[proofplan] The argument passes from the projective intersection $V(F) \cap V(G) \subset \mathbb{P}^n_k$ to the affine intersection $V(F) \cap V(G) \subset \mathbb{A}^{n+1}_k$ on the cone, where the equivalence is between "non-empty in $\mathbb{P}^n_k$" and "contains a non-zero point in $\mathbb{A}^{n+1}_k$". We argue by contradiction: assume the projective intersection is empty, hence the affine intersection is exactly $\{0\}$. Then by [Hilbert's Nullstellensatz](/theorems/2124) the radical of $(F, G)$ equals the ideal of the origin, which is the irrelevant ideal $(X_0, \ldots, X_n)$. This forces every minimal prime over $(F, G)$ to be $(X_0, \ldots, X_n)$, of height $n+1$. But [Krull's Height Theorem](/theorems/???) — the principal ideal theorem in its generalised form — bounds the height of any minimal prime over an ideal generated by $r$ elements by $r$. With $r = 2$ generators and $n + 1 \geq 3$, the bound is violated, giving the contradiction. [/proofplan] [step:Reduce the projective intersection question to a question about the affine cone] Recall the construction of $\mathbb{P}^n_k$ as the quotient $(k^{n+1} \setminus \{0\})/{\sim}$ where $(a_0, \ldots, a_n) \sim (\lambda a_0, \ldots, \lambda a_n)$ for $\lambda \in k^\times$. For a homogeneous polynomial $H \in k[X_0, \ldots, X_n]$ of degree $d$, \begin{align*} H(\lambda a_0, \ldots, \lambda a_n) = \lambda^d \, H(a_0, \ldots, a_n), \end{align*} so $H$ vanishes at the representative $(a_0, \ldots, a_n) \in k^{n+1}$ if and only if it vanishes at every scalar multiple. The set \begin{align*} V_{\mathrm{aff}}(H) := \{(a_0, \ldots, a_n) \in \mathbb{A}^{n+1}_k : H(a_0, \ldots, a_n) = 0\} \end{align*} — the affine cone over the projective variety $V(H)$ — is the union of $\{0\}$ with all representatives of the projective points of $V(H)$: \begin{align*} V_{\mathrm{aff}}(H) = \{0\} \cup \{\lambda \cdot a : \lambda \in k^\times,\, [a] \in V(H)\}. \end{align*} The origin $0 = (0, \ldots, 0)$ is contained in $V_{\mathrm{aff}}(H)$ for every homogeneous $H$ of positive degree (since $H(0) = 0$ when $\deg H \geq 1$). Hence \begin{align*} V_{\mathrm{aff}}(H) = \{0\} \iff V(H) = \varnothing \text{ in } \mathbb{P}^n_k. \end{align*} Applied to $V_{\mathrm{aff}}(F) \cap V_{\mathrm{aff}}(G)$, the condition that this intersection equals $\{0\}$ in $\mathbb{A}^{n+1}_k$ is equivalent to the condition that $V(F) \cap V(G) = \varnothing$ in $\mathbb{P}^n_k$. [/step] [step:Suppose by contradiction that the projective intersection is empty] Suppose for contradiction that $V(F) \cap V(G) = \varnothing$ in $\mathbb{P}^n_k$. By Step 1, \begin{align*} V_{\mathrm{aff}}(F) \cap V_{\mathrm{aff}}(G) = \{0\} \subset \mathbb{A}^{n+1}_k. \end{align*} The intersection $V_{\mathrm{aff}}(F) \cap V_{\mathrm{aff}}(G)$ is the affine variety of the ideal $(F, G) \subset k[X_0, \ldots, X_n]$: \begin{align*} V_{\mathrm{aff}}(F) \cap V_{\mathrm{aff}}(G) = V((F, G)), \end{align*} because the simultaneous vanishing locus of $F$ and $G$ equals the vanishing locus of any ideal they generate. [/step] [step:Apply Hilbert's Nullstellensatz to identify the radical] Since $k$ is algebraically closed, [Hilbert's Nullstellensatz](/theorems/2124) applies to the ideal $(F, G) \subset k[X_0, \ldots, X_n]$ — the polynomial ring over an algebraically closed field. The Nullstellensatz states $I(V(J)) = \sqrt{J}$ for any ideal $J$. Applied to $J = (F, G)$ with $V((F, G)) = \{0\}$: \begin{align*} \sqrt{(F, G)} = I(V((F, G))) = I(\{0\}) = (X_0, X_1, \ldots, X_n). \end{align*} Here we used $I(\{0\}) = (X_0, \ldots, X_n)$: a polynomial $p \in k[X_0, \ldots, X_n]$ vanishes at the origin if and only if its constant term is zero, equivalently if and only if $p$ lies in the maximal ideal $(X_0, \ldots, X_n)$. The ideal $\mathfrak{m}_0 := (X_0, X_1, \ldots, X_n)$ is the irrelevant maximal ideal. It is prime (the quotient $k[X_0, \ldots, X_n]/\mathfrak{m}_0 \cong k$ is a field). Since $\sqrt{(F, G)} = \mathfrak{m}_0$ is itself a prime ideal, $\mathfrak{m}_0$ is the unique minimal prime of $\sqrt{(F, G)}$, and hence of $(F, G)$ as well (a prime is minimal over $J$ if and only if it is minimal over $\sqrt{J}$). [/step] [step:Compute the height of the irrelevant ideal] Recall that the height $\operatorname{ht}(\mathfrak{p})$ of a prime ideal $\mathfrak{p}$ in a Noetherian ring $R$ is the supremum of lengths of strictly ascending chains of primes \begin{align*} \mathfrak{p}_0 \subsetneq \mathfrak{p}_1 \subsetneq \cdots \subsetneq \mathfrak{p}_\ell = \mathfrak{p}. \end{align*} For the ring $R = k[X_0, \ldots, X_n]$, the irrelevant ideal $\mathfrak{m}_0 = (X_0, \ldots, X_n)$ has height $n+1$. This is because $k[X_0, \ldots, X_n]$ has Krull dimension $n+1$ (the dimension of polynomial rings over a field equals the number of variables), and $\mathfrak{m}_0$ is a maximal ideal with residue field $k$, so its height equals the Krull dimension: \begin{align*} \operatorname{ht}(\mathfrak{m}_0) = \dim k[X_0, \ldots, X_n] = n + 1. \end{align*} A concrete chain witnessing height $n+1$ is \begin{align*} (0) \subsetneq (X_0) \subsetneq (X_0, X_1) \subsetneq \cdots \subsetneq (X_0, X_1, \ldots, X_n) = \mathfrak{m}_0, \end{align*} of length $n+1$. [/step] [step:Apply Krull's Height Theorem to derive a contradiction] By [Krull's Height Theorem](/theorems/???) (the generalised principal ideal theorem) applied to the ideal $(F, G) \subset k[X_0, \ldots, X_n]$: every minimal prime over an ideal generated by $r$ elements in a Noetherian ring has height at most $r$. The ring $k[X_0, \ldots, X_n]$ is Noetherian (it is a polynomial ring over a field, so by the Hilbert Basis Theorem all its ideals are finitely generated). The ideal $(F, G)$ is generated by $r = 2$ elements. Hence every minimal prime $\mathfrak{p}$ over $(F, G)$ satisfies \begin{align*} \operatorname{ht}(\mathfrak{p}) \leq 2. \end{align*} By Step 3, the unique minimal prime over $(F, G)$ is $\mathfrak{m}_0 = (X_0, \ldots, X_n)$, with $\operatorname{ht}(\mathfrak{m}_0) = n + 1$ by Step 4. The height bound from Krull's theorem gives \begin{align*} n + 1 = \operatorname{ht}(\mathfrak{m}_0) \leq 2, \end{align*} i.e. $n \leq 1$. This contradicts the hypothesis $n \geq 2$. The assumption $V(F) \cap V(G) = \varnothing$ is therefore false, and we conclude $V(F) \cap V(G) \neq \varnothing$ in $\mathbb{P}^n_k$. [guided] The proof has a clean three-part structure: a geometric translation (project $\to$ cone), an algebraic identification (Nullstellensatz), and a dimension bound (Krull). Tracking the role of each piece: \textbf{Why pass to the cone?} Projective space $\mathbb{P}^n_k$ has dimension $n$, while its affine cone $\mathbb{A}^{n+1}_k$ has dimension $n+1$ — one more dimension. The extra dimension is exactly the scaling degree of freedom we quotient out to make $\mathbb{P}^n_k$. The Nullstellensatz, Krull's Height Theorem, and other commutative-algebra tools work cleanly in the affine ring $k[X_0, \ldots, X_n]$, not directly in the projective coordinate ring. So we lift the geometric question "is $V(F) \cap V(G)$ empty in $\mathbb{P}^n_k$?" to "is $V_{\mathrm{aff}}(F) \cap V_{\mathrm{aff}}(G) = \{0\}$ in $\mathbb{A}^{n+1}_k$?" — equivalent because the cone always contains the origin and otherwise consists of full lines through the origin. \textbf{Why does the Nullstellensatz give us $\sqrt{(F, G)} = \mathfrak{m}_0$?} Hilbert's Nullstellensatz translates between geometry (vanishing loci $V(\cdot)$) and algebra (radical ideals $\sqrt{(\cdot)}$). Applied to $(F, G)$ with the affine vanishing locus equal to $\{0\}$, it states that the radical of $(F, G)$ is the ideal of the origin — and the ideal of the origin is exactly the maximal ideal $\mathfrak{m}_0 = (X_0, \ldots, X_n)$. The hypothesis $k$ algebraically closed is essential here: without it, varieties may have fewer geometric points than the algebra suggests, and the Nullstellensatz can fail. \textbf{Why is $\mathfrak{m}_0$ the unique minimal prime?} Since $\mathfrak{m}_0$ is itself prime, and the radical is the intersection of all primes containing $(F, G)$, having $\sqrt{(F, G)} = \mathfrak{m}_0$ means $\mathfrak{m}_0$ is contained in every prime containing $(F, G)$ — but $\mathfrak{m}_0$ is maximal, so the only prime containing it is itself. Hence $\mathfrak{m}_0$ is the only prime over $(F, G)$, in particular the unique minimal one. \textbf{Why is the height of $\mathfrak{m}_0$ equal to $n+1$?} The ring $k[X_0, \ldots, X_n]$ has Krull dimension $n+1$ — a fact established by induction on the number of variables (the Krull dimension of $R[X]$ is $\dim R + 1$ when $R$ is a Noetherian ring of finite Krull dimension). The maximal ideal $\mathfrak{m}_0$ "sees" the entire dimension because the quotient by it is $k$, the smallest possible residue field. Concretely, the chain $(0) \subsetneq (X_0) \subsetneq (X_0, X_1) \subsetneq \cdots$ realises the height. \textbf{The Krull Height Theorem inequality.} The theorem says that adding $r$ generators to an ideal can drop the codimension by at most $r$. A single non-zero-divisor cuts dimension by $1$ (this is the principal ideal theorem); $r$ elements cut dimension by at most $r$. Equality "minimal prime over $(f_1, \ldots, f_r)$ has height $r$" holds when the $f_i$ form a regular sequence (cut transversally), but the inequality $\leq r$ is unconditional. Two homogeneous polynomials in $\mathbb{P}^n_k$ for $n \geq 2$ would need to cut the cone down to dimension $0$ (the origin), which is height $n+1 \geq 3$, exceeding $r = 2$. Hence they cannot, and they must intersect non-trivially. \textbf{The role of $n \geq 2$.} For $n = 1$, two homogeneous polynomials in $\mathbb{P}^1_k$ may indeed be disjoint: e.g. $F = X_0$ and $G = X_1$ have $V(F) \cap V(G) = \varnothing$ since this would require both coordinates to vanish, but $(0, 0)$ is not a valid point of $\mathbb{P}^1_k$. In that case the cone $V_{\mathrm{aff}}(F) \cap V_{\mathrm{aff}}(G) = \{0\} \subset \mathbb{A}^2_k$ has $\mathfrak{m}_0 = (X_0, X_1)$ of height $2$, exactly matching the bound from $r = 2$ generators — no contradiction. The strict inequality $n + 1 > 2$, equivalent to $n \geq 2$, is exactly when the bound is violated. [/guided] [/step]