# Global sections of invertible sheaf generating the same subspace and morphisms to projective space

This is a question from Hartshorne I'm not sure about. Let $X$ be a scheme over a field $k$ and let $\mathcal{L}$ be an invertible sheaf. Suppose that $t_0, \ldots, t_m$ and $s_0, \ldots, s_n$ are global sections of $\mathcal{L}$ which generate it at every point, and furthermore assume that the $t_i$ and the $s_i$ generate the same subspace $V$ of $\Gamma (X, \mathcal{L})$. If $n \leq m$ then we need to prove that the morphism induced by the $s_i$ to $\mathbb{P}^n$ is equal to the morphism induced by the $t_i$ to $\mathbb{P}^m$, followed by a linear projection from a suitable linear subspace of $\mathbb{P}^m$ to $\mathbb{P}^n$, and finally an automorphism of $\mathbb{P}^n$.

Now, if the $s_i$ happen to be linearly independent then I think this can be solved: without loss, take $t_0, \ldots , t_n$ to generate $V$. Then the invertible matrix that takes $t_i$ to $s_i$ gives the required automorphism of $\mathbb{P}^n$. But what about the general case?

(This is Hartshorne Chap II Ex 7.2)

It's almost the same: suppose $s_0,\dots,s_r$ and $t_0,\dots,t_r$ are basis of $V$, with $r\leq n$. After a projecton, you may suppose $n=m$. Take the invertible $r+1\times r+1$ matrix sending $t_i$ to $s_i$, and extend it to an invertible $n+1\times n+1$ matrix as you want (adding an $n-r\times n-r$ identity block, for example). Hence you may suppose $t_i=s_i$ $\forall i\leq r$. Here comes the trick: it's enough to do the case where $s_{r+1},\dots,s_n$ are all zero (in the general case you will simply apply the special case two times).
To do this, take the $n-r\times r+1$ matrix $A$ expressing the coordinates of $t_{r+1},\dots t_n$ with respect to $t_0,\dots,t_r$. The last automporphism of $\mathbb{P}^n$ you need is expressed by:
$$\begin{bmatrix}\operatorname{Id}_{r+1} & 0\\-A & \operatorname{Id}_{n-r}\end{bmatrix}$$
which takes $t_0,\dots,t_r,t_{r+1},\dots,t_n$ to $t_0,\dots,t_r,0,\dots,0$.
• I must be missing something simple... How can this matrix be invertible if it takes the possibly nonzero element $t_i$ to zero? Commented Feb 14, 2014 at 10:38
• Because it acts on $k^{n+1}$ , not on $V$! It takes $t_i$ to some non zero vector of $k^{n+1}$ that, when evaluated on $t_0,\dots,t_n$, is zero in $V$. Commented Feb 14, 2014 at 11:44
• @user115940 Hi. I have a question. Why can these (invertible) matrices finally give an automorphism of $\mathbb{P}^n_k$? The matrix between $t_i$ and $s_i$ must be with coefficients in $\Gamma(X, \mathscr{O}_X)$, while the one in $\mathrm{PGL}(n,k)$ is only with coefficients in $k$. How to correlate these? Commented Apr 9, 2020 at 11:26