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

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 10.131.5 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

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

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

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

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