The Stacks project

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 $u^{-1} \in R$. $\square$

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. $\square$

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

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. $\square$

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 $x^{nm} y \in B$. $\square$

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

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

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. $\square$

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 $A = A_1 \cap B$. $\square$

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. $\square$

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

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

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 $a$. $\square$

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

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. $\square$

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\} $. $\square$