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Let $Y$ be a scheme. For a locally free $\mathcal{O}_Y$-module $\mathcal{E}$ of rank $n$, the scheme $\mathbb{V}(\mathcal{E})$ is defined as $\mathbb{V}(\mathcal{E}) := \mathbf{Spec}(S^\bullet\mathcal{E})$.

Now if $f: X \rightarrow Y$ is a morphism of schemes, we can consider the locally free $\mathcal{O}_X$-module $f^*\mathcal{E}$, and the associated scheme $\mathbb{V}(f^*\mathcal{E})$.

Are $f^*$ and $\mathbb{V}$ compatible in the sense that there exists a (unique?) morphism $\mathbb{V}(f): \mathbb{V}(f^*\mathcal{E}) \rightarrow \mathbb{V}(\mathcal{E})$ that commutes with the canonical maps $p_X:\mathbb{V}(f^*\mathcal{E}) \rightarrow X$ and $p_Y:\mathbb{V}(\mathcal{E}) \rightarrow Y$? I.e. does $$p_Y\circ\mathbb{V}(f) = f \circ p_X$$ hold?

My problem is that I do not know how to relate the sheaves $f^*\mathcal{E}$ and $\mathcal{E}$, as both live on different spaces, let alone the schemes defined by those.

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Such a morphism is not unique. For instance, if $X = Y = Spec(k)$ and $f$ is the identity, then $V(E) = V(f^*E)$ is just an affine space, and any its automorphism satisfies the required compatibility.

On the other hand, there is a canonical choice of a morphism induced by the natural isomorphism of sheaves of algebras $$ f^*(S^\bullet E) \cong S^\bullet(f^* E). $$

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  • $\begingroup$ I know the isomorphism $f^*(S^\bullet E) = S^\bullet(f^*E)$. But how exactly does it induce a map $V(f^*E) \rightarrow V(E)$? I also thought that maybe $V(f^*E) = X \times_Y V(E)$? Is that true? $\endgroup$ – red_trumpet Sep 24 '18 at 12:38
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    $\begingroup$ The fiber product formula is correct. Maybe the universal property of $V(E)$ is useful --- a morphism from $S$ to $V_X(E)$ is the same as a morphism $\phi : S \to X$ and a morphism of sheaves $\phi^*E \to O_S$. $\endgroup$ – Sasha Sep 24 '18 at 13:25
  • $\begingroup$ Yes, the universal property seems to work. Do you have an reference or is that easy to see? $\endgroup$ – red_trumpet Sep 24 '18 at 14:06
  • $\begingroup$ To prove the property it is enough to assume that $S$ and $X$ are affine, and then the statement is evident (it is about extending a morphism of algebras to a polynomial algebra over the source). $\endgroup$ – Sasha Sep 24 '18 at 14:18

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