Lemma 33.36.3. In Situation 33.36.1 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 $B$.

Lemma 33.36.3. In Situation 33.36.1 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.
$\square$

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