Consider an integral scheme $X$ of finite type over $k$ of dimension $1$. Then $X$ consists of one generic point $\eta$ and the rest of the points are closed points. We call the set of closed points $\mathcal{X}$. Let $K_{X}$ be the constant sheaf with value $K(X)(=\mathcal{O}_{X,\eta})$. I want to show that the map $$K(X)\rightarrow \bigoplus_{x\in\mathcal{X}}K(X)/\mathcal{O}_{X,x}, \qquad [U,f]\mapsto ([U,f] + \mathcal{O}_{X,x})_{x\in\mathcal{X}}$$ is a well-defined map of $k$-vectorspaces.
Background information: By showing that this is a well-defined map of $k$-vectorspaces one can actually show that the quotient sheaf $K_{X}/\mathcal{O}_{X}$ is isomorphic to the direct sum sheaf $\bigoplus_{x\in\mathcal{X}}i_{x,*}(K(X)/\mathcal{O}_{X,x})$ induced by the inclusion $i_{x}:\{x\}\rightarrow X$. It is clear that this particular direct sum sheaf is flasque. We can thus find a canonical flasque resolution of $\mathcal{O}_{X}$ to compute the cohomology groups of $\mathcal{O}_{X}$.
Problems: 1. To show that this map is well-defined we need for every $[U,f]\in K(X)$ that for only finitely many $x\in\mathcal{X}$, $[U,f]\neq 0$. But I don't see why this has to be the case.
2 I struggle with the scalar multiplication inside $K(X)$. I know that $\mathcal{O}_{X,\eta}\cong \mathcal{O}_{\operatorname{Spec}(R),\eta} = \operatorname{Frac}(R)$ for some finitely generated $k$-algebra $R$ by the fact that $X$ is of finite type over $k$. Thus $K(X)$ indeed is a field and has a $k$-structure on it. But I don't understand the structure well enough to show (formally) that the map indeed preserves the scalar multiplication.
Any help would be appreciated!
First, we reduce to the affine case: the complement of an affine open $U\subset X$ is a finite set of closed points, so it's enough to prove that the image of an element in $K(X)$ is nonzero in finitely many $K(X)/\mathcal{O}_{X,x}$ as $x$ ranges over the closed points in $U$. But this is exactly the statement for affine $X$.
Now suppose $X$ is affine with coordinate algebra $k[X]$. For an arbitrary element $f\in k(X)$, pick a representation $f=g/h$ with $g,h\in k[X]$ and $h\neq 0$. For a closed point $x\in X$, $f$ belongs to $\mathcal{O}_{X,x}$ if $h$ does not vanish at $x$. But $h$, being nonzero, vanishes on a finite set of closed points. So $f\notin \mathcal{O}_{X,x}$ for only finitely many $x\in X$, and thus we really do have a map in to the direct sum instead of the direct product.
I'm not totally sure what you're getting at with your second question. If one has $A$ an $R$-algebra, then any localization of $A$ is an $R$-algebra too: if we have an element $\frac{f}{s}\in S^{-1}A$, then $r\cdot \frac{f}{s}=\frac{rf}{s}$, and in this way the image of $A$ in $S^{-1}A$ by the map $a\mapsto \frac{a}{1}$ is naturally a sub-$R$-module. In our case, we have that $k(X)$ is a localization of $\mathcal{O}_{X,x}$ and the natural map $\mathcal{O}_{X,x}\to k(X)$ is an injection, so our quotient $k(X)/\mathcal{O}_{X,x}$ is a quotient of $k$-modules and the quotient map $k(X)\to k(X)/\mathcal{O}_{X,x}$ is a map of $k$-modules. Does this clear things up?