# A question from Algebraic Geometry: scheme over the spectrum of a DVR

I am trying to read a paper entitled " Valuative Criteria for Families of Vector Bundles on Algebraic Varieties" by Stacy G. Langton. There, the author mentions the following-

Let $(R, m ,k)$ be a DVR with $k=\bar k$ and $m= (\pi)$ where $\pi \in R$. Let $X$ be a non-singular projective variety over $k$. Let $X_R = X \times$ Spec $R$ , $X_k$ be the closed fiber of $X_R$ over Spec $R$ and $j$ be the closed immersion $X_k \rightarrow X_R$. Let $\beta$ be the unique generic point of $X_k$. Then $\mathcal O _{X_R},_\beta$ is a DVR with maximal ideal $m_\beta$ generated by $\pi $$\in R. My question is, How to prove that \mathcal O _{X_R},_\beta is a DVR with maximal ideal m_\beta generated by \pi$$\in R$ ?

• I tried to fix the spelling error in the title, but the software wouldn't let me, so I had to add some further description of the question. – user64687 Oct 27 '14 at 10:13

This is just a very rough sketch of a proof -- you'll need to fill in some of the relevant hypotheses and do some commutative algebra to make it rigorous -- but it's possible to give a rigorous proof that's totally unenlightening in this context, so instead I'll explain what's happening here.

Geometrically, $\operatorname{Spec}(k)$ is a single point, which we can call $x_0$. Since $X$ is a scheme over $k$, we can think of $X$ as being a family of fibers, one corresponding to each point of $x_0$. Of course, since there's just the one point of $x_0$, there's just the one fiber, which is all of $X$.

Geometrically, $\operatorname{Spec}(R)$ is the algebro-geometric analogue of a small open ball $B_\varepsilon(x_0)$ around $x_0$ from analysis. It's a one-dimensional object consisting of the point $x_0$ -- a zero-dimensional object -- together with an open neighborhood of it. Unlike the open neighborhoods you're probably familiar with from analysis, this one is very small; in fact, it doesn't even contain any other (closed) points! It really is distinct from $x_0$, though: $x_0 = \operatorname{Spec}(k)$ consists of a single point, whereas $\operatorname{Spec}(R)$ consists of two points, namely $x_0$ and a generic point.

Base changing $X$ from a variety over $k$ to a scheme $X_R$ over $R$ gives us a family $X_R \to R$. The fiber over $x_0$ -- what you're calling $X_k$ -- is just your original space $X$.

Say $X$ is $n$-dimensional. Then $X_R$ is a family of $n$-dimensional fibers over the one-dimensional space $\operatorname{Spec}(R)$, so it's $(n+1)$-dimensional. The fiber $X_k$ is is a copy of $X$, so it's $n$-dimensional. $X_k$ is a codimension-1 closed set of $X$, so its local ring $\mathcal{O}_{X_R, \beta}$ is a DVR. (That's what DVRs are: local rings of codimension-one closed sets.)

The function $\pi$ cuts $x_0$ out of $\operatorname{Spec}(R)$, analogous to how the equation $z = 0$ cuts a single point out of the complex plane. Consequently, $\pi$ also cuts $X_k$ out of $X_R$, just as $z = 0$ cuts $\{0\} \times \mathbb{C}$ out of $\mathbb{C} \times \mathbb{C}$ with coordinates $(z, w)$.

The author presumably intended everything above to be obvious to the reader, and it is if you're familiar with some of the conventions used in algebraic geometry, but unfortunately these basic ideas aren't actually written down anywhere as far as I can tell; learning algebraic geometry almost seems to be some sort of oral tradition passed on directly from person to person without ever being put into a book, at least not in a form that useful to someone who doesn't already know the stuff.

• +1: great answer, and your sociological analysis at the end is very interesting. – Georges Elencwajg Oct 27 '14 at 8:03
• As $X_R=X\times_k Spec(R)$, I wonder if it is possible to compute $\mathcal{O}_{X_R,\beta}$ in terms of $\mathcal{O}_{X,\beta}=K(X)$ and $R$, such as $\mathcal{O}_{X_R,\beta}=K(X)\otimes_k R$ – Diego Oct 27 '14 at 13:04