Differentials

Our goal in this post is to algebraically define derivations. After examining the resulting Kähler differential module and its two fundamental exact sequences, in the last two sections we discuss the naive cotangent complex, which extends the conormal sequence one term further to the left.

Kähler Differential Modules

Definition 1 Let a ring \(A\), an \(A\)-algebra \(E\), and an \(E\)-module \(M\) be given. Then an \(A\)-linear map satisfying the Leibniz rule

\[\dd{(xy)}=y\dd{x}+x\dd{y}\]

for all \(x,y\in E\) is called an \(A\)-derivation, and their collection is denoted by \(\Der_A(E,M)\).

One of the basic properties of derivations is that \(\Der_A(E,M)\) has an \(E\)-module structure. This is because if for any \(x\in E\) and \(d\in \Der_A(E, M)\) we define the \(A\)-linear map \(x d\) by

\[xd: E \rightarrow M;\qquad y\mapsto x\dd{(y)}\]

then for any \(y_1,y_2\in E\), the identity

\[(xd)(y_1y_2)=x\dd{(y_1y_2)}=x(y_1\dd{y_2}+y_2\dd{y_1})=y_1(xd)(y_2)+y_2(xd)(y_1)\]

holds. Furthermore, given any \(A\)-derivation \(d: E \rightarrow M\) and any \(E\)-linear map \(u:M \rightarrow M'\), the composition

\[u\circ d: E \rightarrow M'\]

is also an \(A\)-derivation, which follows from

\[(u\circ d)(y_1y_2)=u(y_1\dd{y_2}+y_2\dd{y_1})=y_1u(\dd{y_2})+y_2u(\dd{y_1})=y_1(u\circ d)(y_2)+y_2(u\circ d)(y_1)\]

That is, \(\Der_A(E, -)\) is a functor from \(\lMod{E}\) to itself.

Lemma 2 The functor \(\Der_A(E, -)\) is representable. That is, between the two functors from \(\lMod{E}\) to itself, a natural isomorphism

\[\Der_A(E,-)\cong\Hom_E(\Omega_{E/A},-)\]

holds for some \(E\)-module \(\Omega_{E/A}\).

The representing object \(\Omega_{E/A}\) is defined as follows.

Definition 3 For an \(A\)-algebra \(E\), the Kähler differential module of \(E\) over \(A\) is the module generated by \(\{\dd{f}\mid f\in E\}\) as an \(E\)-module, subject to the relations

\[\text{$d(xy)=x\dd{y}+y\dd{x}$ for all $x,y\in E$},\qquad \text{$d(ax+by)=a\dd{x}+b\dd{y}$ for all $x,y\in E$ and $a,b\in A$}\]

giving an \(E\)-module denoted by \(\Omega_{E/A}\). Here, defined by \(f\mapsto \dd{f}\), the \(A\)-linear derivation \(d:E \rightarrow \Omega_{E/A}\) is called the universal \(A\)-derivation.

Then one can easily verify that \(\Omega_{E/A}\) satisfies the universal property of Lemma 2.

Meanwhile, we have already constructed the same module in a different way. In [Multilinear Algebra] §Differential Modules, ⁋Definition 9, starting with the multiplication map \(m:E\otimes_AE \rightarrow E\), we took its kernel \(\mathfrak{I}\) to define the \(E\)-module \(\mathfrak{I}/\mathfrak{I}^2\) as the module of \(A\)-differentials, and showed that \(x\mapsto (x\otimes1-1\otimes x)+\mathfrak{I}^2\) plays the role of the universal \(A\)-derivation. Although the starting point differs from Definition 3, which is built directly from generators and relations, the two constructions canonically coincide.

Proposition 4 For an \(A\)-algebra \(E\), let the kernel of the multiplication map \(m:E\otimes_AE \rightarrow E\) be \(\mathfrak{I}\). Then, sending \((x\otimes1-1\otimes x)+\mathfrak{I}^2\) to \(\dd{x}\), there exists a canonical isomorphism

\[\mathfrak{I}/\mathfrak{I}^2\cong\Omega_{E/A}\]

Proof

By [Multilinear Algebra] §Differential Modules, ⁋Proposition 8, for any \(E\)-module \(M\), there is a natural isomorphism \(\Hom_E(\mathfrak{I}/\mathfrak{I}^2, M)\cong\Der_A(E, M)\), and by Lemma 2, \(\Omega_{E/A}\) also represents the same functor. That is, since the two \(E\)-modules represent the same functor on \(\lMod{E}\), they are canonically isomorphic, and this isomorphism identifies the universal \(A\)-derivations on both sides. Therefore, for each \(x\in E\), \((x\otimes1-1\otimes x)+\mathfrak{I}^2\) is sent to \(\dd{x}\).

In the most basic case of a polynomial algebra, this module is the free module spanned by the differentials of the coordinates.

Proposition 5 For a set \(S\) and the polynomial algebra \(R=A[\x_s]_{s\in S}\), \(\Omega_{R/A}\) is, with basis \(\{\dd{\x_s}\}_{s\in S}\), a free \(R\)-module, and for any \(f\in R\),

\[\dd{f}=\sum_{s\in S}\frac{\partial f}{\partial \x_s}\dd{\x_s}\]

holds. Here, since \(f\) contains only finitely many variables, the right-hand side is a finite sum.

Proof

By Definition 3, \(\Omega_{R/A}\) is generated by the elements \(\dd{f}\). Since \(d\) is \(A\)-linear, it suffices to consider the case where \(f\) is a monomial, and in this case applying the Leibniz rule by induction on the degree yields the chain rule above. Therefore, \(\Omega_{R/A}\) is generated by the \(\dd{\x_s}\).

To show that these are linearly independent over \(R\), for each \(t\in S\) consider the partial derivative \(\partial/\partial \x_t:R \rightarrow R\). Since this is an \(A\)-derivation, by Lemma 2 it induces an \(R\)-linear map \(\partial_t:\Omega_{R/A} \rightarrow R\), and by definition \(\partial_t(\dd{\x_s})=\delta_{st}\). Then, for a finite sum \(\sum_sf_s\dd{\x_s}=0\), applying \(\partial_t\) yields \(f_t=0\).

This is the same as the computation in the case where \(M\) is a free \(A\)-module for the symmetric algebra \(\S(M)\) ([Multilinear Algebra] §Differential Modules, ⁋Example 10). Moreover, if we view \(\Omega\) as a functor taking an \(A\)-algebra \(A \rightarrow E\) to \(\Omega_{E/A}\), the following kind of functoriality also holds.

Proposition 6 Suppose we are given a commutative diagram of ring homomorphisms

and, via \(\rho\) and \(\rho'\), regard \(E, E'\) respectively as an \(A\)-algebra and an \(A'\)-algebra. Then there exists a unique \(E\)-linear map \(\Omega_{\varphi/\phi}:\Omega_{E/A} \rightarrow \Omega_{E'/A'}\) making the diagram

commute.

Proof

Since \(d_{E'/A'}\circ \varphi: E \rightarrow \Omega_{E'/A'}\) is an \(A\)-derivation, by Lemma 2, there exists a unique \(E\)-linear map.

Meanwhile, since \(\Omega_{E'/A'}\) is an \(E'\)-module, by [Algebraic Structures] §Change of Scalars, ⁋Proposition 6, we have

\[\Hom_{E'}(\varphi_! \Omega_{E/A},\Omega_{E'/A'})\cong\Hom_E(\Omega_{E/A}, \varphi^\ast\Omega_{E'/A'})\]

Then, strictly speaking, the map \(\Omega_{E/A} \rightarrow \Omega_{E'/A'}\) obtained in Proposition 6 above is \(\Omega_{E/A} \rightarrow \varphi^\ast\Omega_{E'/A'}\), so there exists a corresponding unique \(E'\)-linear homomorphism

\[\Omega_{\varphi/\phi}': \varphi_!\Omega_{E/A}=\Omega_{E/A}\otimes_EE' \rightarrow \Omega_{E'/A'}\]

The most basic consequence of this functoriality is that the Kähler differential module is compatible with localization.

Proposition 7 Suppose an \(A\)-algebra \(E\) and, in \(E\), a multiplicative subset \(S\) are given (§Localization, ⁋Definition 3). Via the canonical homomorphism \(\varphi:E \rightarrow S^{-1}E\), view \(S^{-1}E\) as an \(A\)-algebra. Then the \(S^{-1}E\)-linear homomorphism given by Proposition 6

\[\Omega_{\varphi/\id_A}':\Omega_{E/A}\otimes_ES^{-1}E \longrightarrow \Omega_{S^{-1}E/A}\]

is an isomorphism.

Proof

For any \(S^{-1}E\)-module \(M\), we show that the restriction map along \(\varphi\)

\[\varphi^\ast:\Der_A(S^{-1}E, M) \longrightarrow \Der_A(E, M)\]

is a bijection. Since \(M\) is an \(S^{-1}E\)-module, the elements of \(S\) act invertibly on \(M\), and below, dividing an element of \(M\) by an element of \(S\) means the inverse of this action.

For \(D\in\Der_A(S^{-1}E, M)\), \(x\in E\), and \(s\in S\), applying the Leibniz rule to \(x=s(x/s)\) gives \(D(x)=sD(x/s)+(x/s)D(s)\), so

\[D(x/s)=\bigl(D(x)-(x/s)D(s)\bigr)/s\]

and thus \(D\) is determined by its values on \(E\), whence \(\varphi^\ast\) is injective. For surjectivity, given \(D\in\Der_A(E, M)\), we may take the above formula as a definition and set \(\widetilde{D}(x/s)=\bigl(D(x)-(x/s)D(s)\bigr)/s\). To show that this is well-defined, suppose \(x/s=y/t\). Then some \(u\in S\) satisfies \(u(tx-sy)=0\) (§Localization, ⁋Definition 4), and applying \(D\) to this yields

\[utD(x)+uxD(t)+txD(u)=usD(y)+uyD(s)+syD(u)\]

In \(S^{-1}E\), since \(tx=sy\), the last terms on both sides are equal, and dividing the remaining terms by \(u\) gives

\[tD(x)+xD(t)=sD(y)+yD(s)\]

Meanwhile, from \(tx=sy\) we have \(tx/s=y\) and \(sy/t=x\), so dividing both sides of this equation by \(st\) gives

\[\bigl(D(x)-(x/s)D(s)\bigr)/s=\bigl(D(y)-(y/t)D(t)\bigr)/t\]

which shows that \(\widetilde{D}\) is well-defined. That this preserves scalar multiplication by elements of \(A\) follows directly from the fact that \(D\) is \(A\)-linear and from the definition, and the preservation of addition is verified by applying the definition to \(x/s+y/t=(tx+sy)/(st)\):

\[\widetilde{D}\Bigl(\frac{tx+sy}{st}\Bigr)=\frac{tD(x)+xD(t)+sD(y)+yD(s)}{st}-\Bigl(\frac{x}{s}+\frac{y}{t}\Bigr)\Bigl(\frac{D(s)}{s}+\frac{D(t)}{t}\Bigr)\]

where \(xD(t)/(st)\) and \(yD(s)/(st)\) cancel out, leaving only \(\widetilde{D}(x/s)+\widetilde{D}(y/t)\). The remaining Leibniz rule is verified by

\[\begin{aligned} \widetilde{D}\Bigl(\frac{xy}{st}\Bigr)&=\frac{D(xy)}{st}-\frac{xy}{(st)^2}D(st)\\ &=\frac{xD(y)+yD(x)}{st}-\frac{xy}{(st)^2}\bigl(sD(t)+tD(s)\bigr)\\ &=\frac{x}{s}\Bigl(\frac{D(y)}{t}-\frac{y}{t^2}D(t)\Bigr)+\frac{y}{t}\Bigl(\frac{D(x)}{s}-\frac{x}{s^2}D(s)\Bigr)\\ &=\frac{x}{s}\widetilde{D}\Bigl(\frac{y}{t}\Bigr)+\frac{y}{t}\widetilde{D}\Bigl(\frac{x}{s}\Bigr) \end{aligned}\]

Also, since \(D(1)=D(1\cdot 1)=2D(1)\) implies \(D(1)=0\), we have \(\widetilde{D}(x/1)=D(x)\), and so \(\varphi^\ast\widetilde{D}=D\).

Now since \(\varphi^\ast\) is a bijection natural in \(M\), combining this with Lemma 2 and the adjunction in [Algebraic Structures] §Change of Scalars, ⁋Proposition 6 gives, for any \(S^{-1}E\)-module \(M\),

\[\Hom_{S^{-1}E}\bigl(\Omega_{E/A}\otimes_ES^{-1}E, M\bigr)\cong\Hom_E(\Omega_{E/A}, M)\cong\Der_A(E, M)\cong\Der_A(S^{-1}E, M)\cong\Hom_{S^{-1}E}\bigl(\Omega_{S^{-1}E/A}, M\bigr)\]

and all of these correspondences are natural in \(M\). That is, since the two \(S^{-1}E\)-modules represent the same functor, they are canonically isomorphic; tracing this isomorphism on the generator \(\dd{x}\otimes 1\) gives \(\dd{x}\otimes 1\mapsto \dd{\varphi}(x)\), showing that this is the map \(\Omega_{\varphi/\id_A}'\) given by Proposition 6.

In particular, the universal derivation of \(S^{-1}E\) is obtained from that of \(E\) by \(\dd{(x/s)}=(s\dd{x}-x\dd{s})/s^2\).

Fundamental sequences

Specifically, let \(\phi:A \rightarrow A'\) be \(\id_A:A \rightarrow A\). Then the \(E'\)-linear homomorphism constructed above depends only on the \(A\)-linear map \(\varphi:E \rightarrow E'\), and takes the form

\[\Omega_{\varphi/A}':\Omega_{E/A}\otimes_EE' \rightarrow \Omega_{E'/A}\]

On the other hand, via \(\varphi:E \rightarrow E'\), viewing \(E'\) as an \(E\)-algebra, the module of Kähler differentials of \(E'\) over \(E\), \(\Omega_{E'/E}\), is defined. Here, since the universal \(E\)-derivation \(d_{E'/E}: E' \rightarrow \Omega_{E'/E}\) is also an \(A\)-derivation, by Lemma 2 again there exists, satisfying \(d_{E'/E}=\Omega_\varphi\circ d_{E'/A}\), a unique \(E'\)-linear map

\[\Omega_\varphi:\Omega_{E'/A} \rightarrow \Omega_{E'/E}\]

That is, \(d_{E'/E}\) is identical to the following composition:

\[E' \overset{d_{E'/A}}{\longrightarrow}\Omega_{E'/A}\overset{\Omega_\varphi}{\longrightarrow}\Omega_{E'/E}\]

Proposition 8 (Cotangent sequence) The sequence of \(E'\)-linear maps

\[\Omega_{E/A}\otimes_EE'\overset{\Omega_{\varphi/A}'}{\longrightarrow}\Omega_{E'/A}\overset{\Omega_\varphi}{\longrightarrow}\Omega_{E'/E} \longrightarrow 0\]

is exact.

Proof

Let \(N\) be the image of \(\Omega_{\varphi/A}'\), that is, the \(E'\)-submodule of \(\Omega_{E'/A}\) generated by the elements \(d_{E'/A}\varphi(x)\) (\(x\in E\)), and consider \(C=\Omega_{E'/A}/N\) and the quotient map \(\pi:\Omega_{E'/A} \rightarrow C\).

First, we show that \(\Omega_\varphi\) is surjective. This is because \(\Omega_{E'/E}\) is generated as an \(E'\)-module by the elements \(d_{E'/E}y'\) (\(y'\in E'\)), and by the formula \(d_{E'/E}=\Omega_\varphi\circ d_{E'/A}\) observed above, these all belong to the image of \(\Omega_\varphi\).

Next, we show that the composition of the two maps is \(0\). A generator \(d_{E/A}x\otimes 1\) of \(\Omega_{E/A}\otimes_EE'\) is sent by \(\Omega_{\varphi/A}'\) to \(d_{E'/A}\varphi(x)\), which is then sent by \(\Omega_\varphi\) to \(d_{E'/E}\varphi(x)\). However, since \(d_{E'/E}\) is an \(E\)-derivation, it sends all elements of the form \(\varphi(x)\), that is, elements coming from \(E\), to \(0\). Therefore, \(\Omega_\varphi\circ\Omega_{\varphi/A}'=0\), and in particular, since \(N\subseteq \ker\Omega_\varphi\), \(\Omega_\varphi\) induces an \(E'\)-linear map \(\psi':C \rightarrow \Omega_{E'/E}\).

Now, the proof is complete if we show that \(\ker\Omega_\varphi\subseteq N\). To this end, consider the composition \(\delta=\pi\circ d_{E'/A}:E' \rightarrow C\). Then \(\delta\) is an \(A\)-derivation, and by the definition of \(N\), it sends all elements of \(\varphi(E)\) to \(0\). However, for any \(e\in E\) and \(y'\in E'\),

\[\delta(\varphi(e)y')=\varphi(e)\delta(y')+y'\delta(\varphi(e))=\varphi(e)\delta(y')\]

so \(\delta\) is \(E\)-linear, and hence is an \(E\)-derivation. Then by the universal property of Lemma 2, there exists a unique \(E'\)-linear map \(\psi:\Omega_{E'/E} \rightarrow C\) satisfying \(\psi\circ d_{E'/E}=\delta\). Now, considering the two compositions \(\psi\circ\psi'\) and \(\psi'\circ\psi\), on the generators of \(C\) and \(\Omega_{E'/E}\), respectively, we have

\[\pi(d_{E'/A}y')\overset{\psi'}{\longmapsto} d_{E'/E}y'\overset{\psi}{\longmapsto}\pi(d_{E'/A}y'),\qquad d_{E'/E}y'\overset{\psi}{\longmapsto}\pi(d_{E'/A}y')\overset{\psi'}{\longmapsto}d_{E'/E}y'\]

and so both are the identity. That is, \(\psi'\) is an isomorphism, and therefore \(\ker \Omega_\varphi=N=\im \Omega_{\varphi/A}'\).

Another important exact sequence is obtained especially when \(\varphi:E \rightarrow E'\) is surjective. In this case, by the first isomorphism theorem,

\[E/\ker \varphi\cong E'\]

holds. For convenience, we write \(K=\ker\varphi\). Then we consider the restriction of \(d_{E/A}:E \rightarrow \Omega_{E/A}\) to \(K\),

\[d_{E/A}\vert_K: K \rightarrow \Omega_{E/A}\]

and we can consider the following \(E\)-linear map:

\[K\overset{d\vert_K}{\longrightarrow}\Omega_{E/A}\overset{}{\longrightarrow}\Omega_{E/A}\otimes_EE'\]

Then one can check that the kernel of the above composition contains \(K^2\), and therefore from this, an \(E\)-linear map

\[\bar{d}:K/K^2 \rightarrow \Omega_{E/A}\otimes_EE'\]

is obtained.

Proposition 9 In the above situation, the sequence

\[K/K^2 \overset{\bar{d}}{\longrightarrow}\Omega_{E/A}\otimes_EE' \rightarrow\Omega_{E'/A} \longrightarrow 0\]

is exact.

Proof

The middle map is \(\Omega_{\varphi/A}'\) from Proposition 8 (Cotangent sequence). Let \(N'\) be the submodule generated by the image of \(\bar{d}\) in \(\Omega_{E/A}\otimes_EE'\) as an \(E'\)-submodule, and consider \(C'=(\Omega_{E/A}\otimes_EE')/N'\) and the quotient map \(\pi:\Omega_{E/A}\otimes_EE' \rightarrow C'\).

First, we show that \(\Omega_{\varphi/A}'\) is surjective. This is because \(\varphi\) is surjective, so any generator of \(\Omega_{E'/A}\) can, for some \(x\in E\), be written in the form \(d_{E'/A}\varphi(x)\), which is equal to \(\Omega_{\varphi/A}'(d_{E/A}x\otimes 1)\).

Next, we show that the composition of the two maps is \(0\). For any \(k\in K\), we have \(\bar{d}(k+K^2)=d_{E/A}k\otimes 1\), which is sent by \(\Omega_{\varphi/A}'\) to \(d_{E'/A}\varphi(k)=d_{E'/A}(0)=0\). In particular, since \(N'\subseteq\ker\Omega_{\varphi/A}'\), the map \(\Omega_{\varphi/A}'\) induces an \(E'\)-linear map \(\psi':C' \rightarrow \Omega_{E'/A}\).

Now, the proof is complete if we show that \(\ker\Omega_{\varphi/A}'\subseteq N'\). We define a map \(\delta: E' \rightarrow C'\) by setting, for each \(x'\in E'\) and any of its preimages \(x\in E\), \(\delta(x')=\pi(d_{E/A}x\otimes 1)\). This definition is independent of the choice of preimage, because the difference of two preimages is an element of \(K\), say \(k\), and \(d_{E/A}k\otimes 1\in N'\). Then \(\delta\) is \(A\)-linear, and for any \(x',y'\in E'\) and their preimages \(x,y\),

\[\delta(x'y')=\pi(d_{E/A}(xy)\otimes 1)=\pi((x d_{E/A}y+y d_{E/A}x)\otimes 1)=x'\delta(y')+y'\delta(x')\]

so it is an \(A\)-derivation. Here, the last equality follows from the fact that on \(\Omega_{E/A}\otimes_EE'\), the action of \(E\) coincides, via \(\varphi\), with the action of \(E'\). Then by the universal property of Lemma 2, the condition \(\psi\circ d_{E'/A}=\delta\) uniquely determines an \(E'\)-linear map \(\psi:\Omega_{E'/A} \rightarrow C'\). On the generators, the two compositions \(\psi\circ\psi'\) and \(\psi'\circ\psi\) satisfy

\[\pi(d_{E/A}x\otimes 1)\overset{\psi'}{\longmapsto}d_{E'/A}\varphi(x)\overset{\psi}{\longmapsto}\pi(d_{E/A}x\otimes 1),\qquad d_{E'/A}x'\overset{\psi}{\longmapsto}\pi(d_{E/A}x\otimes 1)\overset{\psi'}{\longmapsto}d_{E'/A}x'\]

and hence both are the identity. That is, \(\psi'\) is an isomorphism, and therefore \(\ker\Omega_{\varphi/A}'=N'=\im\bar{d}\).

This exact sequence is also called the conormal sequence.

Naive cotangent complex

The conormal sequence of Proposition 9 is exact only at the right end, and in general \(\bar{d}\) is not injective. A natural question is what meaning the kernel of \(\bar{d}\) has, and whether this sequence can be extended to the left. In this section, we examine the naive cotangent complex, which is a first answer to this question.

The key idea is to express \(E\) as a quotient of the most tractable algebra, namely a polynomial algebra. If we choose any spanning set of \(E\), say \((t_s)_{s\in S}\) (for instance, all of \(E\)), then by the adjunction of [Algebraic Structures] §Algebras, ⁋Proposition 8, defined by \(\x_s\mapsto t_s\), a surjective \(A\)-algebra homomorphism

\[p: R=A[\x_s]_{s\in S}\longrightarrow E\]

exists. We call such a \(p\) a presentation of \(E\), and write \(\mathfrak{I}=\ker p\). Then we can apply Proposition 9 with \(\varphi=p\), that is, setting \(E\) to be \(R\) and \(E'\) to be \(E\), and the two leftmost terms of its conormal sequence form the following complex.

Definition 10 For a presentation \(p:R \rightarrow E\), the naive cotangent complex \(\operatorname{NL}(p)\) of \(p\) is the two-term complex of \(E\)-modules consisting of the two leftmost terms of the conormal sequence:

\[\operatorname{NL}(p)=\Bigl[\mathfrak{I}/\mathfrak{I}^2\overset{\bar{d}}{\longrightarrow}\Omega_{R/A}\otimes_RE\Bigr]\]

Here, we place \(\mathfrak{I}/\mathfrak{I}^2\) in degree \(1\) and \(\Omega_{R/A}\otimes_RE\) in degree \(0\).

That is, \(\operatorname{NL}(p)\) is a chain complex with only two nonzero terms ([Homological Algebra] §Homology), and its homology consists of

\[H_0\bigl(\operatorname{NL}(p)\bigr)=\coker\bar{d},\qquad H_1\bigl(\operatorname{NL}(p)\bigr)=\ker\bar{d}\]

alone. Since \(\Omega_{R/A}\) is a free \(R\)-module with basis the \(\dd{\x_s}\) by Proposition 5, the degree \(0\) term of \(\operatorname{NL}(p)\) is a free \(E\)-module.

Proposition 11 For any presentation \(p:R \rightarrow E\), a canonical isomorphism

\[H_0\bigl(\operatorname{NL}(p)\bigr)\cong\Omega_{E/A}\]

exists.

Proof

In the conormal sequence

\[\mathfrak{I}/\mathfrak{I}^2\overset{\bar{d}}{\longrightarrow}\Omega_{R/A}\otimes_RE\overset{\Omega_{p/A}'}{\longrightarrow}\Omega_{E/A}\longrightarrow0\]

of Proposition 9, since \(\Omega_{p/A}'\) is surjective and its kernel is \(\im\bar{d}\), we have \(\coker\bar{d}\cong\Omega_{E/A}\) by the first isomorphism theorem.

In other words, \(H_0\) does not depend on the presentation and always recovers the module of Kähler differentials. Therefore, the new information is contained in \(H_1(\operatorname{NL}(p))\), which measures the failure of exactness at the left end of the conormal sequence. Of course, for this discussion to be meaningful, \(H_1\) must also not depend on the choice of presentation.

Independence of the Choice of Presentation

To compare two presentations, let us first observe the following. Suppose we are given two presentations \(p:R=A[\x_s]_{s\in S} \rightarrow E\) and \(p':R'=A[\y_t]_{t\in T} \rightarrow E\). For each \(s\in S\), since \(p'\) is surjective, we can choose \(g_s\in R'\) such that \(p'(g_s)=p(\x_s)\), and \(\x_s\mapsto g_s\) induces an \(A\)-algebra homomorphism \(\varphi:R \rightarrow R'\) satisfying \(p'\circ\varphi=p\). Then since \(\varphi(\mathfrak{I})\subseteq\mathfrak{I}'\), \(\varphi\) induces a morphism

\[\operatorname{NL}(\varphi):\operatorname{NL}(p) \rightarrow \operatorname{NL}(p');\qquad \overline{f}\mapsto\overline{\varphi(f)},\quad \dd{\x_s}\otimes1\mapsto \dd{\varphi}(\x_s)\otimes1\]

between the two complexes. That the square commutes is verified from the identity \(\dd{\varphi}(f)=\sum_s\varphi(\partial f/\partial\x_s)\dd{\varphi}(\x_s)\) given by Proposition 5.

Lemma 12 In the situation above, suppose we are given two \(A\)-algebra homomorphisms \(\varphi,\psi:R \rightarrow R'\) satisfying \(p'\circ\varphi=p=p'\circ\psi\). Then for any \(f\in R\), the congruence

\[\varphi(f)-\psi(f)\equiv\sum_{s\in S}\varphi\left(\frac{\partial f}{\partial \x_s}\right)\bigl(\varphi(\x_s)-\psi(\x_s)\bigr)\pmod{\mathfrak{I}'^2}\]

holds.

Proof

First, since \(p'(\varphi(f)-\psi(f))=p(f)-p(f)=0\), we have \(\varphi(f)-\psi(f)\in\mathfrak{I}'\), and for the same reason both sides of the above congruence are elements of \(\mathfrak{I}'\), so the expression makes sense. Let \(P\) be the set of \(f\) satisfying the given congruence. It is clear that \(P\) contains the elements of \(A\) (both sides are \(0\)), contains each variable \(\x_s\) (\(\partial\x_s/\partial\x_{s'}\) is \(1\) when \(s=s'\) and \(0\) otherwise), and is closed under \(A\)-linear combinations. Therefore, if we show that \(P\) is closed under multiplication, then \(P=R\).

Let \(f,g\in P\), and for convenience let us write \(\alpha=\varphi(f)-\psi(f)\) and \(\beta=\varphi(g)-\psi(g)\in\mathfrak{I}'\). Then

\[\varphi(fg)-\psi(fg)=\varphi(f)\varphi(g)-\psi(f)\psi(g)=\psi(f)\beta+\varphi(g)\alpha=\varphi(f)\beta+\varphi(g)\alpha-\alpha\beta\]

where the last equality is rewritten using \(\psi(f)\beta=\varphi(f)\beta-\alpha\beta\). Since \(\alpha\beta\in\mathfrak{I}'^2\), we have

\[\varphi(fg)-\psi(fg)\equiv\varphi(f)\beta+\varphi(g)\alpha\pmod{\mathfrak{I}'^2}\]

Now, substituting the assumption that \(f,g\in P\) into \(\alpha,\beta\) and using the Leibniz rule \(\partial(fg)/\partial\x_s=f(\partial g/\partial\x_s)+g(\partial f/\partial\x_s)\), we obtain \(fg\in P\) from the fact that products of elements of \(\mathfrak{I}'\) vanish in \(\mathfrak{I}'^2\).

Proposition 13 In the situation of Lemma 12, the two morphisms \(\operatorname{NL}(\varphi),\operatorname{NL}(\psi):\operatorname{NL}(p) \rightarrow \operatorname{NL}(p')\) are chain homotopic. In particular, the morphisms they induce on homology coincide.

Proof

Since \(\Omega_{R/A}\otimes_RE\) is a free \(E\)-module with basis \(\dd{\x_s}\otimes1\), the \(E\)-linear map

\[h:\Omega_{R/A}\otimes_RE \rightarrow \mathfrak{I}'/\mathfrak{I}'^2;\qquad \dd{\x_s}\otimes1\mapsto\overline{\varphi(\x_s)-\psi(\x_s)}\]

is well-defined. Here, the fact that \(\varphi(\x_s)-\psi(\x_s)\in\mathfrak{I}'\) was seen in the proof of Lemma 12, and since \(\mathfrak{I}'\) annihilates \(\mathfrak{I}'/\mathfrak{I}'^2\), \(\mathfrak{I}'/\mathfrak{I}'^2\) is an \(E=R'/\mathfrak{I}'\)-module. We now show that \(h\) is a chain homotopy for \(\operatorname{NL}(\varphi)-\operatorname{NL}(\psi)\).

First, in degree \(0\), for the generator \(\dd{\x_s}\otimes 1\), we have

\[\bar{d}'\bigl(h(\dd{\x_s}\otimes1)\bigr)=d\bigl(\varphi(\x_s)-\psi(\x_s)\bigr)\otimes1=\bigl(\operatorname{NL}(\varphi)-\operatorname{NL}(\psi)\bigr)(\dd{\x_s}\otimes1)\]

Next, in degree \(1\), for any \(f\in\mathfrak{I}\), since \(\dd{f}=\sum_s(\partial f/\partial\x_s)\dd{\x_s}\), we have

\[h\bigl(\bar{d}(\overline{f})\bigr)=\sum_sp\left(\frac{\partial f}{\partial\x_s}\right)\cdot\overline{\varphi(\x_s)-\psi(\x_s)}=\overline{\sum_s\varphi\left(\frac{\partial f}{\partial\x_s}\right)\bigl(\varphi(\x_s)-\psi(\x_s)\bigr)}\]

The last equality holds because the action of \(E=R'/\mathfrak{I}'\) on \(\mathfrak{I}'/\mathfrak{I}'^2\) is given via \(p'\), and \(p'\circ\varphi=p\). Then by Lemma 12, this is equal to \(\overline{\varphi(f)-\psi(f)}=\bigl(\operatorname{NL}(\varphi)-\operatorname{NL}(\psi)\bigr)(\overline{f})\). Therefore \(h\) is a chain homotopy, and the fact that chain homotopic morphisms induce the same morphism on homology was seen in [Homological Algebra] §Homology.

Theorem 14 For two presentations \(p:R \rightarrow E\) and \(p':R' \rightarrow E\), the complexes \(\operatorname{NL}(p)\) and \(\operatorname{NL}(p')\) are homotopy equivalent. In particular, there exists a canonical isomorphism

\[H_i\bigl(\operatorname{NL}(p)\bigr)\cong H_i\bigl(\operatorname{NL}(p')\bigr),\qquad i=0,1\]

Proof

As seen above, choose \(\varphi:R \rightarrow R'\) such that \(p'\circ\varphi=p\) and \(\varphi':R' \rightarrow R\) such that \(p\circ\varphi'=p'\). Then since both \(\varphi'\circ\varphi\) and \(\id_R\) satisfy \(p\circ(\varphi'\circ\varphi)=p=p\circ\id_R\), by Proposition 13 we have

\[\operatorname{NL}(\varphi')\circ\operatorname{NL}(\varphi)=\operatorname{NL}(\varphi'\circ\varphi)\simeq\operatorname{NL}(\id_R)=\id_{\operatorname{NL}(p)}\]

and symmetrically, \(\operatorname{NL}(\varphi)\circ\operatorname{NL}(\varphi')\simeq\id_{\operatorname{NL}(p')}\). That is, \(\operatorname{NL}(\varphi)\) is a homotopy equivalence and induces an isomorphism on homology. This isomorphism is canonical because any other choice of \(\varphi\) induces the same morphism on homology by Proposition 13.

Therefore, we may forget the choice of presentation and write \(\operatorname{NL}_{E/A}\), under the proviso of disregarding homotopy equivalence. Its homology

\[H_0(\operatorname{NL}_{E/A})\cong\Omega_{E/A},\qquad H_1(\operatorname{NL}_{E/A})\]

is an invariant of the \(A\)-algebra \(E\). Let us see what \(H_1\) measures in the following example.

Example 15 Consider the algebra \(E=A[\x]/(f)\) presented by a single polynomial. Here, we assume that \(f\) is a nonzerodivisor in \(A[\x]\). For the presentation \(p:R=A[\x] \rightarrow E\), we have \(\mathfrak{I}=(f)\), and since \(f\) is a nonzerodivisor, \(\mathfrak{I}/\mathfrak{I}^2\) is a free \(E\)-module with basis \(\overline{f}\). Indeed, if \(af\in(f^2)\), then from \(af=bf^2\) we obtain \(a=bf\in(f)\). Therefore,

\[\operatorname{NL}_{E/A}=\Bigl[E\overline{f}\overset{\bar{d}}{\longrightarrow}E\dd{\x}\Bigr],\qquad \bar{d}(\overline{f})=\overline{Df}\dd{\x}\]

where \(Df\) is the derivative of \(f\). Then

\[H_0(\operatorname{NL}_{E/A})=\Omega_{E/A}\cong E/(\overline{Df}),\qquad H_1(\operatorname{NL}_{E/A})\cong\ann_E(\overline{Df})\]

For instance, if \(A=\mathbb{K}\) is a field of characteristic not \(2\) and \(f=\x^2\), then \(\overline{Df}=2\overline{\x}\), so \(H_1\cong\ann_E(2\overline{\x})=(\overline{\x})\neq0\). On the other hand, when the ideal generated by \(f\) and \(Df\) is all of \(A[\x]\), that is, when there exist \(u,v\in A[\x]\) such that \(uf+vDf=1\), then \(\overline{Df}\) is a unit in \(E\), so \(H_0=H_1=0\). If \(A=\mathbb{K}\) is a field, then since \(A[\x]\) is a PID, this condition is equivalent to \(f\) being a separable polynomial. That is, \(H_1\) contains information about the multiplicity of the relations in the presentation, which is not visible from \(\Omega_{E/A}\) alone.

Remark 16 The naive cotangent complex is, just as its name suggests, a shadow of a more refined object. Instead of a single presentation by a polynomial algebra, Quillen and André used simplicial resolutions to define the cotangent complex \(L_{E/A}\), which has homology in all degrees. The degree \(0,1\) part of this complex is precisely \(\operatorname{NL}_{E/A}\) defined above. Since such homotopy-theoretic constructions are beyond the scope of this category, here we content ourselves with the two-term truncation \(\operatorname{NL}_{E/A}\).

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References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.

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