Morphism of Sheaves of Rings

algebraic-geometrysheaf-theory

Let $f: X \to Y$ be a morphism of ringed spaces, $\mathcal{G}$ a $\mathcal{O}_Y$-module, $\mathcal{F}$ a $\mathcal{O}_X$-module.

It is well known that the fuctors $f^*, f_*$ are adjunct via the adjunction formula

$$Hom_{\mathcal{O}_X}(f^*\mathcal{G},\mathcal{F})= Hom_{\mathcal{O}_Y}(\mathcal{G}, f_*\mathcal{F})$$

I have following question:

If we take $\mathcal{G}:= \mathcal{O}_Y$ and $\mathcal{F}:= f^*(\mathcal{O}_Y)$ then the adjunction relation becomes

$$Hom_{\mathcal{O}_X}(f^*(\mathcal{O}_Y),f^*(\mathcal{O}_Y))= Hom_{\mathcal{O}_Y}(\mathcal{O}_Y, f_*f^*(\mathcal{O}_Y))$$

And I often read that with this formula the identity $id_{f^*\mathcal{G}}$ on the left side corresponds to the sheaf morphism $f^{\#}: \mathcal{O}_Y \to f_*f^*(\mathcal{O}_Y)=f^* \mathcal{O}_X$ on the right hand side which coinsides with sheaf morphism from the morphism of ringed spaces $f = (f, f^{\#}):(X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$.

What I don't understand is that via the adjunction formula $f^{\#}$ has to be a morphism of $\mathcal{O}_Y$-modules, so for every open $U \subset Y$ the induced morphism $f^{\#}_U: \mathcal{O}_Y(U) \to f^* \mathcal{O}_X(U) =\mathcal{O}_X(f^{-1}(U))$ has to be a morphism of $\mathcal{O}_Y(U)$-modules

but on the other hand it is well known that $f^{\#}$ as a morphism arrising from morphism of ringed spaces $f$ is a morphism of sheaves of rings on $Y$, so it gives for each open $U \subset Y$ a ring morphism $f^{\#}_U: \mathcal{O}_Y(U) \to \mathcal{O}_X(f^{-1}(U))$

and a ring morphism $\phi: R \to A$ is in general never a $R$-module morphism.

So I don't understand why we can say that $f^{\#}$ is a morphism of $\mathcal{O}_Y$-modules (this gives the adjunction formula) where each $f^{\#}_U$ is a ring morphism since on the other hand $f^{\#}$ is a morphism of sheaves of rings on $Y$.

References for this construction: e. g.

  • Bosch "Commutative Algebra and Algebraic Geometry" (page 269)

  • Görtz, Wedhorn "Algebraic Geometry" (page 181)

Best Answer

The way you view $f_\ast \mathcal{F}$ as an $\mathcal{O}_Y$-module is via the morphism $f^\sharp\colon \mathcal{O}_Y\to f_\ast \mathcal{O}_X$. So it all boils down to observing that, given an $R$-algebra $A$, the structure morphism $R\to A$ is an $R$-module morphism.

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