## 10.149 Conormal modules and universal thickenings

It turns out that one can define the first infinitesimal neighbourhood not just for a closed immersion of schemes, but already for any formally unramified morphism. This is based on the following algebraic fact.

Lemma 10.149.1. Let $R \to S$ be a formally unramified ring map. There exists a surjection of $R$-algebras $S' \to S$ whose kernel is an ideal of square zero with the following universal property: Given any commutative diagram

$\xymatrix{ S \ar[r]_ a & A/I \\ R \ar[r]^ b \ar[u] & A \ar[u] }$

where $I \subset A$ is an ideal of square zero, there is a unique $R$-algebra map $a' : S' \to A$ such that $S' \to A \to A/I$ is equal to $S' \to S \to A/I$.

Proof. Choose a set of generators $z_ i \in S$, $i \in I$ for $S$ as an $R$-algebra. Let $P = R[\{ x_ i\} _{i \in I}]$ denote the polynomial ring on generators $x_ i$, $i \in I$. Consider the $R$-algebra map $P \to S$ which maps $x_ i$ to $z_ i$. Let $J = \mathop{\mathrm{Ker}}(P \to S)$. Consider the map

$\text{d} : J/J^2 \longrightarrow \Omega _{P/R} \otimes _ P S$

see Lemma 10.131.9. This is surjective since $\Omega _{S/R} = 0$ by assumption, see Lemma 10.148.2. Note that $\Omega _{P/R}$ is free on $\text{d}x_ i$, and hence the module $\Omega _{P/R} \otimes _ P S$ is free over $S$. Thus we may choose a splitting of the surjection above and write

$J/J^2 = K \oplus \Omega _{P/R} \otimes _ P S$

Let $J^2 \subset J' \subset J$ be the ideal of $P$ such that $J'/J^2$ is the second summand in the decomposition above. Set $S' = P/J'$. We obtain a short exact sequence

$0 \to J/J' \to S' \to S \to 0$

and we see that $J/J' \cong K$ is a square zero ideal in $S'$. Hence

$\xymatrix{ S \ar[r]_1 & S \\ R \ar[r] \ar[u] & S' \ar[u] }$

is a diagram as above. In fact we claim that this is an initial object in the category of diagrams. Namely, let $(I \subset A, a, b)$ be an arbitrary diagram. We may choose an $R$-algebra map $\beta : P \to A$ such that

$\xymatrix{ S \ar[r]_1 & S \ar[r]_ a & A/I \\ R \ar[r] \ar@/_/[rr]_ b \ar[u] & P \ar[u] \ar[r]^\beta & A \ar[u] }$

is commutative. Now it may not be the case that $\beta (J') = 0$, in other words it may not be true that $\beta$ factors through $S' = P/J'$. But what is clear is that $\beta (J') \subset I$ and since $\beta (J) \subset I$ and $I^2 = 0$ we have $\beta (J^2) = 0$. Thus the “obstruction” to finding a morphism from $(J/J' \subset S', 1, R \to S')$ to $(I \subset A, a, b)$ is the corresponding $S$-linear map $\overline{\beta } : J'/J^2 \to I$. The choice in picking $\beta$ lies in the choice of $\beta (x_ i)$. A different choice of $\beta$, say $\beta '$, is gotten by taking $\beta '(x_ i) = \beta (x_ i) + \delta _ i$ with $\delta _ i \in I$. In this case, for $g \in J'$, we obtain

$\beta '(g) = \beta (g) + \sum \nolimits _ i \delta _ i \frac{\partial g}{\partial x_ i}.$

Since the map $\text{d}|_{J'/J^2} : J'/J^2 \to \Omega _{P/R} \otimes _ P S$ given by $g \mapsto \frac{\partial g}{\partial x_ i}\text{d}x_ i$ is an isomorphism by construction, we see that there is a unique choice of $\delta _ i \in I$ such that $\beta '(g) = 0$ for all $g \in J'$. (Namely, $\delta _ i$ is $-\overline{\beta }(g)$ where $g \in J'/J^2$ is the unique element with $\frac{\partial g}{\partial x_ j} = 1$ if $i = j$ and $0$ else.) The uniqueness of the solution implies the uniqueness required in the lemma. $\square$

In the situation of Lemma 10.149.1 the $R$-algebra map $S' \to S$ is unique up to unique isomorphism.

Definition 10.149.2. Let $R \to S$ be a formally unramified ring map.

1. The universal first order thickening of $S$ over $R$ is the surjection of $R$-algebras $S' \to S$ of Lemma 10.149.1.

2. The conormal module of $R \to S$ is the kernel $I$ of the universal first order thickening $S' \to S$, seen as an $S$-module.

We often denote the conormal module $C_{S/R}$ in this situation.

Lemma 10.149.3. Let $I \subset R$ be an ideal of a ring. The universal first order thickening of $R/I$ over $R$ is the surjection $R/I^2 \to R/I$. The conormal module of $R/I$ over $R$ is $C_{(R/I)/R} = I/I^2$.

Proof. Omitted. $\square$

Lemma 10.149.4. Let $A \to B$ be a formally unramified ring map. Let $\varphi : B' \to B$ be the universal first order thickening of $B$ over $A$.

1. Let $S \subset A$ be a multiplicative subset. Then $S^{-1}B' \to S^{-1}B$ is the universal first order thickening of $S^{-1}B$ over $S^{-1}A$. In particular $S^{-1}C_{B/A} = C_{S^{-1}B/S^{-1}A}$.

2. Let $S \subset B$ be a multiplicative subset. Then $S' = \varphi ^{-1}(S)$ is a multiplicative subset in $B'$ and $(S')^{-1}B' \to S^{-1}B$ is the universal first order thickening of $S^{-1}B$ over $A$. In particular $S^{-1}C_{B/A} = C_{S^{-1}B/A}$.

Note that the lemma makes sense by Lemma 10.148.4.

Proof. With notation and assumptions as in (1). Let $(S^{-1}B)' \to S^{-1}B$ be the universal first order thickening of $S^{-1}B$ over $S^{-1}A$. Note that $S^{-1}B' \to S^{-1}B$ is a surjection of $S^{-1}A$-algebras whose kernel has square zero. Hence by definition we obtain a map $(S^{-1}B)' \to S^{-1}B'$ compatible with the maps towards $S^{-1}B$. Consider any commutative diagram

$\xymatrix{ B \ar[r] & S^{-1}B \ar[r] & D/I \\ A \ar[r] \ar[u] & S^{-1}A \ar[r] \ar[u] & D \ar[u] }$

where $I \subset D$ is an ideal of square zero. Since $B'$ is the universal first order thickening of $B$ over $A$ we obtain an $A$-algebra map $B' \to D$. But it is clear that the image of $S$ in $D$ is mapped to invertible elements of $D$, and hence we obtain a compatible map $S^{-1}B' \to D$. Applying this to $D = (S^{-1}B)'$ we see that we get a map $S^{-1}B' \to (S^{-1}B)'$. We omit the verification that this map is inverse to the map described above.

With notation and assumptions as in (2). Let $(S^{-1}B)' \to S^{-1}B$ be the universal first order thickening of $S^{-1}B$ over $A$. Note that $(S')^{-1}B' \to S^{-1}B$ is a surjection of $A$-algebras whose kernel has square zero. Hence by definition we obtain a map $(S^{-1}B)' \to (S')^{-1}B'$ compatible with the maps towards $S^{-1}B$. Consider any commutative diagram

$\xymatrix{ B \ar[r] & S^{-1}B \ar[r] & D/I \\ A \ar[r] \ar[u] & A \ar[r] \ar[u] & D \ar[u] }$

where $I \subset D$ is an ideal of square zero. Since $B'$ is the universal first order thickening of $B$ over $A$ we obtain an $A$-algebra map $B' \to D$. But it is clear that the image of $S'$ in $D$ is mapped to invertible elements of $D$, and hence we obtain a compatible map $(S')^{-1}B' \to D$. Applying this to $D = (S^{-1}B)'$ we see that we get a map $(S')^{-1}B' \to (S^{-1}B)'$. We omit the verification that this map is inverse to the map described above. $\square$

Lemma 10.149.5. Let $R \to A \to B$ be ring maps. Assume $A \to B$ formally unramified. Let $B' \to B$ be the universal first order thickening of $B$ over $A$. Then $B'$ is formally unramified over $A$, and the canonical map $\Omega _{A/R} \otimes _ A B \to \Omega _{B'/R} \otimes _{B'} B$ is an isomorphism.

Proof. We are going to use the construction of $B'$ from the proof of Lemma 10.149.1 although in principle it should be possible to deduce these results formally from the definition. Namely, we choose a presentation $B = P/J$, where $P = A[x_ i]$ is a polynomial ring over $A$. Next, we choose elements $f_ i \in J$ such that $\text{d}f_ i = \text{d}x_ i \otimes 1$ in $\Omega _{P/A} \otimes _ P B$. Having made these choices we have $B' = P/J'$ with $J' = (f_ i) + J^2$, see proof of Lemma 10.149.1.

Consider the canonical exact sequence

$J'/(J')^2 \to \Omega _{P/A} \otimes _ P B' \to \Omega _{B'/A} \to 0$

see Lemma 10.131.9. By construction the classes of the $f_ i \in J'$ map to elements of the module $\Omega _{P/A} \otimes _ P B'$ which generate it modulo $J'/J^2$ by construction. Since $J'/J^2$ is a nilpotent ideal, we see that these elements generate the module altogether (by Nakayama's Lemma 10.20.1). This proves that $\Omega _{B'/A} = 0$ and hence that $B'$ is formally unramified over $A$, see Lemma 10.148.2.

Since $P$ is a polynomial ring over $A$ we have $\Omega _{P/R} = \Omega _{A/R} \otimes _ A P \oplus \bigoplus P\text{d}x_ i$. We are going to use this decomposition. Consider the following exact sequence

$J'/(J')^2 \to \Omega _{P/R} \otimes _ P B' \to \Omega _{B'/R} \to 0$

see Lemma 10.131.9. We may tensor this with $B$ and obtain the exact sequence

$J'/(J')^2 \otimes _{B'} B \to \Omega _{P/R} \otimes _ P B \to \Omega _{B'/R} \otimes _{B'} B \to 0$

If we remember that $J' = (f_ i) + J^2$ then we see that the first arrow annihilates the submodule $J^2/(J')^2$. In terms of the direct sum decomposition $\Omega _{P/R} \otimes _ P B = \Omega _{A/R} \otimes _ A B \oplus \bigoplus B\text{d}x_ i$ given we see that the submodule $(f_ i)/(J')^2 \otimes _{B'} B$ maps isomorphically onto the summand $\bigoplus B\text{d}x_ i$. Hence what is left of this exact sequence is an isomorphism $\Omega _{A/R} \otimes _ A B \to \Omega _{B'/R} \otimes _{B'} B$ as desired. $\square$

## Comments (2)

Comment #1149 by Matthieu Romagny on

Last sentence of the statement of Lemma Tag 04EB : "such that $S'\to A\to A/I$ is equal to $S'\to S\to A/I$".

## 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 04EA. Beware of the difference between the letter 'O' and the digit '0'.