I'm trying to pin down the various ways we can extend a property of commutative rings to a corresponding property for schemes. Let $P$ be a property of commutative rings. We could define a scheme $(X,\mathcal{O}_X)$ to have property $P$ in one of the following ways:

- $\mathcal{O}_X(U)$ has $P$ for every open subset $U\subset X$.
- $\mathcal{O}_X(U)$ has $P$ for every
*affine*open subset $U\subset X$. - there exists an affine open cover ${U_i}$ of $X$ such that $\mathcal{O}_X(U_i)$ has $P$ for all $i$.
- for each $x\in U\subset X$ with $U$ an open subset, there exists an affine open $V\subset X$ with $x\in V\subset U$ such that $\mathcal{O}_X(V)$ has $P$.
- $\mathcal{O}_{X,x}$ has $P$ for all $x\in X$.

Evidently, $(1)\Rightarrow (2)\Rightarrow (3)$. If the property $P$ is stable under inversion of single elements (that is, $A$ has $P$ $\Rightarrow$ $A[1/s]$ has $P$ for any element $s\in A$) then $(3)\Rightarrow (4)$. Furthermore, if the property $P$ is stable under arbitrary localizations (that is, $A$ has $P$ $\Rightarrow$ $S^{-1} A$ has $P$ for any multiplicative subset $S$ of $A$) then $(4)\Rightarrow (5)$.

Thus, for many properties of commutative rings $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (4)\Rightarrow (5)$.

Now we need to consider going in the other direction. I believe it can be shown that if $P$ is a local property in the sense that $A$ has $P$ iff $A_{\mathfrak{p}}$ has $P$ for each $\mathfrak{p}\in$ Spec$(A)$ then $(5)\Rightarrow (2)$. Now, it seems to me that if $P$ is a local property in this sense then the property is stable under localization by arbitrary multiplicative subsets. Thus if $P$ is a local property in this sense then $(2) \Leftrightarrow (3) \Leftrightarrow (4) \Leftrightarrow (5)$.

Finally, here is another notion of a property being local. Suppose that $P$ is such that $A$ has $P$ implies that $A[1/s]$ has $P$ for each $s\in A$ and that on the other hand, if $s_1,\ldots,s_n \in A$ are such that Spec$(A)=D(s_1)\cup D(s_2) \cup \cdots \cup D(s_n)$ then $A[1/s_i]$ has $P$ for all $i=1,..,n$ implies that $A$ has $P$. Then I believe it can be shown that $(4)\Rightarrow (2)$ and thus for such a property we have that $(2) \Leftrightarrow (3) \Leftrightarrow (4)$.

Does this seem right to you? I haven't seen any books on algebraic geometry discuss this question to my satisfaction and I am nervous that there may be some holes in my proofs, so if anyone knows off the top of their head that what I have described seems right then I would be happy to hear from you. Do you have any further comments to make about this process of extending a property of commutative rings to schemes?