# Classification of local and semi-local rings in function fields

Let $C$ be a non-singular algebraic curve over an algebraically closed field $k$, and $F$ a function field of this curve. It is well-known that non-trivial discrete valuation rings of $F$ correspond to points of $C$.

Is there any classification of local and semi-local rings in $F$? What examples of semi-local rings in $F$ do you know?

• Cross-posted: math.stackexchange.com/questions/1283525/… May 17, 2015 at 14:11
• I found some classification of semi-local rings by local rings. A semi-local ring $\neq F$ may be expressed as the intersection of a finite number of local rings, no two of which have a place in common, in one and only one way. See jstor.org/stable/1969773?seq=1#page_scan_tab_contents (Rosenlicht M. Equivalence Relations on Algebraic Curves, theorem 3). May 21, 2015 at 11:26

This isn't a full answer, just a sketch.

Well, presumably you want all the elements of $k$ to be units in your local and semi-local rings (ie $k \subseteq R$)? Otherwise things can get very complicated.

I think such a classification does indeed though exist, here's how I would proceed:

Take $R$ to be a local or semi-local (Noetherian?) ring in $F$ that has all elements of $k$ as units. Consider the normalization $S$ of $R$. This normalization should be semi-local and an intersection of those DVRs. Then you can re-obtain $R$ from $S$ by doing a pullback in the category of rings.

Is there a "geometric" intuition underlying the notion of normal varieties?

and

Obtaining non-normal varieties by pushout

The relevant thing is that you noticed that $S = \bigcap R_i$ where the $R_i$s are DVRs, then you would pick ideal an ideal $I \subseteq S$ (which can just be thought of as picking an ideal in each $R_i$) and forming the pullback of the diagram

$$\{ S = \bigcap R_i \to \prod S/I \leftarrow A \}$$ where $A$ is some subring of the (usually Artinian) ring $S/I$.

• Of course $k \subsetneq R$. For simplicity, let rings be Noetherian. Karl, thank you for the first answer the question. May 21, 2015 at 11:34