The Stacks project

Lemma 107.23.2. Let $S$ be a scheme and $s \in S$ a point. Let $f : X \to S$ and $g_ i : Y_ i \to S$, $i = 1, 2$ be families of curves. Let $c_ i : X \to Y_ i$ be morphisms over $S$. Assume there is an isomorphism $Y_{1, s} \cong Y_{2, s}$ of fibres compatible with $c_{1, s}$ and $c_{2, s}$. If $c_{1, s, *}\mathcal{O}_{X_ s} = \mathcal{O}_{Y_{1, s}}$ and $R^1c_{1, s, *}\mathcal{O}_{X_ s} = 0$, then there exist an open neighbourhood $U$ of $s$ and an isomorphism $Y_{1, U} \cong Y_{2, U}$ of families of curves over $U$ compatible with the given isomorphism of fibres and with $c_1$ and $c_2$.

Proof. Recall that $\mathcal{O}_{S, s} = \mathop{\mathrm{colim}}\nolimits \mathcal{O}_ S(U)$ where the colimit is over the system of affine neighbourhoods $U$ of $s$. Thus the category of algebraic spaces of finite presentation over the local ring is the colimit of the categories of algebraic spaces of finite presentation over the affine neighbourhoods of $s$. See Limits of Spaces, Lemma 68.7.1. In this way we reduce to the case where $S$ is the spectrum of a local ring and $s$ is the closed point.

Assume $S = \mathop{\mathrm{Spec}}(A)$ where $A$ is a local ring and $s$ is the closed point. Write $A = \mathop{\mathrm{colim}}\nolimits A_ j$ with $A_ j$ local Noetherian (say essentially of finite type over $\mathbf{Z}$) and local transition homomorphisms. Set $S_ j = \mathop{\mathrm{Spec}}(A_ j)$ with closed point $s_ j$. We can find a $j$ and families of curves $X_ j \to S_ j$, $Y_{j, i} \to S_ j$, see Lemma 107.5.3 and Limits of Stacks, Lemma 100.3.5. After possibly increasing $j$ we can find morphisms $c_{j, i} : X_ j \to Y_{j, i}$ whose base change to $s$ is $c_ i$, see Limits of Spaces, Lemma 68.7.1. Since $\kappa (s) = \mathop{\mathrm{colim}}\nolimits \kappa (s_ j)$ we can similarly assume there is an isomorphism $Y_{j, 1, s_ j} \cong Y_{j, 2, s_ j}$ compatible with $c_{j, 1, s_ j}$ and $c_{j, 2, s_ j}$. Finally, the assumptions $c_{1, s, *}\mathcal{O}_{X_ s} = \mathcal{O}_{Y_{1, s}}$ and $R^1c_{1, s, *}\mathcal{O}_{X_ s} = 0$ are inherited by $c_{j, 1, s_ j}$ because $\{ s_ j \to s\} $ is an fpqc covering and $c_{1, s}$ is the base of $c_{j, 1, s_ j}$ by this covering (details omitted). In this way we reduce the lemma to the case discussed in the next paragraph.

Assume $S$ is the spectrum of a Noetherian local ring $\Lambda $ and $s$ is the closed point. Consider the scheme theoretic image $Z$ of

\[ (c_1, c_2) : X \longrightarrow Y_1 \times _ S Y_2 \]

The statement of the lemma is equivalent to the assertion that $Z$ maps isomorphically to $Y_1$ and $Y_2$ via the projection morphisms. Since taking the scheme theoretic image of this morphism commutes with flat base change (Morphisms of Spaces, Lemma 65.30.12, we may replace $\Lambda $ by its completion (More on Algebra, Section 15.42).

Assume $S$ is the spectrum of a complete Noetherian local ring $\Lambda $. Observe that $X$, $Y_1$, $Y_2$ are schemes in this case (More on Morphisms of Spaces, Lemma 74.43.5). Denote $X_ n$, $Y_{1, n}$, $Y_{2, n}$ the base changes of $X$, $Y_1$, $Y_2$ to $\mathop{\mathrm{Spec}}(\Lambda /\mathfrak m^{n + 1})$. Recall that the arrow

\[ \mathcal{D}\! \mathit{ef}_{X_ s \to Y_{2, s}} \cong \mathcal{D}\! \mathit{ef}_{X_ s \to Y_{1, s}} \longrightarrow \mathcal{D}\! \mathit{ef}_{X_ s} \]

is an equivalence, see Deformation Problems, Lemma 91.10.6. Thus there is an isomorphism of formal objects $(X_ n \to Y_{1, n}) \cong (X_ n \to Y_{2, n})$ of $\mathcal{D}\! \mathit{ef}_{X_ s \to Y_{1, s}}$. Finally, by Grothendieck's algebraization theorem (Cohomology of Schemes, Lemma 30.28.3) this produces an isomorphism $Y_1 \to Y_2$ compatible with $c_1$ and $c_2$. $\square$


Comments (0)


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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0E89. Beware of the difference between the letter 'O' and the digit '0'.