[proofplan] A morphism $\mathbb{P}^n_k \to \mathbb{P}^1_k$ is, by the standard description of morphisms between projective spaces, a pair $[F : G]$ of homogeneous polynomials in $n+1$ variables of equal degree $d$ with no common zero in $\mathbb{P}^n_k$. We argue by induction on $d$. The base case $d = 0$ is immediate: $F, G$ are constants in $k$, not both zero, and $[F : G]$ is the constant point $[F : G] \in \mathbb{P}^1_k$. For the inductive step, suppose $d \geq 1$. Then [Projective Hypersurfaces Always Intersect](/theorems/2157) (which uses $n \geq 2$ centrally) forces a common zero of $F$ and $G$ in $\mathbb{P}^n_k$, contradicting the no-common-zero hypothesis. The "no $d \geq 1$" possibility forces $d = 0$, and the morphism is constant. [/proofplan] [step:Recall the description of morphisms $\mathbb{P}^n_k \to \mathbb{P}^1_k$ as pairs of homogeneous polynomials] A morphism $\varphi : \mathbb{P}^n_k \to \mathbb{P}^1_k$ is given globally by a pair of homogeneous polynomials of equal degree with no common zero. Concretely, there exist \begin{align*} F, G \in k[X_0, X_1, \ldots, X_n] \end{align*} homogeneous of equal degree $d \geq 0$ such that: \begin{itemize} \item $V(F) \cap V(G) = \varnothing$ in $\mathbb{P}^n_k$ — no point of $\mathbb{P}^n_k$ is a common zero, \item $\varphi([a_0 : \cdots : a_n]) = [F(a_0, \ldots, a_n) : G(a_0, \ldots, a_n)]$ for every $[a_0 : \cdots : a_n] \in \mathbb{P}^n_k$. \end{itemize} The condition that $F$ and $G$ have the same degree $d$ ensures that the right-hand side respects the homogeneous-coordinate equivalence $(a_0, \ldots, a_n) \sim (\lambda a_0, \ldots, \lambda a_n)$: \begin{align*} [F(\lambda a) : G(\lambda a)] = [\lambda^d F(a) : \lambda^d G(a)] = [F(a) : G(a)], \end{align*} the last equality because $[s F(a) : s G(a)] = [F(a) : G(a)]$ for any $s \in k^\times$ — projective coordinates are defined up to global scaling. The no-common-zero condition $V(F) \cap V(G) = \varnothing$ ensures that for every $[a_0 : \cdots : a_n] \in \mathbb{P}^n_k$, the pair $(F(a), G(a))$ is not $(0, 0)$, so it represents a valid point of $\mathbb{P}^1_k$. This is the standard "global description" of morphisms between projective spaces — a morphism out of $\mathbb{P}^n_k$ to $\mathbb{P}^m_k$ corresponds to a tuple of $m+1$ homogeneous polynomials of equal degree with no common zero in $\mathbb{P}^n_k$. The proof of this description requires that $\mathbb{P}^n_k$ has a small set of "global" sections of line bundles, but for our purposes we may take it as given. [/step] [step:Consider the case $d = 0$ and show the morphism is constant] Suppose $d = 0$. Then $F, G \in k[X_0, \ldots, X_n]$ are homogeneous of degree $0$ — that is, constants $F = c_F \in k$ and $G = c_G \in k$. The no-common-zero condition becomes $(c_F, c_G) \neq (0, 0)$, so $[c_F : c_G]$ is a well-defined point of $\mathbb{P}^1_k$. For every $[a_0 : \cdots : a_n] \in \mathbb{P}^n_k$, \begin{align*} \varphi([a_0 : \cdots : a_n]) = [F(a) : G(a)] = [c_F : c_G]. \end{align*} The morphism takes the constant value $p_0 := [c_F : c_G] \in \mathbb{P}^1_k$ on every input. Hence $\varphi$ is constant. [/step] [step:Suppose $d \geq 1$ and apply the projective intersection theorem to derive a contradiction] Suppose $d \geq 1$, so $F$ and $G$ are non-constant homogeneous polynomials of equal degree $d$. By the hypothesis $n \geq 2$, the variety $\mathbb{P}^n_k$ is a projective space of dimension $\geq 2$, and $V(F), V(G)$ are hypersurfaces of $\mathbb{P}^n_k$ (defined by single non-constant homogeneous polynomials). The hypotheses of [Projective Hypersurfaces Always Intersect](/theorems/2157) are: \begin{itemize} \item $k$ algebraically closed — given, \item $n \geq 2$ — given, \item $F, G$ non-constant homogeneous polynomials in $n+1$ variables — verified, since $d \geq 1$. \end{itemize} The theorem concludes \begin{align*} V(F) \cap V(G) \neq \varnothing \quad \text{in } \mathbb{P}^n_k. \end{align*} But this contradicts the no-common-zero condition from Step 1, which requires $V(F) \cap V(G) = \varnothing$ for the formula $[F : G]$ to define a morphism on all of $\mathbb{P}^n_k$. The assumption $d \geq 1$ is therefore impossible. [guided] The crux of the argument is a direct application of the previous theorem. The setup demands $V(F) \cap V(G) = \varnothing$ for the morphism to be globally defined; the previous theorem says this is impossible for $n \geq 2$ unless one of $F, G$ is a constant. So we are forced into the constant case. \textbf{What goes wrong if $V(F) \cap V(G) \neq \varnothing$?} Geometrically, a common zero $p \in V(F) \cap V(G) \subset \mathbb{P}^n_k$ would have $\varphi(p) = [F(p) : G(p)] = [0 : 0]$, which is not a valid point of $\mathbb{P}^1_k$ — the homogeneous coordinates $(0, 0)$ are excluded from the equivalence-class definition of $\mathbb{P}^1_k$. So the formula breaks down at $p$, and the morphism is not defined globally — but $\varphi$ was assumed to be a morphism of projective varieties, hence defined everywhere. Contradiction. \textbf{Why does the previous theorem need $n \geq 2$?} The argument in [Projective Hypersurfaces Always Intersect](/theorems/2157) uses Krull's Height Theorem to bound the height of any minimal prime over $(F, G) \subset k[X_0, \ldots, X_n]$ by $2$. The empty intersection scenario requires the radical of $(F, G)$ to be the irrelevant ideal $\mathfrak{m}_0 = (X_0, \ldots, X_n)$, of height $n+1$. The contradiction is $n + 1 \leq 2$, i.e. $n \leq 1$. In dimension $n = 1$, hypersurfaces can be disjoint — for example $V(X_0)$ and $V(X_1)$ in $\mathbb{P}^1_k$ are the two points $[0 : 1]$ and $[1 : 0]$, which are distinct. \textbf{The case $n = 1$ is genuinely different.} For $\mathbb{P}^1_k \to \mathbb{P}^1_k$, non-constant morphisms exist in abundance — they are the rational maps $[X_0 : X_1] \mapsto [F(X_0, X_1) : G(X_0, X_1)]$ for any non-constant $F, G$ of equal degree with no common zero (in $\mathbb{P}^1_k$, "no common zero" is a meaningful condition because $V(F)$ is finite). These are the standard endomorphisms of $\mathbb{P}^1_k$, including the identity, the Frobenius, and so on. So the dimension condition $n \geq 2$ is genuinely used. \textbf{Why does this fail in projective spaces with $n \geq 2$?} Because the source projective space is too "rigid" — it has too few global homogeneous polynomials to construct non-constant morphisms to lower-dimensional projective space. Geometrically, $\mathbb{P}^n_k$ for $n \geq 2$ admits no global rational function (no non-constant element of $k(\mathbb{P}^n_k)$ that is a morphism to $\mathbb{A}^1_k$), and the same rigidity prevents non-constant morphisms to $\mathbb{P}^1_k$ — any such morphism would have to "split" $\mathbb{P}^n_k$ along the levels of a ratio $F/G$, but no such global splitting is possible without a common zero of $F, G$. [/guided] [/step] [step:Conclude that the only possibility is $d = 0$, hence $\varphi$ constant] Combining Steps 2 and 3: \begin{itemize} \item If $d = 0$, then $\varphi$ is constant (Step 2). \item If $d \geq 1$, we reach a contradiction (Step 3), so this case is impossible. \end{itemize} Since $d \geq 0$ is the only constraint on the degree of homogeneous polynomials, and the case $d \geq 1$ has been excluded, we must have $d = 0$. Hence $\varphi$ is constant, completing the proof. [/step]