Höchste Zeit, unsern Kurs zu ändern

This is the continuation of the last post. First I want to prove a couple of algebraic results that were glossed over last time.

Proposition 1 Noetherian value ring is a DVR.

Proof: By the above correspondence and Noetherianity there is an minimal element $l$ of the value group which is $> 0$. Take an arbitrary positive element $k$ of the value group and consider $k-1$, $k-2$, \ldots. Some $k-n$ must be $l$, otherwise $l$ is not minimal, that gives an isomorphism with ${\mathbb Z}$. ☐

Lemma 2 (Atiyah-MacDonald, Theorem~5.21) Let ${\mathcal O} \subset K$ be an integral local ring. Then there exists a valuation ring that dominates it. In fact, it is the maximal local ring that contains ${\mathcal O}$ with respect to the domination order.

Here is the easier dimension 1 version of this statement.

Proposition 3 Let $R$ be a dimension 1 local ring with maximal ideal $\mathfrak{m}$. Take the integral closure of $R$ in $Q(R)$ and let $\mathfrak{p}$ be a maximal ideal in $R^{int}$ such that $\mathfrak{p} \cap R = m$. Then the localisation $R^{int}_p$ is a valuation ring.

Here are some notes on the valuative criterion of separatedness, annoyingly detailed at times.

Let $f: X \rightarrow Y$ be a morphism. Then by the universal property of the fibre product there is a diagonal morphism $\Delta: X \rightarrow X \times_Y X$ such that $p_1 \circ \Delta = p_2 \circ \Delta = id_X$.

The morphism $f$ is is called separated if $\Delta$ is a closed embedding.

Proposition 1 If $X, Y$ are affine, then $f$ is separated.

Proof: Let $X = \mathrm{Spec}{}S, Y = \mathrm{Spec}{}R$. Passing to the corresponding maps of rings we get $\Delta \circ p_1 = \Delta \circ p_2 = id$, $\Delta: S \otimes_R S \rightarrow S$. Then $\Delta$ is defined by $\Delta(s_1 \otimes s_2) = s_1 s_2$. This map is surjective, hence it corresponds to a closed embedding on $\mathrm{Spec}$-s. ☐

Example.  Let $X$ be the double affine line (over a field), i.e. $X = {\mathbb A}^1 \sqcup {\mathbb A}^1/\sim$ where $x \sim y \textrm{ iff } x \neq 0 \textrm{ or } y \neq 0$. There is a surjective map $p: X \rightarrow {\mathbb A}^1$. Let $p^{-1}(0)=\{a_1, a_2\}$. Let $f_1, f_2$ be two different sections of $p$, $f_1(0)=a_1, f_2(0)=a_2$. Let $\varphi$ be the morphism such that $p_1 \circ \varphi = f_1, p_2 \circ \varphi = f_2$ that exists by the universal property of the fibre product.

Then $\varphi$ coincides with $\Delta$ on ${\mathbb A}^1 \setminus \{0\}$. In particular, $\varphi.(A^1 \setminus \{0\}) \subset \Delta(X)$. Since $\varphi$ is continuous,

$\displaystyle \overline{\varphi(A^1 \setminus \{0\})} \subset \overline{\Delta(X)} $

but $\varphi(0) = (a_1, a_2) \in \overline{\varphi.(A^1 \setminus \{0\})}$ (again by continuity of $\varphi$) and $(a_1, a_2)$ is clearly not in $\Delta(X)$.