The Stacks project

Proof. Let $u_1, \ldots , u_ m \in V' \subset V \subset U$ be as in Lemma 40.14.4. Since $U \setminus V$ is closed in $U$, $j$ an immersion, and $V(I) \cup j(U)$ is closed in $\mathop{\mathrm{Spec}}(A)$, we can find an ideal $J \subset I$ such that $V(J) = V(I) \cup j(U \setminus V)$. For example we can take the ideal of elements of $I$ which vanish on $j(U \setminus V)$. Thus we can replace $(U, R, s, t, c)$, $j : U \to \mathop{\mathrm{Spec}}(A)$, and $I$ by $(V', R', s', t', c')$, $j|_{V'} : V' \to \mathop{\mathrm{Spec}}(A)$, and $J$. In other words, we may assume that $U$ is affine and that $s$ and $t$ are finite locally free. Take any $f \in I$ which does not vanish at all the points in the $R$-orbits of $u_1, \ldots , u_ m$ (Algebra, Lemma 10.15.2). Consider

Since $f \in I$ and since $V(I) \cup j(U)$ is closed we see that $U \cap D(f) \to D(f)$ is a closed immersion. Hence $f^ ng$ is the image of an element $h \in I$ for some $n > 0$. We claim that $h$ works. Namely, we have seen in Groupoids, Lemma 39.23.2 that $g$ is an $R$-invariant function, hence $D(g) \subset U$ is $R$-invariant. Since $f$ does not vanish on the orbit of $u_ j$, the function $g$ does not vanish at $u_ j$. Moreover, we have $V(g) \supset V(j^\sharp (f))$ and hence $j^{-1}D(h) = D(g)$. $\square$