## 10.131 Differentials

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

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

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

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

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

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

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 10.131.4. 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

10.131.4.1
\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 10.131.5. 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

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

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

Proof. This is clear from the 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 $\text{d}(i)$, $i \in I$.)

Proof. We urge the reader to find their own (hopefully different) proof of this lemma. Consider the map of presentations above. 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

$\theta = \sum s_ j'[a_ j', b_ j'] + \sum s_ k'[f_ k', g_ k'] + \sum s_ l'[r_ l']$

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 substracting 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 $\sum s'_ l [\beta (r'_ l)]$. Pick $a' \in S'$. 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. $\square$

Lemma 10.131.7. 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 (10.131.4.1) 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.

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 10.131.9. 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

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

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 10.131.6.

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

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

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 10.131.11. 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

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

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

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

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. $\square$

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

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

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 10.131.14. $\square$

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. $\square$

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 10.131.12 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.

Comment #6053 by Jonas Ehrhard on

This might be nitpicking, but shouldn't it be $\text{d} \varphi(r) = 0$ just in front of definition 10.131.2?

Comment #6189 by on

Nitpick or not, it has to be correct. Thanks and fixed here.

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