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.


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 $$ f^*(S^\bullet E) \cong S^\bullet(f^* E). $$

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.