This follows a previous question where I want to show that if we have a morphism $\varphi\colon \mathbb P^n\to \mathbb P^m$ with $n>m$, then it is constant. We know that this morphism is induced by a global generated invertible sheaf $\mathcal L$ generated global sections $s_0=\varphi^*x_0,\dots,s_m=\varphi^*x_m$. We also know $\mathcal L\cong \mathcal O_{\mathbb P^n}(d)$ for some $d\in\mathbb Z$. I did the case $d\leq 0$ but unfortunately I can't seem to do the case $d>0.$
The previous "proof" I wrote used dimension (see my previous question) and assumed that $\dim \Gamma(X,\mathcal L)=m+1$, that is $\{s_0,\dots,s_m\}$ is a basis which is either not true or requires justification. I guess the idea is to say $\mathcal O_{\mathbb P^n}(d)\cong \varphi^*\mathcal O_{\mathbb P^m}(1)$ and argue somehow.
Someone also gave an answer to this question in my previous post which I did not understand. The idea is taking a fiber $F$, observing that $\mathcal O_X(d)|_F$ is ample and trivial (why ?), so $F$ must be finite (why ?) so $\varphi$ must be finite (why not only quasi-finite ?) and we get a contradiction from dimension.
Any help ?
So the claim is as follows: for every $d>0$, there do not exist global sections $s_0,\ldots,s_m$ of $\mathcal{L}=\mathcal{O}_{\mathbb{P}^n}(d)$ such that they generate $\mathcal{L}$ at every point.
To prove it, notice that it is equivalent to the following: for any polynomials $p_0,\ldots,p_m$ of $k[x_0,\ldots,x_n]$, homogeneous of the same degree $d>0$, we do not have $\sqrt{(p_0,\ldots,p_m)}=(x_0,\ldots,x_n)$.
Now, this second claim is clear, because $\sqrt{(p_0,\ldots,p_m)}$ is contained in $(x_0,\ldots,x_n)$, so its height is at most $m+1$ (the number of generators). But $(x_0,\ldots,x_n)$ has height $n+1> m+1$…
For the other proof (still assuming $d>0$), let $F$ be a closed fibre of $\varphi$. Then $\mathcal{M} := (\varphi^{\ast}\mathcal{O}(1))_{|F}$ is trivial, since $\varphi: F \rightarrow \mathbb{P}^m$ factors through a closed point.
But $\mathcal{M} \cong \mathcal{O}_{\mathbb{P}^n}(d)_{|F}$ for some $d>0$, hence is ample.
So $F$ is a proper $k$-scheme of finite type such that the trivial line bundle is ample.
But the definition shows that there are global functions $g_1,\ldots,g_r$ on $F$ duch that the $D(g_i)$ are affine and cover $F$.
But $F$ is proper over $k$, so the $g_i$ are locally constant and the $D(g_i)$ are reunions of connected components of $F$. It follows that the connected components of $F$ are affine and proper, hence finite, and F$ is finite.
Hence $\varphi$ is proper and quasi-finite, hence finite and we got our contradiction by dimension theory.