[proofplan] The morphism $\varphi$ is given by a tuple $[F_0 : F_1 : \cdots : F_m]$ of homogeneous polynomials in $X_0, \ldots, X_n$ of equal degree $d$ with no common zero in $\mathbb{P}^n_k$. We argue by contradiction: assume $d \geq 1$. The no-common-zero condition translates to "the affine intersection $V_{\mathrm{aff}}(F_0) \cap \cdots \cap V_{\mathrm{aff}}(F_m)$ in $\mathbb{A}^{n+1}_k$ equals $\{0\}$", which by [Hilbert's Nullstellensatz](/theorems/2124) forces the radical $\sqrt{(F_0, \ldots, F_m)}$ to equal the irrelevant maximal ideal $\mathfrak{m}_0 = (X_0, \ldots, X_n)$ of height $n+1$. By [Krull's Height Theorem](/theorems/???), every minimal prime over an ideal generated by $m+1$ elements has height at most $m+1$. The contradiction is $n + 1 \leq m + 1$, i.e. $n \leq m$, contradicting $n > m$. Hence $d = 0$ and $\varphi$ is constant. [/proofplan]

[step:Recall the description of morphisms $\mathbb{P}^n_k \to \mathbb{P}^m_k$] A morphism $\varphi : \mathbb{P}^n_k \to \mathbb{P}^m_k$ is given globally by an $(m+1)$-tuple of homogeneous polynomials of equal degree with no common zero. There exist \begin{align*} F_0, F_1, \ldots, F_m \in k[X_0, X_1, \ldots, X_n] \end{align*} homogeneous of common degree $d \geq 0$ such that: \begin{itemize} \item $V(F_0) \cap V(F_1) \cap \cdots \cap V(F_m) = \varnothing$ in $\mathbb{P}^n_k$ — no point of $\mathbb{P}^n_k$ is a simultaneous zero of all $F_i$, \item $\varphi([a_0 : \cdots : a_n]) = [F_0(a) : F_1(a) : \cdots : F_m(a)]$ for all $[a] \in \mathbb{P}^n_k$. \end{itemize} The equal-degree condition ensures the formula descends to projective coordinates: scaling $a$ by $\lambda \in k^\times$ scales each $F_i(a)$ by $\lambda^d$, leaving the ratio unchanged. The no-common-zero condition ensures $(F_0(a), \ldots, F_m(a)) \neq (0, \ldots, 0)$, so the right-hand side is a valid point of $\mathbb{P}^m_k$. [/step]

[step:Translate the no-common-zero condition to an ideal-theoretic condition on the affine cone] For each $i$, let $V_{\mathrm{aff}}(F_i) \subset \mathbb{A}^{n+1}_k$ be the affine cone over the projective hypersurface $V(F_i)$. Each $V_{\mathrm{aff}}(F_i)$ contains the origin (since $F_i$ is homogeneous of positive degree when $d \geq 1$, or constant when $d = 0$ — we handle the cases separately). For $d \geq 1$: each $F_i$ is homogeneous of positive degree, so $F_i(0, \ldots, 0) = 0$, and $0 \in V_{\mathrm{aff}}(F_i)$. The simultaneous vanishing locus \begin{align*} V_{\mathrm{aff}}(F_0) \cap V_{\mathrm{aff}}(F_1) \cap \cdots \cap V_{\mathrm{aff}}(F_m) \subset \mathbb{A}^{n+1}_k \end{align*} contains the origin. Its non-origin part corresponds to the projective intersection $V(F_0) \cap \cdots \cap V(F_m) \subset \mathbb{P}^n_k$ — points of $\mathbb{A}^{n+1}_k$ are non-zero representatives, projective points are equivalence classes under scaling, and the cone $V_{\mathrm{aff}}(F_i)$ is closed under scaling. By the no-common-zero hypothesis from Step 1, the projective intersection $V(F_0) \cap \cdots \cap V(F_m)$ is empty. Hence the affine cone intersection consists of only the origin: \begin{align*} V_{\mathrm{aff}}(F_0) \cap \cdots \cap V_{\mathrm{aff}}(F_m) = \{0\} \subset \mathbb{A}^{n+1}_k. \end{align*} Equivalently, $V((F_0, F_1, \ldots, F_m)) = \{0\}$ in $\mathbb{A}^{n+1}_k$, where $(F_0, \ldots, F_m) \subset k[X_0, \ldots, X_n]$ is the ideal generated by the $F_i$. [/step]

[step:Apply Hilbert's Nullstellensatz to identify the radical] Since $k$ is algebraically closed, [Hilbert's Nullstellensatz](/theorems/2124) gives \begin{align*} \sqrt{(F_0, F_1, \ldots, F_m)} = I(V((F_0, \ldots, F_m))) = I(\{0\}) = (X_0, X_1, \ldots, X_n) =: \mathfrak{m}_0. \end{align*} The middle equality $I(\{0\}) = (X_0, \ldots, X_n)$ holds because a polynomial vanishes at the origin if and only if its constant term is zero, equivalently if and only if it lies in $\mathfrak{m}_0$. The ideal $\mathfrak{m}_0$ is maximal in $k[X_0, \ldots, X_n]$ (the quotient is the field $k$), hence prime. Since $\sqrt{(F_0, \ldots, F_m)} = \mathfrak{m}_0$ is itself prime and maximal, it is the unique prime containing $(F_0, \ldots, F_m)$, and in particular the unique minimal prime over $(F_0, \ldots, F_m)$. [/step]

[step:Compute the height of $\mathfrak{m}_0$ and derive a contradiction via Krull's Height Theorem]The irrelevant maximal ideal $\mathfrak{m}_0 = (X_0, X_1, \ldots, X_n)$ has height $n+1$ in $k[X_0, \ldots, X_n]$, witnessed by the explicit chain of primes \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 $n+1$ (each $(X_0, \ldots, X_i)$ is prime since the quotient $k[X_0, \ldots, X_n]/(X_0, \ldots, X_i) \cong k[X_{i+1}, \ldots, X_n]$ is an integral domain). This matches the Krull dimension of the polynomial ring: $\dim k[X_0, \ldots, X_n] = n+1$, with $\mathfrak{m}_0$ a maximal ideal of maximal height. Now apply [Krull's Height Theorem](/theorems/???) (the generalised principal ideal theorem) to the ideal $(F_0, F_1, \ldots, F_m) \subset k[X_0, \ldots, X_n]$. The hypotheses are: \begin{itemize} \item The ring $k[X_0, \ldots, X_n]$ is Noetherian — a polynomial ring over a field, by the Hilbert Basis Theorem. \item The ideal $(F_0, \ldots, F_m)$ is generated by $m + 1$ elements. \end{itemize} The conclusion is: every minimal prime $\mathfrak{p}$ over $(F_0, \ldots, F_m)$ satisfies \begin{align*} \operatorname{ht}(\mathfrak{p}) \leq m + 1. \end{align*} By Step 3, the unique minimal prime is $\mathfrak{m}_0$ with $\operatorname{ht}(\mathfrak{m}_0) = n+1$. Combining: \begin{align*} n + 1 = \operatorname{ht}(\mathfrak{m}_0) \leq m + 1, \end{align*} i.e. $n \leq m$. This contradicts the hypothesis $n > m$.[/step]

[guided]The argument is a direct generalisation of the proof of [Projective Hypersurfaces Always Intersect](/theorems/2157), with the number $2$ of generating polynomials replaced by $m+1$. \textbf{Why is the dimension count $n + 1 \leq m + 1$?} The affine cone over $\mathbb{P}^n_k$ is $\mathbb{A}^{n+1}_k$, of dimension $n+1$. The simultaneous vanishing of $m+1$ homogeneous polynomials (one for each homogeneous coordinate of $\mathbb{P}^m_k$) defines a closed subset that, by Krull, has codimension at most $m+1$ in the cone. The "no common zero" requirement says this subset is just the origin — codimension $n+1$ in the cone. So we need $n + 1 \leq m + 1$, the same as $n \leq m$. When $n > m$, this fails: there are not enough polynomials to cut the cone all the way down to a point. \textbf{Geometric intuition.} Each polynomial $F_i$ defines a hypersurface (codimension-$1$ subvariety) in $\mathbb{P}^n_k$. Generically, $m+1$ hypersurfaces intersect in codimension $m+1$. For $\mathbb{P}^n_k$ of dimension $n$, this intersection has dimension $n - (m+1) = n - m - 1$. When $n > m$, this is $\geq 0$, so the intersection is non-empty. The Krull theorem makes this rigorous as an upper bound on codimension, regardless of whether the polynomials are in "general position". \textbf{The $n = m$ case.} When $n = m$, the inequality $n+1 \leq m+1$ holds with equality, so the Krull bound does not give a contradiction, and indeed there exist non-constant morphisms $\mathbb{P}^n_k \to \mathbb{P}^n_k$ — for example, the identity map $[X_0 : \cdots : X_n] \mapsto [X_0 : \cdots : X_n]$, which is given by the $n+1$ coordinate polynomials of degree $1$ with no common zero. Or any automorphism of $\mathbb{P}^n_k$, given by a linear change of coordinates. \textbf{The $n < m$ case.} When $n < m$, even more morphisms are possible (e.g. linear inclusions $\mathbb{P}^n_k \hookrightarrow \mathbb{P}^m_k$ given by homogeneous coordinates of degree $1$). The phenomenon "morphisms must be constant" is special to $n > m$. \textbf{Why is the no-common-zero condition strict?} Without it, the formula $[F_0 : \cdots : F_m]$ would not extend to a morphism on all of $\mathbb{P}^n_k$ — at a common zero, we would get $[0 : \cdots : 0]$, undefined in $\mathbb{P}^m_k$. So a morphism, by definition, requires no common zero, and we are forced into the contradiction.[/guided]

[step:Conclude that the only possibility is $d = 0$, hence $\varphi$ constant] The contradiction in Step 4 was reached under the assumption $d \geq 1$. Hence $d = 0$. When $d = 0$, each $F_i = c_i \in k$ is a constant. The no-common-zero condition reduces to "$(c_0, c_1, \ldots, c_m) \neq (0, 0, \ldots, 0)$" — there is at least one non-zero constant — so $[c_0 : c_1 : \cdots : c_m] \in \mathbb{P}^m_k$ is a well-defined point. For every $[a_0 : \cdots : a_n] \in \mathbb{P}^n_k$, \begin{align*} \varphi([a]) = [F_0(a) : \cdots : F_m(a)] = [c_0 : c_1 : \cdots : c_m]. \end{align*} The morphism takes the constant value $p_0 := [c_0 : \cdots : c_m] \in \mathbb{P}^m_k$ on every input. Hence $\varphi$ is constant, completing the proof. [/step]