Lemma 37.71.3. Let $A \to B$ be a ring map. Say $B$ is generated as an $A$-module by $b_1, \ldots , b_ d \in B$. Set $h = \sum x_ ib_ i \in B[x_1, \ldots , x_ d]$. Then $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is universally open if and only if the image of $D(h)$ in $\mathop{\mathrm{Spec}}(A[x_1, \ldots , x_ d])$ is open.

**Proof.**
If $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is universally open, then of course the image of $D(h)$ is open. Conversely, assume the image $U$ of $D(h)$ is open. Let $A \to A'$ be a ring map. It suffices to show that the image of any principal open $D(g) \subset \mathop{\mathrm{Spec}}(A' \otimes _ A B)$ in $\mathop{\mathrm{Spec}}(A')$ is open. We may write $g = \sum _{i = 1, \ldots , d} a'_ i \otimes b_ i$ for some $a'_ i \in A'$. Let $A[x_1, \ldots , x_ n] \to A'$ be the $A$-algebra homomorphism sending $x_ i$ to $a'_ i$. This also induces a $B$-algebra homomorphism $B[x_1, \ldots , x_ n] \to A' \otimes _ A B$ sending $h$ to $g$. Since

is cartesian the image of $D(g)$ in $\mathop{\mathrm{Spec}}(A')$ is equal to the inverse image of $U$ in $\mathop{\mathrm{Spec}}(A')$ and hence open. $\square$

## 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Your email address will not be published.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)