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. $\square$

## Post a comment

Your email address will not be published.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)