I was reading through some notes for an exam and one exericse asks me to prove the following
There is a unique sheaf of rings making a topological set $X$ with discrete topology a ringed space.
I tried doing it but I feel I'm missing something, using the definition of presheaf and than of sheaf doesn't seem to bring me any result.
How can I solve such a problem? I leave you my definitions of presheaf and sheaf.
A presheaf $\mathfrak{F}$ (of rings) on a topological space X consists of the data:
- for every open set $U\subset X$ a ring $\mathfrak{F}(U)$ (think of this as the ring of functions on U),
- for every inclusion $U\subset V$ of open sets in X a ring homomorphism $\rho_{V,U}:\mathfrak{F}(V)\to \mathfrak{F}(U)$ called the restriction map (think of this as the usual restriction of functions to a subset),
such that
- $\mathfrak{F}(\varnothing)=0$
- $\rho_{U,U}$ is the identity map of $\mathfrak{F}(U)$ for all U,
- for any inclusion $U\subset V\subset W$ of open sets in X we have $\rho_{V,U}\circ \rho_{W,V}=\rho_{W,U}$
The elements of $\mathfrak{F}(U)$ are usually called the sections of $\mathfrak{F}$ over U, and the restriction maps $\rho_{V,U}$
are written as $\varphi\to \varphi|U$.
A presheaf $\mathfrak{F}$ is called a sheaf of rings if it satisfies the following gluing property:
- if U$\subset X$ is
an open set, $\{U_i : i \in I\}$ an arbitrary open cover of U and $\varphi_i\in \mathfrak{F}(U_i)$ sections for all i such that $\varphi_i|_{U_i\cap U_j}=\varphi_j|_{U_i\cap U_j}$
for all $i, j \in I$, then there is a unique $\varphi \in \mathfrak{F}(U)$ such that $\varphi|_{U_i} = \varphi_i$ for all i.
EDIT: added picture of the exercise text for reference in case I explained myself wrong
EDIT2: This is what is given as the definition of a K-ringed space:
A ringed spaces equipped with a sheaf of rings such that the elements of $\mathfrak{O}_X(U)$ are actual functions from U to a fixed ring K;
EDIT3: It turns out that the actual definition is
A ringed spaces equipped with a sheaf of rings such that the elements of $\mathfrak{O}_X(U)$ are actual functions from U to a fixed ring K and $\mathfrak{O}_X(U)$ is not only a subring of the ring of functions from $U\to K$ but a $sub-K-algebra$ of it;
What does this change?
Best Answer
Being a $K$-ringed space has a very specific meaning, namely that the associated sheaf $\mathcal{O}_X(U)$ is a $K$-subalgebra of the $K$-algebra $\operatorname{Mor}(U, K)$. In particular this means that $\mathcal{O}_X(U)$ contains the constant functions.
A set $X$ with the discrete topology has the trivial open covering $$X = \bigcup_{x \in X} x, $$ where we identify $x$ with the set $\{x\}$. The sheaf axioms then give that we can decompose any sheaf $\mathcal{F}$ over an open subset $U$ as $$\mathcal{F}(U) \cong \prod_{x \in U} \mathcal{F}(x).$$ After all the intersection of $x \cap y$ for any distinct elements is empty. For $\mathcal{F}(x)$ to sheaf for a $K$-ringed space we need it to be a $K$-subalgebra of $K_x := \operatorname{Mor}(x, K)$ and must contain the constant functions. But all elements of $\operatorname{Mor}(x, K)$ are constant functions and the set can be identified with $K$. We must thus have $\mathcal{F}(x) = K_x$ for all $x \in X$. It follows that for a $K$-ringed space with the specified conditions we must have $$\mathcal{O}_X(U) \cong \prod_{u \in U} K.$$ It is easy to see existence of a presheaf defined that way and the sheaf axioms hold because the restriction maps come from dropping terms in the product.