The Stacks project

Lemma 37.38.6. Let $Y \to T \to S$ be finite type morphisms of Noetherian schemes. Let $t \in T$ map to $s \in S$ and let $\sigma : \mathcal{O}_{T, t} \to \mathcal{O}_{S, s}^\wedge $ be a local $\mathcal{O}_{S, s}$-algebra map. There exists a finite type morphism $(T', t') \to (T, t)$ such that $\sigma $ factors through $\mathcal{O}_{T, t} \to \mathcal{O}_{T', t'}$ and such that for every local $\mathcal{O}_{S, s}$-algebra map $\sigma ' : \mathcal{O}_{T, t} \to \mathcal{O}_{S, s}^\wedge $ which factors through $\mathcal{O}_{T, t} \to \mathcal{O}_{T', t'}$ the closed immersions

\[ Y \times _{T, \sigma } \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}^\wedge ) = Y_\sigma \longleftarrow Y_ t \longrightarrow Y_{\sigma '} = Y \times _{T, \sigma '} \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}^\wedge ) \]

have isomorphic conormal algebras.

Proof. A useful observation is that $\kappa (s) = \kappa (t)$ by the existence of $\sigma $. Observe that the statement makes sense as the fibres of $Y_\sigma $ and $Y_{\sigma '}$ over $s \in \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}^\wedge )$ are both canonically isomorphic to $Y_ t$. We will think of the property “$\sigma '$ factors through $\mathcal{O}_{T, t} \to \mathcal{O}_{T', t'}$” as a constraint on $\sigma '$. If we have several such constraints, say given by $(T'_ i, t'_ i) \to (T, t)$, $i = 1, \ldots , n$ then we can combined them by considering $(T'_1 \times _ T \ldots \times _ T T'_ n, (t'_1, \ldots , t'_ n)) \to (T, t)$. We will use this without further mention in the following.

By Lemma 37.38.5 we can assume that any $\sigma '$ as in the statement of the lemma is the same as $\sigma $ modulo $\mathfrak m_ s^2$. Note that the conormal algebra of $Y_ t$ in $Y_\sigma $ is just the quasi-coherent graded $\mathcal{O}_{Y_ t}$-algebra

\[ \bigoplus \nolimits _{n \geq 0} \mathfrak m_ s^ n\mathcal{O}_{Y_\sigma }/ \mathfrak m_ s^{n + 1}\mathcal{O}_{Y_\sigma } \]

and similarly for $Y_{\sigma '}$. Since $\sigma $ and $\sigma '$ agree modulo $\mathfrak m_ s^2$ we see that these two algebras are the same in degrees $0$ and $1$. On the other hand, these conormal algebras are generated in degree $1$ over degree $0$. Hence if there is an isomorphism extending the isomorphism just constructed in degrees $0$ and $1$, then it is unique.

We may assume $S$ and $T$ are affine. Let $Y = Y_1 \cup \ldots \cup Y_ n$ be an affine open covering. If we can construct $(T_ i', t'_ i) \to (T, t)$ as in the lemma such that the desired isomorphism (see previous paragraph) exists for $Y_ i \to T \to S$ and $\sigma $, then these glue by uniqueness to prove the result for $Y \to T$. Thus we may assume $Y$ is affine.

Write $S = \mathop{\mathrm{Spec}}(R)$, $T = \mathop{\mathrm{Spec}}(C)$, and $Y = \mathop{\mathrm{Spec}}(B)$. Choose a presentation $B = C[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$. Denote $R^\wedge = \mathcal{O}_{S, s}^\wedge $. Let $a_{kj} \in R^\wedge [x_1, \ldots , x_ n]$ be polynomials such that

\[ \sum \nolimits _{j = 1, \ldots , m} a_{kj}\sigma (f_ j) = 0,\quad \text{for }k = 1, \ldots , K \]

is a set of generators for the module of relations among the $\sigma (f_ j) \in R^\wedge [x_1, \ldots , x_ n]$. Thus we have an exact sequence
\begin{equation} \label{more-morphisms-equation-resolution} R^\wedge [x_1, \ldots , x_ n]^{\oplus K} \to R^\wedge [x_1, \ldots , x_ n]^{\oplus m} \to R^\wedge [x_1, \ldots , x_ n] \to B \otimes _{C, \sigma } R^\wedge \to 0 \end{equation}

Let $c$ be an integer which works in the Artin-Rees lemma for both the first and the second map in this sequence and the ideal $\mathfrak m_{R^\wedge }R^\wedge [x_1, \ldots , x_ n]$ as defined in More on Algebra, Section 15.4. Write

\[ a_{kj} = \sum \nolimits _{I \in \Omega } a_{kj, I} x^ I \quad \text{and}\quad f_ j = \sum \nolimits _{I \in \Omega } f_{j, I} x^ I \]

in multiindex notation where $a_{kj, I} \in R^\wedge $, $f_{j, I} \in C$, and $\Omega $ a finite set of multiindices. Then we see that

\[ \sum \nolimits _{j = 1, \ldots , m,\ I, I' \in \Omega ,\ I + I' = I''} a_{kj, I} \sigma (f_{j, I'}) = 0,\quad I''\text{ a multiindex} \]

in $R^\wedge $. Thus we take

\[ C' = C[t_{jk, I}]/ \left( \sum \nolimits _{j = 1, \ldots , m,\ I, I' \in \Omega ,\ I + I' = I''} t_{kj, I} f_{j, I'},\ I''\text{ a multiindex}\right) \]

Then $\sigma $ factors through a map $\tilde\sigma : C' \to R^\wedge $ sending $t_{kj, I}$ to $a_{jk, I}$. Thus $T' = \mathop{\mathrm{Spec}}(C')$ comes with a point $t' \in T'$ such that $\sigma $ factors through $\mathcal{O}_{T, t} \to \mathcal{O}_{T', t'}$. Let $t_{kj} = \sum t_{kj, I} x^ I$ in $C'[x_1, \ldots , x_ n]$. Then we see that we have a complex
\begin{equation} \label{more-morphisms-equation-resolution-new} C'[x_1, \ldots , x_ n]^{\oplus K} \to C'[x_1, \ldots , x_ n]^{\oplus m} \to C'[x_1, \ldots , x_ n] \to B \otimes _ C C' \to 0 \end{equation}

which is exact at $C'[x_1, \ldots , x_ n]$ and whose base change by $\tilde\sigma $ gives (

By Lemma 37.38.5 we can find a further morphism $(T'', t'') \to (T', t')$ such that $\tilde\sigma $ factors through $\mathcal{O}_{T', t'} \to \mathcal{O}_{T'', t''}$ and such that if $\sigma ' : C \to R^\wedge $ factors through $\mathcal{O}_{T'', t''}$, then the induced map $\tilde\sigma ' : C' \to R^\wedge $ agrees modulo $\mathfrak m_ s^{c + 1}$ with $\tilde\sigma $. Thus if $\sigma '$ is such a map, then we obtain a complex

\[ R^\wedge [x_1, \ldots , x_ n]^{\oplus K} \to R^\wedge [x_1, \ldots , x_ n]^{\oplus m} \to R^\wedge [x_1, \ldots , x_ n] \to B \otimes _{C, \sigma '} R^\wedge \to 0 \]

over $R^\wedge [x_1, \ldots , x_ n]$ by applying $\tilde\sigma '$ to the polynomials $t_{kj}$ and $f_ j$. In other words, this is the base change of the complex ( by $\tilde\sigma '$. The matrices defining this complex are congruent modulo $\mathfrak m_ s^{c + 1}$ to the matrices defining the complex ( because $\tilde\sigma $ and $\tilde\sigma '$ are congruent modulo $\mathfrak m_ s^{c + 1}$. Since ( is exact, we can apply More on Algebra, Lemma 15.4.2 to conclude that

\[ \text{Gr}_{\mathfrak m_ s}(B \otimes _{C, \sigma '} R^\wedge ) \cong \text{Gr}_{\mathfrak m_ s}(B \otimes _{C, \sigma } R^\wedge ) \]

as desired. $\square$

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