Definition 10.131.1. Let $\varphi : R \to S$ be a ring map and let $M$ be an $S$-module. A *derivation*, or more precisely an *$R$-derivation* into $M$ is a map $D : S \to M$ which is additive, annihilates elements of $\varphi (R)$, and satisfies the *Leibniz rule*: $D(ab) = aD(b) + bD(a)$.

## 10.131 Differentials

In this section we define the module of differentials of a ring map.

Note that $D(ra) = rD(a)$ if $r \in R$ and $a \in S$. An equivalent definition is that an $R$-derivation is an $R$-linear map $D : S \to M$ which satisfies the Leibniz rule. The set of all $R$-derivations forms an $S$-module: Given two $R$-derivations $D, D'$ the sum $D + D' : S \to M$, $a \mapsto D(a)+D'(a)$ is an $R$-derivation, and given an $R$-derivation $D$ and an element $c\in S$ the scalar multiple $cD : S \to M$, $a \mapsto cD(a)$ is an $R$-derivation. We denote this $S$-module

Also, if $\alpha : M \to N$ is an $S$-module map, then the composition $\alpha \circ D$ is an $R$-derivation into $N$. In this way the assignment $M \mapsto \text{Der}_ R(S, M)$ is a covariant functor.

Consider the following map of free $S$-modules

defined by the rules

with obvious notation. Let $\Omega _{S/R}$ be the cokernel of this map. There is a map $\text{d} : S \to \Omega _{S/R}$ which maps $a$ to the class $\text{d}a$ of $[a]$ in the cokernel. This is an $R$-derivation by the relations imposed on $\Omega _{S/R}$, in other words

where $a,b,f,g \in S$ and $r \in R$.

Definition 10.131.2. The pair $(\Omega _{S/R}, \text{d})$ is called the *module of Kähler differentials* or the *module of differentials* of $S$ over $R$.

Lemma 10.131.3. The module of differentials of $S$ over $R$ has the following universal property. The map

is an isomorphism of functors.

**Proof.**
By definition an $R$-derivation is a rule which associates to each $a \in S$ an element $D(a) \in M$. Thus $D$ gives rise to a map $[D] : \bigoplus S[a] \to M$. However, the conditions of being an $R$-derivation exactly mean that $[D]$ annihilates the image of the map in the displayed presentation of $\Omega _{S/R}$ above.
$\square$

Lemma 10.131.4. Suppose that $R \to S$ is surjective. Then $\Omega _{S/R} = 0$.

**Proof.**
You can see this either because all $R$-derivations clearly have to be zero, or because the map in the presentation of $\Omega _{S/R}$ is surjective.
$\square$

Suppose that

is a commutative diagram of rings. In this case there is a natural map of modules of differentials fitting into the commutative diagram

To construct the map just use the obvious map between the presentations for $\Omega _{S/R}$ and $\Omega _{S'/R'}$. Namely,

The result is simply that $f\text{d}g \in \Omega _{S/R}$ is mapped to $\varphi (f)\text{d}\varphi (g)$.

Lemma 10.131.5. Let $I$ be a directed set. Let $(R_ i \to S_ i, \varphi _{ii'})$ be a system of ring maps over $I$, see Categories, Section 4.21. Then we have

where $R \to S = \mathop{\mathrm{colim}}\nolimits (R_ i \to S_ i)$.

**Proof.**
This is clear from the defining presentation of $\Omega _{S/R}$ and the functoriality of this described above.
$\square$

Lemma 10.131.6. In diagram (10.131.4.1), suppose that $S \to S'$ is surjective with kernel $I \subset S$. Then $\Omega _{S/R} \to \Omega _{S'/R'}$ is surjective with kernel generated as an $S$-module by the elements $\text{d}a$, where $a \in S$ is such that $\varphi (a) \in \beta (R')$. (This includes in particular the elements $\text{d}(i)$, $i \in I$.)

**First proof.**
Consider the map of presentations (10.131.4.2). Clearly the right vertical map of free modules is surjective. Thus the map is surjective. Suppose that some element $\eta $ of $\Omega _{S/R}$ maps to zero in $\Omega _{S'/R'}$. Write $\eta $ as the image of $\sum s_ i[a_ i]$ for some $s_ i, a_ i \in S$. Then we see that $\sum \varphi (s_ i)[\varphi (a_ i)]$ is the image of an element

in the upper left corner of the diagram. Since $\varphi $ is surjective, the terms $s_ j'[a_ j', b_ j']$ and $s_ k'[f_ k', g_ k']$ are in the image of elements in the lower right corner. Thus, modifying $\eta $ and $\theta $ by subtracting the images of these elements, we may assume $\theta = \sum s_ l'[r_ l']$. In other words, we see $\sum \varphi (s_ i)[\varphi (a_ i)]$ is of the form $\sum s'_ l [\beta (r'_ l)]$. Next, we may assume that we have some $a' \in S'$ such that $a' = \varphi (a_ i)$ for all $i$ and $a' = \beta (r_ l')$ for all $l$. This is clear from the direct sum decomposition of the upper right corner of the diagram. Choose $a \in S$ with $\varphi (a) = a'$. Then we can write $a_ i = a + x_ i$ for some $x_ i \in I$. Thus we may assume that all $a_ i$ are equal to $a$ by using the relations that are allowed. But then we may assume our element is of the form $s[a]$. We still know that $\varphi (s)[a'] = \sum \varphi (s_ l')[\beta (r_ l')]$. Hence either $\varphi (s) = 0$ and we're done, or $a' = \varphi (a)$ is in the image of $\beta $ and we're done as well. $\square$

**Second proof.**
We will use the universal property of modules of differentials given in Lemma 10.131.3 without further mention.

In (10.131.4.1) let $R'' = S \times _{S'} R'$. Then we have following diagram:

Let $M$ be an $S$-module. It follows immediately from the definitions that an $R$-derivation $D : S \to M$ is an $R''$-derivation if and only if it annihilates the elements in the image of $R'' \to S$. The universal property translates this into the statement that the natural map $\Omega _{S/R} \to \Omega _{S/R''}$ is surjective with kernel generated as an $S$-module by the image of $R''$.

From the previous paragraph we see that it suffices to show that $\Omega _{S/R} \to \Omega _{S'/R'}$ is an isomorphism when $S \to S'$ is surjective and $R = S \times _{S'} R'$. Let $M'$ be an $S'$-module. Observe that any $R'$-derivation $D' : S' \to M'$ gives an $R$-derivation by precomposing with $S \to S'$. Conversely, suppose $M$ is an $S$-module and $D : S \to M$ is an $R$-derivation. If $i \in I$, then there exist an $a \in R$ with $\alpha (a) = i$ (as $R = S \times _{S'} R'$). It follows that $D(i) = 0$ and hence $0 = D(is) = iD(s)$ for all $s \in S$. Thus the image of $D$ is contained in the submodule $M' \subset M$ of elements annihilated by $I$ and moreover the induced map $S \to M'$ factors through an $R'$-derivation $S' \to M'$. It is an exercise to use the universal property to see that this means $\Omega _{S/R} \to \Omega _{S'/R'}$ is an isomorphism; details omitted. $\square$

Lemma 10.131.7. Let $A \to B \to C$ be ring maps. Then there is a canonical exact sequence

of $C$-modules.

**Proof.**
We get a diagram (10.131.4.1) by putting $R = A$, $S = C$, $R' = B$, and $S' = C$. By Lemma 10.131.6 the map $\Omega _{C/A} \to \Omega _{C/B}$ is surjective, and the kernel is generated by the elements $\text{d}(c)$, where $c \in C$ is in the image of $B \to C$. The lemma follows.
$\square$

Lemma 10.131.8. Let $\varphi : A \to B$ be a ring map.

If $S \subset A$ is a multiplicative subset mapping to invertible elements of $B$, then $\Omega _{B/A} = \Omega _{B/S^{-1}A}$.

If $S \subset B$ is a multiplicative subset then $S^{-1}\Omega _{B/A} = \Omega _{S^{-1}B/A}$.

**Proof.**
To show the equality of (1) it is enough to show that any $A$-derivation $D : B \to M$ annihilates the elements $\varphi (s)^{-1}$. This is clear from the Leibniz rule applied to $1 = \varphi (s) \varphi (s)^{-1}$. To show (2) note that there is an obvious map $S^{-1}\Omega _{B/A} \to \Omega _{S^{-1}B/A}$. To show it is an isomorphism it is enough to show that there is a $A$-derivation $\text{d}'$ of $S^{-1}B$ into $S^{-1}\Omega _{B/A}$. To define it we simply set $\text{d}'(b/s) = (1/s)\text{d}b - (1/s^2)b\text{d}s$. Details omitted.
$\square$

Lemma 10.131.9. In diagram (10.131.4.1), suppose that $S \to S'$ is surjective with kernel $I \subset S$, and assume that $R' = R$. Then there is a canonical exact sequence of $S'$-modules

The leftmost map is characterized by the rule that $f \in I$ maps to $\text{d}f \otimes 1$.

**Proof.**
The middle term is $\Omega _{S/R} \otimes _ S S/I$. For $f \in I$ denote $\overline{f}$ the image of $f$ in $I/I^2$. To show that the map $\overline{f} \mapsto \text{d}f \otimes 1$ is well defined we just have to check that $\text{d} f_1f_2 \otimes 1 = 0$ if $f_1, f_2 \in I$. And this is clear from the Leibniz rule $\text{d} f_1f_2 \otimes 1 = (f_1 \text{d}f_2 + f_2 \text{d} f_1 )\otimes 1 = \text{d}f_2 \otimes f_1 + \text{d}f_1 \otimes f_2 = 0$. A similar computation show this map is $S' = S/I$-linear.

The map $\Omega _{S/R} \otimes _ S S' \to \Omega _{S'/R}$ is the canonical $S'$-linear map associated to the $S$-linear map $\Omega _{S/R} \to \Omega _{S'/R}$. It is surjective because $\Omega _{S/R} \to \Omega _{S'/R}$ is surjective by Lemma 10.131.6.

The composite of the two maps is zero because $\text{d}f$ maps to zero in $\Omega _{S'/R}$ for $f \in I$. Note that exactness just says that the kernel of $\Omega _{S/R} \to \Omega _{S'/R}$ is generated as an $S$-submodule by the submodule $I\Omega _{S/R}$ together with the elements $\text{d}f$, with $f \in I$. We know by Lemma 10.131.6 that this kernel is generated by the elements $\text{d}(a)$ where $\varphi (a) = \beta (r)$ for some $r \in R$. But then $a = \alpha (r) + a - \alpha (r)$, so $\text{d}(a) = \text{d}(a - \alpha (r))$. And $a - \alpha (r) \in I$ since $\varphi (a - \alpha (r)) = \varphi (a) - \varphi (\alpha (r)) = \beta (r) - \beta (r) = 0$. We conclude the elements $\text{d}f$ with $f \in I$ already generate the kernel as an $S$-module, as desired. $\square$

Lemma 10.131.10. In diagram (10.131.4.1), suppose that $S \to S'$ is surjective with kernel $I \subset S$, and assume that $R' = R$. Moreover, assume that there exists an $R$-algebra map $S' \to S$ which is a right inverse to $S \to S'$. Then the exact sequence of $S'$-modules of Lemma 10.131.9 turns into a short exact sequence

which is even a split short exact sequence.

**Proof.**
Let $\beta : S' \to S$ be the right inverse to the surjection $\alpha : S \to S'$, so $S = I \oplus \beta (S')$. Clearly we can use $\beta : \Omega _{S'/R} \to \Omega _{S/R}$, to get a right inverse to the map $\Omega _{S/R} \otimes _ S S' \to \Omega _{S'/R}$. On the other hand, consider the map

It is easy to show that $D$ is an $R$-derivation (omitted). Moreover $x D(s) = 0$ if $x \in I, s \in S$. Hence, by the universal property $D$ induces a map $\tau : \Omega _{S/R} \otimes _ S S' \to I/I^2$. We omit the verification that it is a left inverse to $\text{d} : I/I^2 \to \Omega _{S/R} \otimes _ S S'$. Hence we win. $\square$

Lemma 10.131.11. Let $R \to S$ be a ring map. Let $I \subset S$ be an ideal. Let $n \geq 1$ be an integer. Set $S' = S/I^{n + 1}$. The map $\Omega _{S/R} \to \Omega _{S'/R}$ induces an isomorphism

**Proof.**
This follows from Lemma 10.131.9 and the fact that $\text{d}(I^{n + 1}) \subset I^ n\Omega _{S/R}$ by the Leibniz rule for $\text{d}$.
$\square$

Lemma 10.131.12. Suppose that we have ring maps $R \to R'$ and $R \to S$. Set $S' = S \otimes _ R R'$, so that we obtain a diagram (10.131.4.1). Then the canonical map defined above induces an isomorphism $\Omega _{S/R} \otimes _ R R' = \Omega _{S'/R'}$.

**Proof.**
Let $\text{d}' : S' = S \otimes _ R R' \to \Omega _{S/R} \otimes _ R R'$ denote the map $\text{d}'( \sum a_ i \otimes x_ i ) = \sum \text{d}(a_ i) \otimes x_ i$. It exists because the map $S \times R' \to \Omega _{S/R} \otimes _ R R'$, $(a, x)\mapsto \text{d}a \otimes _ R x$ is $R$-bilinear. This is an $R'$-derivation, as can be verified by a simple computation. We will show that $(\Omega _{S/R} \otimes _ R R', \text{d}')$ satisfies the universal property. Let $D : S' \to M'$ be an $R'$-derivation into an $S'$-module. The composition $S \to S' \to M'$ is an $R$-derivation, hence we get an $S$-linear map $\varphi _ D : \Omega _{S/R} \to M'$. We may tensor this with $R'$ and get the map $\varphi '_ D : \Omega _{S/R} \otimes _ R R' \to M'$, $\varphi '_ D(\eta \otimes x) = x\varphi _ D(\eta )$. It is clear that $D = \varphi '_ D \circ \text{d}'$.
$\square$

The multiplication map $S \otimes _ R S \to S$ is the $R$-algebra map which maps $a \otimes b$ to $ab$ in $S$. It is also an $S$-algebra map, if we think of $S \otimes _ R S$ as an $S$-algebra via either of the maps $S \to S \otimes _ R S$.

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 10.131.4 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

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. $\square$

Lemma 10.131.14. If $S = R[x_1, \ldots , x_ n]$, then $\Omega _{S/R}$ is a finite free $S$-module with basis $\text{d}x_1, \ldots , \text{d}x_ n$.

**Proof.**
We first show that $\text{d}x_1, \ldots , \text{d}x_ n$ generate $\Omega _{S/R}$ as an $S$-module. To prove this we show that $\text{d}g$ can be expressed as a sum $\sum g_ i \text{d}x_ i$ for any $g \in R[x_1, \ldots , x_ n]$. We do this by induction on the (total) degree of $g$. It is clear if the degree of $g$ is $0$, because then $\text{d}g = 0$. If the degree of $g$ is $> 0$, then we may write $g$ as $c + \sum g_ i x_ i$ with $c\in R$ and $\deg (g_ i) < \deg (g)$. By the Leibniz rule we have $\text{d}g = \sum g_ i \text{d} x_ i + \sum x_ i \text{d}g_ i$, and hence we win by induction.

Consider the $R$-derivation $\partial / \partial x_ i : R[x_1, \ldots , x_ n] \to R[x_1, \ldots , x_ n]$. (We leave it to the reader to define this; the defining property being that $\partial / \partial x_ i (x_ j) = \delta _{ij}$.) By the universal property this corresponds to an $S$-module map $l_ i : \Omega _{S/R} \to R[x_1, \ldots , x_ n]$ which maps $\text{d}x_ i$ to $1$ and $\text{d}x_ j$ to $0$ for $j \not= i$. Thus it is clear that there are no $S$-linear relations among the elements $\text{d}x_1, \ldots , \text{d}x_ n$. $\square$

Lemma 10.131.15. Suppose $R \to S$ is of finite presentation. Then $\Omega _{S/R}$ is a finitely presented $S$-module.

**Proof.**
Write $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$. Write $I = (f_1, \ldots , f_ m)$. According to Lemma 10.131.9 there is an exact sequence of $S$-modules

The result follows from the fact that $I/I^2$ is a finite $S$-module (generated by the images of the $f_ i$), and that the middle term is finite free by Lemma 10.131.14. $\square$

Lemma 10.131.16. Suppose $R \to S$ is of finite type. Then $\Omega _{S/R}$ is finitely generated $S$-module.

**Proof.**
This is very similar to, but easier than the proof of Lemma 10.131.15.
$\square$

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