Differentials

edit_note

This post is being revised. The text below is the version as of 2026-06-10, so some statements may be out of date and references to this post from other posts may not line up.

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

The goal of this post is to define differentials algebraically.

Kähler Differential Modules

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

\[d(xy)=y\,dx+x\,dy\]

for all \(x,y\in E\) \(A\)-derivations, and denote the set of all such maps by \(\Der_A(E,M)\).

One of the basic properties of derivations is that \(\Der_A(E,M)\) carries an \(E\)-module structure: for any \(x\in E\) and \(d\in \Der_A(E, M)\), defining the \(A\)-linear map \(x d\) by the formula

\[xd: E \rightarrow M;\qquad y\mapsto x\,d(y)\]

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

\[(xd)(y_1y_2)=x\,d(y_1y_2)=x\, (y_1\,dy_2+y_2\,dy_1)=y_1(xd)(y_2)+y_2(xd)(y_1).\]

Moreover, 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 again an \(A\)-derivation, as can be seen from the following equation:

\[(u\circ d)(y_1y_2)=u(y_1\,dy_2+y_2\,dy_1)=y_1u(dy_2)+y_2u(dy_1)=y_1(u\circ d)(y_2)+y_2(u\circ d)(y_1).\]

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

Lemma 2 The functor \(\Der_A(E, -)\) is representable. That is, there exists an \(E\)-module \(\Omega_{E/A}\) such that there is a natural isomorphism between the two functors from \(\lMod{E}\) to itself:

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

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 \(E\)-module generated by \(\{df\mid f\in E\}\) subject to the following relations

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

and is denoted by \(\Omega_{E/A}\). The \(A\)-linear derivation \(d:E \rightarrow \Omega_{E/A}\) defined by \(f\mapsto df\) is called the universal \(A\)-derivation.

Then one can easily verify that \(\Omega_{E/A}\) satisfies the desired universal property (Lemma 2). Moreover, thinking of \(\Omega_{E/A}\) as a functor taking an \(A\)-algebra \(A \rightarrow E\) and producing \(\Omega_{E/A}\), the following functoriality also holds.

Proposition 4 Suppose we are given the commutative diagram of ring homomorphisms

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

commute.

Proof

Since \(d_{E'/A'}\circ \phi\) is an \(A\)-derivation, this is immediate from Lemma 2.

On the other hand, 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 the map \(\Omega_{E/A} \rightarrow \Omega_{E'/A'}\) obtained from Proposition 4 above is, strictly speaking, a map \(\Omega_{E/A} \rightarrow \varphi^\ast\Omega_{E'/A'}\), and thus there exists a unique corresponding \(E'\)-linear homomorphism

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

Fundamental sequences

In particular, let us take \(\phi:A \rightarrow A'\) to 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, viewing \(E'\) as an \(E\)-algebra via \(\varphi:E \rightarrow E'\), the Kähler differential module \(\Omega_{E'/E}\) of \(E'\) over \(E\) is defined. The universal \(E\)-derivation \(d_{E'/E}: E \rightarrow \Omega_{E'/E}\) is also an \(A\)-derivation, so by Lemma 2 again it coincides with the following:

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

Proposition 5 (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

Another important exact sequence is obtained in the special case where \(\varphi:E \rightarrow E'\) is surjective. In this case, the first isomorphism theorem yields

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

For convenience, write \(K=\ker\varphi\). Then, considering the restriction of \(d_{E/A}:E \rightarrow \Omega_{E/A}\) to \(K\), denoted \(d_{E/A}\vert_K: K \rightarrow \Omega_{E/A}\), we may consider the following \(E\)-linear map

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

One can verify that the kernel of the above composition contains \(K^2\), and thus we obtain an \(E\)-linear map

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

Proposition 6 In the above situation, the following sequence

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

is exact.