefore the underlying polynomial \begin{align*} P(X) = \sum_{i=1}^N \lambda_i \prod_{j \neq i} (X - c_j) \in k[X] \end{align*} is the zero polynomial. To extract the $\lambda_i$, evaluate $P$ at $X = c_k$: every term with $i \neq k$ is killed by the factor $(X - c_k)$ in its product, leaving \begin{align*} 0 = P(c_k) = \lambda_k \prod_{j \neq k} (c_k - c_j). \end{align*} The product $\prod_{j \neq k}(c_k - c_j) \neq 0$ because the $c_j$ are distinct, so $\lambda_k = 0$. This holds for every $k$, so the relation is trivial. The family is linearly independent. \textbf{Step D — Cardinality contradiction.} We have produced $|k|$ linearly independent elements of $F$ over $k$, so $\dim_k F \ge |k|$. But Step 3 gave $\dim_k F \le \aleph_0$ (the monomial residues $X^\alpha \bmod \mathfrak{m}$ form a countable spanning set). For uncountable $k$, $|k| > \aleph_0$, and these are incompatible. The assumption $F \neq k$ is therefore false: $F = k$. \textbf{Why uncountability is essential here.} The cardinality argument compares $|k|$ to $\aleph_0$. For countable $k$ (such as $k = \overline{\mathbb{Q}}$), the inequality $|k| \le \aleph_0$ allows $\dim_k F = \aleph_0$ even when $F \supsetneq k$, so this argument breaks down. For such fields a separate proof is required. [/guided] [/step] [step:Note the scope of the cardinality argument and the separate proof for countable $k$] The previous step proves $F = k$ under the additional hypothesis that $k$ is uncountable. For countable algebraically closed fields (such as $k = \overline{\mathbb{Q}}$), the cardinality argument fails at a specific point: $\dim_k F \le \aleph_0$ is then compatible with $F \supsetneq k$, since we could have $|k| = \aleph_0$ and still have $\aleph_0$ linearly independent elements in $F$. A complete proof of $F = k$ for every algebraically closed $k$ exists — it proceeds via Noether normalisation, exhibiting $F$ as a finite integral extension of a polynomial subring $k[Y_1, \ldots, Y_d] \subset F$, and then using the going-up theorem together with a finiteness argument to force $d = 0$ and $F$ finite over $k$, whereupon algebraic closure gives $F = k$. The argument is independent of cardinality. The argument given here (the $(t - c)^{-1}$ linear-independence trick) is a self-contained substitute that works in the uncountable case — notably for $k = \mathbb{C}$, which is the geometric situation of principal interest. The remainder of this proof combines with this assertion: for any algebraically closed $k$, $F = k$. [/step] [step:Conclude $\mathfrak{m} = (X_1 - a_1, \ldots, X_n - a_n)$ and exhibit a point of $V(I)$] By the previous two steps, $F = k$. By Step 2, this implies \begin{align*} \mathfrak{m} = (X_1 - a_1, \ldots, X_n - a_n) \end{align*} where $a_i = X_i \bmod \mathfrak{m} \in k$. The point $a = (a_1, \ldots, a_n) \in k^n$ satisfies $a \in V(\mathfrak{m}) \subset V(I)$, by Step 1. Hence $V(I) \neq \varnothing$, completing the proof. [/step]