Skyscrapper sheaf and generalization of points

Let $A$ be a commutative ring with unity and $X=\operatorname{Spec}(A)$ together with its structure sheaf $\mathcal{O}_X$. Let $\operatorname{Sky}(\mathcal{O}_{X,x})$ be the skyscrapper sheaf and suppose that $\operatorname{Sky}(\mathcal{O}_{X,x})$ is a quasicoherent sheaf of $\mathcal{O}_X$-modules. I want to show that $x$ does no admit a non trivial generalizations.

Suppose that there is some $y \neq x$ such that $y \in \overline{\{ x \}}$. Then the stalk $\operatorname{Sky}(\mathcal{O}_{X,x})_{y} \cong \mathcal{O}_{X,x}$ by definition of skyscrapper sheaf. Since this sheaf is quasicoherent and $X$ is affine it must be isomorphic to the sheaf associated to the $A$-module $\mathcal{O}_{X,x}$.

Let $x=\mathfrak{p}$, and $y=\mathfrak{m}$ (we may assume $y$ to be a maximal ideal) then it is clear that, $$A_{\mathfrak{p}} \cong A_{\mathfrak{p}} \otimes_{A} A_{\mathfrak{m}}\cong \mathcal{O}_{X,x} \otimes_{A} A_{\mathfrak{m}},$$

Which respect to which ring is this isomorphism of modules is taken $A_{\mathfrak{p}}$? Is this already a contradiction? Was my reasoning wrong at some point?

• What is your definition of $\operatorname{Sky}(-)$? – MooS Nov 25 '16 at 6:11

You miss the term generalization here. No non-trivial generalization means, that there is no point $y \neq x$ with $x \in \overline{\{y\}}$. For an affine scheme, this translates into $x$ coming from a minimal prime ideal.
Since $X$ is affine, if $\operatorname{Sky}(\mathcal O_{X,x})$ is quasi-coherent, we have
$$\operatorname{Sky}(\mathcal O_{X,x}) = (\Gamma(X,\operatorname{Sky}(\mathcal O_{X,x})))^\sim = (\mathcal O_{X,x})^\sim = (A_{\mathfrak p})^\sim$$
Now it is very easy to show that $(A_{\mathfrak p})^\sim$ is a skyscraper sheaf if and only if $\mathfrak p$ is a minimal prime ideal.