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The exercise is given below.

$5.1.24.$ Let $X$ be an algebraic variety over a field $k$. Let $f : X \to Y := \mathbb{P}^{n}_{k}$ be a morphism of algebraic $k-$varieties. Let $\mathcal{L} = f^{*}\mathcal{O}_{Y}(1)$. Let $\sigma$ be an automorphism of $X$ such that $\sigma^{*}\mathcal{L} \simeq \mathcal{L}$. Show that there exists an automorphism $\sigma'$ of $Y$ such that $\sigma' \circ f = f \circ \sigma$.

From what I know, a $k-$ automorphism of $ \mathbb{P}^{n}_{k} $ comes from an automorphism of the graded $k-$ algebra $k[X_{1}, ..., X_{n}]$. With the hypothesis, if $g := f \circ \sigma $ we have $f^{*}\mathcal{O}_{Y}(1) = g^{*} \mathcal{O}_{Y}(1)$. From here I really don't know what to do. Any help would be appreciated.

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    $\begingroup$ Please don't use images for questions because such questions don't show up in search results. $\endgroup$ Jul 9, 2023 at 8:26
  • $\begingroup$ Hi, you're right : I edit my post. $\endgroup$
    – Analyse300
    Jul 10, 2023 at 7:54
  • $\begingroup$ The annoying thing for me is that the construction of $f^{*}$ is not explicit. $\endgroup$
    – Analyse300
    Jul 13, 2023 at 8:44

1 Answer 1

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Let $V = H^0(X, \mathcal L)$. Then the map $f$ corresponds to the quotient $$V \otimes \mathcal O_X \to \mathcal L \to 0,$$ so we can think of $f$ as a map $f: X \to \mathbb P(V)$¹.

Now choose an isomorphism $\varphi: \sigma^* \mathcal L \to \mathcal L$. This induces an isomorphism on global sections $\phi: V \to V$, which in turn induces an isomorphism $\sigma': \mathbb P(V) \to \mathbb P(V)$.

Now it remains to check that $\sigma'$ is compatible with $\sigma$. Can you do this on your own?


¹ I'm using Grothendieck's version of $\mathbb P(V)$ here, i.e. a point in $\mathbb P(V)$ is an equivalence class of quotients $V \to k \to 0$. Two quotients are equivalent iff they have the same kernel. The map $f$ then maps each closed point $x \in X$ to the quotient $[V \to \mathcal L_x \otimes_{\mathcal O_{X,x}} k(x) \to 0]$.

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  • $\begingroup$ Thanks for your answer. I don't understand what is $\mathbb{P}(V)$. Do you have references for that? $\endgroup$
    – Analyse300
    Jul 15, 2023 at 9:14
  • $\begingroup$ @Analyse300 That is the projective space over $V$. If you learned that projective space $\mathbb P^n$ consists of lines through the origin $L \subset k^{n+1}$, then in my notation, $\mathbb P(V)$ parametrises lines in the vector space $V^*$. A lines $L \subset V^*$ is the same as a quotient $V \to L^* \to 0$. $\endgroup$ Mar 12 at 9:05

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