# Morphism of Sheaves of Rings

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)

## 1 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.

• ok so the $\mathcal{O}_Y$-module structure for every morphism $m: \mathcal{G} \to f_*\mathcal{F}$ is settled locally for every open $U \subset Y$ on $m_U:\mathcal{G}(U) \to f_*\mathcal{F}(U)$ via $rn \mapsto m_U(rn) := f^{\#}(r) \cdot m_U(n)$ where $n \in \mathcal{G}(U)$ and $r \in \mathcal{O}_Y(U)$, right? – KarlPeter Feb 25 at 14:07
• And in case of $m= f^{\#}$ locally the ring morphism is "swapped" to the $\mathcal{O}_Y(U)$ module stricure by $f^{\#}_U(r) = f^{\#}_U(r \cdot 1) = f^{\#}_U(r) \cdot f^{\#}_U(1)$? – KarlPeter Feb 25 at 14:13
• @KarlPeter I don't quiet understand what you want to say. The $\mathcal{O}_Y$-module structure on $f_\ast \mathcal{F}$ is given as follows: For an open $U$ of $Y$, $(f_\ast \mathcal{F})(U)=\mathcal{F}(f^{-1}(U))$ is an $\mathcal{O}_X(f^{-1}(U))$-module (as $\mathcal{F}$ is by assumption an $\mathcal{O}_X$-module). Now you have $f^\sharp_U\colon \mathcal{O}_Y(U)\to \mathcal{O}_X(f^{-1}(U))$ making $\mathcal{O}_X(f^{-1}(U))$ into an $\mathcal{O}_Y(U)$-algebra. In particular, via this algebra structure any module over $\mathcal{O}_X(f^{-1}(U))$ may be viewed as an $\mathcal{O}_Y(U)$-module. – user363120 Feb 25 at 16:10