Lemma 10.131.13 (00RW)—The Stacks project

Lemma 10.131.13. Let $R \to S$ be a ring map. Let $J = \mathop{\mathrm{Ker}}(S \otimes _ R S \to S)$ be the kernel of the multiplication map. There is a canonical isomorphism of $S$-modules $\Omega _{S/R} \to J/J^2$, $a \text{d} b \mapsto a \otimes b - ab \otimes 1$.

First proof. Apply Lemma 10.131.10 to the commutative diagram

\[ \xymatrix{ S \otimes _ R S \ar[r] & S \\ S \ar[r] \ar[u] & S \ar[u] } \]

where the left vertical arrow is $a \mapsto a \otimes 1$. We get the exact sequence $0 \to J/J^2 \to \Omega _{S \otimes _ R S/S} \otimes _{S \otimes _ R S} S \to \Omega _{S/S} \to 0$. By Lemma 10.131.4 the term $\Omega _{S/S}$ is $0$, and we obtain an isomorphism between the other two terms. We have $\Omega _{S \otimes _ R S/S} = \Omega _{S/R} \otimes _ S (S \otimes _ R S)$ by Lemma 10.131.12 as $S \to S \otimes _ R S$ is the base change of $R \to S$ and hence

\[ \Omega _{S \otimes _ R S/S} \otimes _{S \otimes _ R S} S = \Omega _{S/R} \otimes _ S (S \otimes _ R S) \otimes _{S \otimes _ R S} S = \Omega _{S/R} \]

We omit the verification that the map is given by the rule of the lemma. $\square$

Second proof. First we show that the rule $a \text{d} b \mapsto a \otimes b - ab \otimes 1$ is well defined. In order to do this we have to show that $\text{d}r$ and $a\text{d}b + b \text{d}a - d(ab)$ map to zero. The first because $r \otimes 1 - 1 \otimes r = 0$ by definition of the tensor product. The second because

\[ (a \otimes b - ab \otimes 1) + (b \otimes a - ba \otimes 1) - (1 \otimes ab - ab \otimes 1) = (a \otimes 1 - 1\otimes a)(1\otimes b - b \otimes 1) \]

is in $J^2$.

We construct a map in the other direction. We may think of $S \to S \otimes _ R S$, $a \mapsto a \otimes 1$ as the base change of $R \to S$. Hence we have $\Omega _{S \otimes _ R S/S} = \Omega _{S/R} \otimes _ S (S \otimes _ R S)$, by Lemma 10.131.12. At this point the sequence of Lemma 10.131.9 gives a map

\[ J/J^2 \to \Omega _{S \otimes _ R S/ S} \otimes _{S \otimes _ R S} S = (\Omega _{S/R} \otimes _ S (S \otimes _ R S))\otimes _{S \otimes _ R S} S = \Omega _{S/R}. \]

We leave it to the reader to see it is the inverse of the map above. $\square$

Comments (2)

Comment #1694 by Jeroen Sijsling on November 17, 2015 at 18:09

A shorter proof seems available by using 02HP: there one takes $R = S'$ to be $S$ and $S$ to be $S \otimes_R S$ . This is possible because as mentioned the maps $S \otimes_R S \to S$ and $S \to S \otimes_R S$ are in fact $S$ -algebra maps, with moreover the latter a right inverse of the former.

We get the exact sequence $0 \to J / J^2 \to \Omega_{S \otimes_R S / S} \otimes_{S \otimes_R S} S \to \Omega_{S / S} \to 0$ . By 00RP the term $\Omega_{S / S}$ is trivial, and one obtains an isomorphism between the other two terms. One then rewrites $\Omega_{S \otimes_R S / S} \otimes_{S \otimes_R S} S$ as $\Omega_{S / R}$ as in the current proof.

Comment #1742 by Johan on December 15, 2015 at 18:51

Dear Jeroen, I have added your proof as a first proof and left in the original as well. Thanks! For changes see here.

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