I was reading Basic Algebraic geometry 2 by Shafarevich,
In the definition of presheaf, $\mathscr F(\emptyset)=1$
To illustrate this definition they have given an following example:
suppose first that $A$ has no zerodivisors, and write $K$ for its field of fractions. In this case $A$ is a subfield of $K$. For an open set $U\subset\operatorname{Spec}A$ we denote $\mathscr O(U)$ the set of elements $u \in K$ such that for any point $x \in U$ we have an expression $ u=\frac{a}{b}$ with a,b $\in A$ and $b(x)\neq 0$ that is, b is not an element of the prime ideal $x$. Now $\mathscr O(U)$ is obviously a ring. Since all the rings $\mathscr O(U)$ are contained in K, we can compare them as subsets of one set. If $U\subset V$ then clearly $\mathscr O(V)$ $\subset$ $\mathscr O(U)$. We write $\rho_{U}^{V}$ for the inclusion $\mathscr O(V)$ $\hookrightarrow$ $\mathscr O(U)$.
A trivial verification shows that we get a presheaf of rings.
According to definition of presheaf $\mathscr O(\emptyset)$ should be 1 but I am getting $\mathscr O(\emptyset)=K$
I could not figure out why: can anyone kindly explain what is wrong with this?
Thanks in advance