Situation 33.37.1. Here we are given a commutative diagram of rings

where $K$ is a field and $A$, $B$ are subrings of $K$ with fraction field $K$. Finally, $R = A \times _ K B = A \cap B$.

This section contains some algebraic preliminaries to proving that a finite set of codimension $1$ points of a separated scheme is contained in an affine open.

Situation 33.37.1. Here we are given a commutative diagram of rings

\[ \xymatrix{ A \ar[r] & K \\ R \ar[u] \ar[r] & B \ar[u] } \]

where $K$ is a field and $A$, $B$ are subrings of $K$ with fraction field $K$. Finally, $R = A \times _ K B = A \cap B$.

Lemma 33.37.2. In Situation 33.37.1 assume that $B$ is a valuation ring. Then for every unit $u$ of $A$ either $u \in R$ or $u^{-1} \in R$.

**Proof.**
Namely, if the image $c$ of $u$ in $K$ is in $B$, then $u \in R$. Otherwise, $c^{-1} \in B$ (Algebra, Lemma 10.50.4) and $u^{-1} \in R$.
$\square$

The following lemma explains the meaning of the condition “$A \otimes B \to K$ is surjective” which comes up quite a bit in the following.

Lemma 33.37.3. In Situation 33.37.1 assume $A$ is a Noetherian ring of dimension $1$. The following are equivalent

$A \otimes B \to K$ is not surjective,

there exists a discrete valuation ring $\mathcal{O} \subset K$ containing both $A$ and $B$.

**Proof.**
It is clear that (2) implies (1). On the other hand, if $A \otimes B \to K$ is not surjective, then the image $C \subset K$ is not a field hence $C$ has a nonzero maximal ideal $\mathfrak m$. Choose a valuation ring $\mathcal{O} \subset K$ dominating $C_\mathfrak m$. By Algebra, Lemma 10.119.12 applied to $A \subset \mathcal{O}$ the ring $\mathcal{O}$ is Noetherian. Hence $\mathcal{O}$ is a discrete valuation ring by Algebra, Lemma 10.50.18.
$\square$

Lemma 33.37.4. In Situation 33.37.1 assume

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

$B$ is a discrete valuation ring,

Then we have the following two possibilities

If $A^*$ is not contained in $R$, 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 $K = A \otimes _ R B$.

If $A^*$ is contained in $R$, then $B$ dominates one of the local rings of $A$ at a maximal ideal and $A \otimes B \to K$ is not surjective.

**Proof.**
Assumption (a) implies there is a unit $u$ of $A$ whose image in $K$ lies in the maximal ideal of $B$. Then $u$ is a nonzerodivisor of $R$ and for every $a \in A$ there exists an $n$ such that $u^ n a \in R$. It follows that $A = R_ u$.

Let $\mathfrak m_ A$ be the Jacobson radical of $A$. Let $x \in \mathfrak m_ A$ be a nonzero element. Since $\dim (A) = 1$ we see that $K = A_ x$. After replacing $x$ by $x^ n u^ m$ for some $n \geq 1$ and $m \in \mathbf{Z}$ we may assume $x$ maps to a unit of $B$. We see that for every $b \in B$ we have that $x^ nb$ in the image of $R$ for some $n$. Thus $B = R_ x$.

Let $z \in R$. If $z \not\in \mathfrak m_ A$ and $z$ does not map to an element of $\mathfrak m_ B$, then $z$ is invertible. Thus $x + u$ is invertible in $R$. Hence $\mathop{\mathrm{Spec}}(R) = D(x) \cup D(u)$. We have seen above that $D(u) = \mathop{\mathrm{Spec}}(A)$ and $D(x) = \mathop{\mathrm{Spec}}(B)$.

Case (b). If $x \in \mathfrak m_ A$, then $1 + x$ is a unit and hence $1 + x \in R$, i.e, $x \in R$. Thus we see that $\mathfrak m_ A \subset R \subset A$. In fact, in this case $A$ is integral over $R$. Namely, write $A/\mathfrak m_ A = \kappa _1 \times \ldots \times \kappa _ n$ as a product of fields. Say $x = (c_1, \ldots , c_ r, 0, \ldots , 0)$ is an element with $c_ i \not= 0$. Then

\[ x^2 - x(c_1, \ldots , c_ r, 1, \ldots , 1) = 0 \]

Since $R$ contains all units we see that $A/\mathfrak m_ A$ is integral over the image of $R$ in it, and hence $A$ is integral over $R$. It follows that $R \subset A \subset B$ as $B$ is integrally closed. Moreover, if $x \in \mathfrak m_ A$ is nonzero, then $K = A_ x = \bigcup x^{-n}A = \bigcup x^{-n}R$. Hence $x^{-1} \not\in B$, i.e., $x \in \mathfrak m_ B$. We conclude $\mathfrak m_ A \subset \mathfrak m_ B$. Thus $A \cap \mathfrak m_ B$ is a maximal ideal of $A$ thereby finishing the proof. $\square$

Lemma 33.37.5. Let $B$ be a semi-local Noetherian domain of dimension $1$. Let $B'$ be the integral closure of $B$ in its fraction field. Then $B'$ is a semi-local Dedekind domain. Let $x$ be a nonzero element of the Jacobson radical of $B'$. Then for every $y \in B'$ there exists an $n$ such that $x^ n y \in B$.

**Proof.**
Let $\mathfrak m_ B$ be the Jacobson radical of $B$. The structure of $B'$ results from Algebra, Lemma 10.120.18. Given $x, y \in B'$ as in the statement of the lemma consider the subring $B \subset A \subset B'$ generated by $x$ and $y$. Then $A$ is finite over $B$ (Algebra, Lemma 10.36.5). Since the fraction fields of $B$ and $A$ are the same we see that the finite module $A/B$ is supported on the set of closed points of $B$. Thus $\mathfrak m_ B^ n A \subset B$ for a suitable $n$. Moreover, $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(A)$ is surjective (Algebra, Lemma 10.36.17), hence $A$ is semi-local as well. It also follows that $x$ is in the Jacobson radical $\mathfrak m_ A$ of $A$. Note that $\mathfrak m_ A = \sqrt{\mathfrak m_ B A}$. Thus $x^ m y \in \mathfrak m_ B A$ for some $m$. Then $x^{nm} y \in B$.
$\square$

Lemma 33.37.6. In Situation 33.37.1 assume

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

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

$A \otimes B \to K$ 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 33.37.4 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 $R = R_ r$ 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 33.37.5. 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 33.37.5 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. $\square$

Lemma 33.37.7. Let $K$ be a field. Let $A_1, \ldots , A_ r \subset K$ be Noetherian semi-local rings of dimension $1$ with fraction field $K$. If $A_ i \otimes A_ j \to K$ is surjective for all $i \not= j$, then there exists a Noetherian semi-local domain $A \subset K$ of dimension $1$ contained in $A_1, \ldots , A_ r$ such that

$A \to A_ i$ induces an open immersion $j_ i : \mathop{\mathrm{Spec}}(A_ i) \to \mathop{\mathrm{Spec}}(A)$,

$\mathop{\mathrm{Spec}}(A)$ is the union of the opens $j_ i(\mathop{\mathrm{Spec}}(A_ i))$,

each closed point of $\mathop{\mathrm{Spec}}(A)$ lies in exactly one of these opens.

**Proof.**
Namely, we can take $A = A_1 \cap \ldots \cap A_ r$. First we note that (3), once (1) and (2) have been proven, follows from the assumption that $A_ i \otimes A_ j \to K$ is surjective since if $\mathfrak m \in j_ i(\mathop{\mathrm{Spec}}(A_ i)) \cap j_ j(\mathop{\mathrm{Spec}}(A_ j))$, then $A_ i \otimes A_ j \to K$ ends up in $A_\mathfrak m$. To prove (1) and (2) we argue by induction on $r$. If $r > 1$ by induction we have the results (1) and (2) for $B = A_2 \cap \ldots \cap A_ r$. Then we apply Lemma 33.37.6 to see they hold for $A = A_1 \cap B$.
$\square$

Lemma 33.37.8. Let $A$ be a domain with fraction field $K$. Let $B_1, \ldots , B_ r \subset K$ be Noetherian $1$-dimensional semi-local domains whose fraction fields are $K$. If $A \otimes B_ i \to K$ are surjective for $i = 1, \ldots , r$, then there exists an $x \in A$ such that $x^{-1}$ is in the Jacobson radical of $B_ i$ for $i = 1, \ldots , r$.

**Proof.**
Let $B_ i'$ be the integral closure of $B_ i$ in $K$. Suppose we find a nonzero $x \in A$ such that $x^{-1}$ is in the Jacobson radical of $B'_ i$ for $i = 1, \ldots , r$. Then by Lemma 33.37.5, after replacing $x$ by a power we get $x^{-1} \in B_ i$. Since $\mathop{\mathrm{Spec}}(B'_ i) \to \mathop{\mathrm{Spec}}(B_ i)$ is surjective we see that $x^{-1}$ is then also in the Jacobson radical of $B_ i$. Thus we may assume that each $B_ i$ is a semi-local Dedekind domain.

If $B_ i$ is not local, then remove $B_ i$ from the list and add back the finite collection of local rings $(B_ i)_\mathfrak m$. Thus we may assume that $B_ i$ is a discrete valuation ring for $i = 1, \ldots , r$.

Let $v_ i : K \to \mathbf{Z}$, $i = 1, \ldots , r$ be the corresponding discrete valuations (see Algebra, Lemma 10.120.17). We are looking for a nonzero $x \in A$ with $v_ i(x) < 0$ for $i = 1, \ldots , r$. We will prove this by induction on $r$.

If $r = 1$ and the result is wrong, then $A \subset B$ and the map $A \otimes B \to K$ is not surjective, contradiction.

If $r > 1$, then by induction we can find a nonzero $x \in A$ such that $v_ i(x) < 0$ for $i = 1, \ldots , r - 1$. If $v_ r(x) < 0$ then we are done, so we may assume $v_ r(x) \geq 0$. By the base case we can find $y \in A$ nonzero such that $v_ r(y) < 0$. After replacing $x$ by a power we may assume that $v_ i(x) < v_ i(y)$ for $i = 1, \ldots , r - 1$. Then $x + y$ is the element we are looking for. $\square$

Lemma 33.37.9. Let $A$ be a Noetherian local ring of dimension $1$. Let $L = \prod A_\mathfrak p$ where the product is over the minimal primes of $A$. Let $a_1, a_2 \in \mathfrak m_ A$ map to the same element of $L$. Then $a_1^ n = a_2^ n$ for some $n > 0$.

**Proof.**
Write $a_1 = a_2 + x$. Then $x$ maps to zero in $L$. Hence $x$ is a nilpotent element of $A$ because $\bigcap \mathfrak p$ is the radical of $(0)$ and the annihilator $I$ of $x$ contains a power of the maximal ideal because $\mathfrak p \not\in V(I)$ for all minimal primes. Say $x^ k = 0$ and $\mathfrak m^ n \subset I$. Then

\[ a_1^{k + n} = a_2^{k + n} + {n + k \choose 1} a_2^{n + k - 1} x + {n + k \choose 2} a_2^{n + k - 2} x^2 + \ldots + {n + k \choose k - 1} a_2^{n + 1} x^{k - 1} = a_2^{n + k} \]

because $a_2 \in \mathfrak m_ A$. $\square$

Lemma 33.37.10. Let $A$ be a Noetherian local ring of dimension $1$. Let $L = \prod A_\mathfrak p$ and $I = \bigcap \mathfrak p$ where the product and intersection are over the minimal primes of $A$. Let $f \in L$ be an element of the form $f = i + a$ where $a \in \mathfrak m_ A$ and $i \in IL$. Then some power of $f$ is in the image of $A \to L$.

**Proof.**
Since $A$ is Noetherian we have $I^ t = 0$ for some $t > 0$. Suppose that we know that $f = a + i$ with $i \in I^ kL$. Then $f^ n = a^ n + na^{n - 1}i \bmod I^{k + 1}L$. Hence it suffices to show that $na^{n - 1}i$ is in the image of $I^ k \to I^ kL$ for some $n \gg 0$. To see this, pick a $g \in A$ such that $\mathfrak m_ A = \sqrt{(g)}$ (Algebra, Lemma 10.60.8). Then $L = A_ g$ for example by Algebra, Proposition 10.60.7. On the other hand, there is an $n$ such that $a^ n \in (g)$. Hence we can clear denominators for elements of $L$ by multiplying by a high power of $a$.
$\square$

Lemma 33.37.11. Let $A$ be a Noetherian local ring of dimension $1$. Let $L = \prod A_\mathfrak p$ where the product is over the minimal primes of $A$. Let $K \to L$ be an integral ring map. Then there exist $a \in \mathfrak m_ A$ and $x \in K$ which map to the same element of $L$ such that $\mathfrak m_ A = \sqrt{(a)}$.

**Proof.**
By Lemma 33.37.10 we may replace $A$ by $A/(\bigcap \mathfrak p)$ and assume that $A$ is a reduced ring (some details omitted). We may also replace $K$ by the image of $K \to L$. Then $K$ is a reduced ring. The map $\mathop{\mathrm{Spec}}(L) \to \mathop{\mathrm{Spec}}(K)$ is surjective and closed (details omitted). Hence $\mathop{\mathrm{Spec}}(K)$ is a finite discrete space. It follows that $K$ is a finite product of fields.

Let $\mathfrak p_ j$, $j = 1, \ldots , m$ be the minimal primes of $A$. Set $L_ j$ be the fraction field of $A_ j$ so that $L = \prod _{j = 1, \ldots , m} L_ j$. Let $A_ j$ be the normalization of $A/\mathfrak p_ j$. Then $A_ j$ is a semi-local Dedekind domain with at least one maximal ideal, see Algebra, Lemma 10.120.18. Let $n$ be the sum of the numbers of maximal ideals in $A_1, \ldots , A_ m$. For such a maximal ideal $\mathfrak m \subset A_ j$ we consider the function

\[ v_{\mathfrak m} : L \to L_ j \to \mathbf{Z} \cup \{ \infty \} \]

where the second arrow is the discrete valuation corresponding to the discrete valuation ring $(A_ j)_{\mathfrak m}$ extended by mapping $0$ to $\infty $. In this way we obtain $n$ functions $v_1, \ldots , v_ n : L \to \mathbf{Z} \cup \{ \infty \} $. We will find an element $x \in K$ such that $v_ i(x) < 0$ for all $i = 1, \ldots , n$.

First we claim that for each $i$ there exists an element $x \in K$ with $v_ i(x) < 0$. Namely, suppose that $v_ i$ corresponds to $\mathfrak m \subset A_ j$. If $v_ i(x) \geq 0$ for all $x \in K$, then $K$ maps into $(A_ j)_{\mathfrak m}$ inside the fraction field $L_ j$ of $A_ j$. The image of $K$ in $L_ j$ is a field over $L_ j$ is algebraic by Algebra, Lemma 10.36.18. Combined we get a contradiction with Algebra, Lemma 10.50.8.

Suppose we have found an element $x \in K$ such that $v_1(x) < 0, \ldots , v_ r(x) < 0$ for some $r < n$. If $v_{r + 1}(x) < 0$, then $x$ works for $r + 1$. If not, then choose some $y \in K$ with $v_{r + 1}(y) < 0$ as is possible by the result of the previous paragraph. After replacing $x$ by $x^ n$ for some $n > 0$, we may assume $v_ i(x) < v_ i(y)$ for $i = 1, \ldots , r$. Then $v_ j(x + y) = v_ j(x) < 0$ for $j = 1, \ldots , r$ by properties of valuations and similarly $v_{r + 1}(x + y) = v_{r + 1}(y) < 0$. Arguing by induction, we find $x \in K$ with $v_ i(x) < 0$ for $i = 1, \ldots , n$.

In particular, the element $x \in K$ has nonzero projection in each factor of $K$ (recall that $K$ is a finite product of fields and if some component of $x$ was zero, then one of the values $v_ i(x)$ would be $\infty $). Hence $x$ is invertible and $x^{-1} \in K$ is an element with $\infty > v_ i(x^{-1}) > 0$ for all $i$. It follows from Lemma 33.37.5 that for some $e < 0$ the element $x^ e \in K$ maps to an element of $\mathfrak m_ A/\mathfrak p_ j \subset A/\mathfrak p_ j$ for all $j = 1, \ldots , m$. Observe that the cokernel of the map $\mathfrak m_ A \to \prod \mathfrak m_ A/\mathfrak p_ j$ is annihilated by a power of $\mathfrak m_ A$. Hence after replacing $e$ by a more negative $e$, we find an element $a \in \mathfrak m_ A$ whose image in $\mathfrak m_ A/\mathfrak p_ j$ is equal to the image of $x^ e$. The pair $(a, x^ e)$ satisfies the conclusions of the lemma. $\square$

Lemma 33.37.12. Let $A$ be a ring. Let $\mathfrak p_1, \ldots , \mathfrak p_ r$ be a finite set of a primes of $A$. Let $S = A \setminus \bigcup \mathfrak p_ i$. Then $S$ is a multiplicative system and $S^{-1}A$ is a semi-local ring whose maximal ideals correspond to the maximal elements of the set $\{ \mathfrak p_ i\} $.

**Proof.**
If $a, b \in A$ and $a, b \in S$, then $a, b \not\in \mathfrak p_ i$ hence $ab \not\in \mathfrak p_ i$, hence $ab \in S$. Also $1 \in S$. Thus $S$ is a multiplicative subset of $A$. By the description of $\mathop{\mathrm{Spec}}(S^{-1}A)$ in Algebra, Lemma 10.17.5 and by Algebra, Lemma 10.15.2 we see that the primes of $S^{-1}A$ correspond to the primes of $A$ contained in one of the $\mathfrak p_ i$. Hence the maximal ideals of $S^{-1}A$ correspond one-to-one with the maximal (w.r.t. inclusion) elements of the set $\{ \mathfrak p_1, \ldots , \mathfrak p_ r\} $.
$\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)