The Stacks project

Lemma 33.36.6. In Situation 33.36.1 assume

  1. $A$ is a Noetherian semi-local domain of dimension $1$,

  2. $B$ is a Noetherian semi-local domain of dimension $1$,

  3. $A \otimes B \to K$ is surjective.

Then $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(R)$ and $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(R)$ are open immersions covering $\mathop{\mathrm{Spec}}(R)$ and $K = A \otimes _ R B$.

Proof. Special case: $B$ is integrally closed in $K$. This means that $B$ is a Dedekind domain (Algebra, Lemma 10.120.17) whence all of its localizations at maximal ideals are discrete valuation rings. Let $\mathfrak m_1, \ldots , \mathfrak m_ r$ be the maximal ideals of $B$. We set

\[ R_1 = A \times _ K B_{\mathfrak m_1} \]

Observing that $A \otimes _{R_1} B_{\mathfrak m_1} \to K$ is surjective we conclude from Lemma 33.36.4 that $A$ and $B_{\mathfrak m_1}$ define open subschemes covering $\mathop{\mathrm{Spec}}(R_1)$ and that $K = A \otimes _{R_1} B_{\mathfrak m_1}$. In particular $R_1$ is a semi-local Noetherian ring of dimension $1$. By induction we define

\[ R_{i + 1} = R_ i \times _ K B_{\mathfrak m_{i + 1}} \]

for $i = 1, \ldots , r - 1$. Observe that $R = R_ n$ because $B = B_{\mathfrak m_1} \cap \ldots \cap B_{\mathfrak m_ r}$ (see Algebra, Lemma 10.157.6). It follows from the inductive procedure that $R \to A$ defines an open immersion $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(R)$. On the other hand, the maximal ideals $\mathfrak n_ i$ of $R$ not in this open correspond to the maximal ideals $\mathfrak m_ i$ of $B$ and in fact the ring map $R \to B$ defines an isomorphisms $R_{\mathfrak n_ i} \to B_{\mathfrak m_ i}$ (details omitted; hint: in each step we added exactly one maximal ideal to $\mathop{\mathrm{Spec}}(R_ i)$). It follows that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(R)$ is an open immersion as desired.

General case. Let $B' \subset K$ be the integral closure of $B$. See Lemma 33.36.5. Then the special case applies to $R' = A \times _ K B'$. Pick $x \in R'$ which is not contained in the maximal ideals of $A$ and is contained in the maximal ideals of $B'$ (see Algebra, Lemma 10.15.4). By Lemma 33.36.5 there exists an integer $n$ such that $x^ n \in R = A \times _ K B$. Replace $x$ by $x^ n$ so $x \in R$. For every $y \in R'$ there exists an integer $n$ such that $x^ n y \in R$. On the other hand, it is clear that $R'_ x = A$. Thus $R_ x = A$. Exchanging the roles of $A$ and $B$ we also find an $y \in R$ such that $B = R_ y$. Note that inverting both $x$ and $y$ leaves no primes except $(0)$. Thus $K = R_{xy} = R_ x \otimes _ R R_ y$. This finishes the proof. $\square$


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