Lemma 33.36.8. 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 33.36.5, 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. $\square$

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