## 10.130 Differentials

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

Definition 10.130.1. 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: $D(ab) = aD(b) + bD(a)$.

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

$\text{Der}_ R(S, M).$

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

$\bigoplus \nolimits _{(a, b)\in S^2} S[(a, b)] \oplus \bigoplus \nolimits _{(f, g)\in S^2} S[(f, g)] \oplus \bigoplus \nolimits _{r\in R} S[r] \longrightarrow \bigoplus \nolimits _{a\in S} S[a]$

defined by the rules

$[(a, b)] \longmapsto [a + b] - [a] - [b],\quad [(f, g)] \longmapsto [fg] -f[g] - g[f],\quad [r] \longmapsto [\varphi (r)]$

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

$\text{d}(a + b) = \text{d}a + \text{d}b, \quad \text{d}(fg) = f\text{d}g + g\text{d}f, \quad \text{d}r = 0$

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

Definition 10.130.2. The pair $(\Omega _{S/R}, \text{d})$ is called the module of Kähler differentials or the module of differentials of $S$ over $R$.

Lemma 10.130.3. The module of differentials of $S$ over $R$ has the following universal property. The map

$\mathop{\mathrm{Hom}}\nolimits _ S(\Omega _{S/R}, M) \longrightarrow \text{Der}_ R(S, M), \quad \alpha \longmapsto \alpha \circ \text{d}$

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 10.130.4. 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

$\Omega _{S/R} = \mathop{\mathrm{colim}}\nolimits _ i \Omega _{S_ i/R_ i}.$

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

Proof. This is clear from the presentation of $\Omega _{S/R}$ given above. $\square$

Lemma 10.130.5. 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

10.130.5.1
\begin{equation} \label{algebra-equation-functorial-omega} \vcenter { \xymatrix{ S \ar[r]_\varphi & S' \\ R \ar[r]^\psi \ar[u]^\alpha & R' \ar[u]_\beta } } \end{equation}

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

$\xymatrix{ \Omega _{S/R} \ar[r] & \Omega _{S'/R'} \\ S \ar[u]^{\text{d}} \ar[r]^{\varphi } & S' \ar[u]_{\text{d}} }$

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

$\xymatrix{ \bigoplus S'[(a', b')] \oplus \bigoplus S'[(f', g')] \oplus \bigoplus S'[r'] \ar[r] & \bigoplus S' [a'] \\ \\ \bigoplus S[(a, b)] \oplus \bigoplus S[(f, g)] \oplus \bigoplus S[r] \ar[r] \ar[uu]^{ \begin{matrix} [(a, b)] \mapsto [(\varphi (a), \varphi (b))] \\ [(f, g)] \mapsto [(\varphi (f), \varphi (g))] \\ [r]\mapsto [\psi (r)] \end{matrix} } & \bigoplus S[a] \ar[uu]_{[a] \mapsto [\varphi (a)]} }$

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

Lemma 10.130.6. In diagram (10.130.5.1), 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 $\text{d}(i)$, $i \in I$.)

Proof. Consider the map of presentations above. Clearly the right vertical map of free modules is surjective. Thus the map is surjective. A diagram chase shows that the following elements generate the kernel as an $S$-module for sure: $i\text{d}a, i\in I, a \in S$, and $\text{d}a$, with $a \in S$ such that $\varphi (a) = \beta (r')$ for some $r' \in R'$. Note that $\varphi (i) = \varphi (ia) = 0 = \beta (0)$, and that $\text{d}(ia) = i\text{d}a + a \text{d}i$. Hence $i\text{d}a = \text{d}(ia) - a \text{d}i$ is an $S$-linear combination of elements of the second kind. $\square$

Lemma 10.130.7. Let $A \to B \to C$ be ring maps. Then there is a canonical exact sequence

$C \otimes _ B \Omega _{B/A} \to \Omega _{C/A} \to \Omega _{C/B} \to 0$

of $C$-modules.

Proof. We get a diagram (10.130.5.1) by putting $R = A$, $S = C$, $R' = B$, and $S' = C$. By Lemma 10.130.6 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 10.130.8. Let $\varphi : A \to B$ be a ring map.

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

2. 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 10.130.9. In diagram (10.130.5.1), 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

$I/I^2 \longrightarrow \Omega _{S/R} \otimes _ S S' \longrightarrow \Omega _{S'/R} \longrightarrow 0$

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 10.130.6.

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 10.130.6 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 10.130.10. In diagram (10.130.5.1), 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 10.130.9 turns into a short exact sequence

$0 \longrightarrow I/I^2 \longrightarrow \Omega _{S/R} \otimes _ S S' \longrightarrow \Omega _{S'/R} \longrightarrow 0$

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

$D : S \longrightarrow I/I^2, \quad x \longmapsto x - \beta (\alpha (x))$

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 10.130.11. 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

$\Omega _{S/R} \otimes _ S S/I^ n \longrightarrow \Omega _{S'/R} \otimes _{S'} S/I^ n.$

Proof. This follows from Lemma 10.130.9 and the fact that $\text{d}(I^{n + 1}) \subset I^ n\Omega _{S/R}$ by the Leibniz rule for $\text{d}$. $\square$

Lemma 10.130.12. 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 (10.130.5.1). 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 $\text{d}'( \sum a_ i \otimes x_ i ) = \text{d}(a_ i) \otimes x_ i$. 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 10.130.13. 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 10.130.10 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 10.130.5 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 10.130.12 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 10.130.12. At this point the sequence of Lemma 10.130.9 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. $\square$

Lemma 10.130.14. 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 10.130.15. 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 10.130.9 there is an exact sequence of $S$-modules

$I/I^2 \to \Omega _{R[x_1, \ldots , x_ n]/R} \otimes _{R[x_1, \ldots , x_ n]} S \to \Omega _{S/R} \to 0$

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 10.130.14. $\square$

Lemma 10.130.16. 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 10.130.15. $\square$

Comment #439 by Leeroy on

I think there is an error in the proof of lemma 10.125.9 : it should say $df_2 \otimes f_1 + df_1 \otimes f_2$ instead of $df_2 \otimes f_1 + df_2 \otimes f_1$.

Comment #1510 by Girish Kulkarni on

I think there is an error in the proof of lemma 10.128.13: In the first paragraph, a term in the last line is $ba \otimes 1$, it should be $1 \otimes ba$.

Comment #3607 by Herman Rohrbach on

I feel that the discussion following diagram 00RQ could be simplified a little bit, by using the universal property instead of the construction of the differentials. If $d': S' \rightarrow \Omega_{S'/R'}$ is the universal derivation, then $d'\phi: S \rightarrow \Omega_{S'/R'}$ is an $R$-derivation and by the universal property of $d: S \rightarrow \Omega_{S/R}$, there exists a corresponding $S$-linear map $\Omega_{S/R} \rightarrow \Omega_{S'/R'}$, and it is still completely clear that the map is defined by $fdg \mapsto \phi(f)d'\phi(g)$.

Comment #3718 by on

@#3607: Yes, initially this section was written starting with just the presentation of the module of differentials and not using the universal property. So there are a few things that could/should be cleaned up in this section. Feel free to edit and send it to me (just do a little bit first to see if I agree with the approach).

Comment #4205 by Pierre on

In lemma 00RW, at some point the conclusion is that $\Omega _{S \otimes _ R S/S} = \Omega _{S/R} \otimes _ S (S \otimes _ R S)$ which is said to follow from lemma 00RV, but in lemma 00RV the tensor product is over the base change $R\to S$, and in the 00RW the tensor product is not over the base change but over the induced map $S\to S\otimes_R S$. Is that a problem or am I missing something?

Comment #4387 by on

Dear Pierre, I think that what you are missing is that given an $S$-module $M$ we have $M \otimes_R S = M \otimes_S (S \otimes_R S)$. I hope that clarifies things. In other words, in the statement of Lemma 10.130.12 we could have added the equality $\Omega_{S/R} \otimes_R R' = \Omega_{S/R} \otimes_S S'$.

Comment #5105 by Jordan Levin on

There is a nice discussion of the exact sequences using only the universal properties in a down-to-Earth way in the book "Commutative Algebra" by Singh on page 249.

Comment #5312 by on

@#5105: I don't have access to this book myself. If another person concurs with you I will add a reference.

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