Constant morphism from $\mathbb{P}^m_k$ to $\mathbb{P}^n_k$ as schemes Let $k$ be a field and $m>n \in \mathbb{N}$. Then any morphism of schemes $\mathbb{P}^m_k \rightarrow \mathbb{P}^n_k$ is constant.
What I know is that every morphism from a scheme $X$ to $\mathbb{P}^n_k$ corresponds $1:1$ to isomorphism classes of invertible sheaves $\mathcal{L}$ on $X$ and generating sections $s_0,...,s_n \in \Gamma(X,\mathcal{L})$.
What I'm thinking is also that $\mathbb{P}^m_k$ cannot be generated by $n+1$ sections.
 A: You've left out a very important part of the definition of a line bundle $\mathcal{L}$ and sections $s_0,\cdots,s_n$ defining a morphism $X\to \Bbb P^n$: the sections $s_0,\cdots, s_n$ cannot have any common zeroes. This will be important to reach a contradiction later.
By the classification of line bundles on projective space, $\mathcal{L}\cong \mathcal{O}
_{\Bbb P^m}(d)$ for some integer $d$. If $d<0$, then $\mathcal{L}$ has no global sections and does not define a morphism to $\Bbb P^n$ at all. If $d=0$, then it only has the constant global sections and must define a constant morphism. Therefore, if $\mathcal{L}$ and a choice of sections $s_0,\cdots,s_n$ were to define a nonconstant map, $d$ must be at least one.
We know that the global sections of $\mathcal{O}_{\Bbb P^m}(d)$ are the homogeneous polynomials of degree $d$ in $m+1$ variables. Now we show that for any choice of $n+1$ homogeneous polynomials of positive degree in $m+1$ variables, they must have a common zero, which will be a contradiction. By dimension theory, the subscheme of $\Bbb P^m$ cut out by the zero locus of the $n+1$ homogeneous polynomials has dimension at least $m-(n+1)$ and is nonempty if this quantity is non-negative. But as $m>n$, it must be the case that $m-(n+1) \geq 0$, so this locus is nonempty, which is a contradiction.
