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I want to make sure my formulation of Kähler differentials is correct. Here it is. Please let me know if you see something that is incorrect.

Let $A$ is a $k$-algebra via a ring homomorphism $f:k\to A$. A $k$-\textit{derivation}, or a derivation over $k$ is a map $D:A\to M$ such that $D(a+b)=Da+Db$, $D(ab)=aDb+bDa$, and $D\circ f=0$. The set of all $k$-derivations of $A$ into $M$ is written as ${\rm Der}_k(A,M)$.

Proposition: Let $k$ be ring and $A$ an $k$-algebra via homomorphism $f:k\to A$. Then there exists an $A$-module $M_0$ and a derivation $d\in {\rm Der}_k(A,M_0)$ such that for any $A$-module $M$ and any $D\in {\rm Der}_k(A,M)$, there is unique $A$-module homomorphism $f:M_0\to M$ such that $D=f\circ d$.

Proof: Define $\mu:A\otimes_k A\to A$ by $\mu(x\otimes y)=xy$. Note that $\mu$ is a homomorphim of $k$-algebras. Set $I={\rm Ker\;}\mu$, $\Omega_{A/k}=I/I^2$, and $B=A\otimes_k A/I^2$. Note $\mu$ induces $\mu':B\to A$ with $\mu'(x\otimes y {\rm \;mod\;}I^2)=xy$. Define $k$-algebra homomorphisms $\lambda_i:A\to B$ for $i=1,2$ by $\lambda_1(a)=a\otimes 1 {\rm \;mod\; }I^2$ and $\lambda_2(a)=1\otimes a{\rm\;mod\;}I^2$ Observe that the following diagram of $k$-algebras commutes for both $i=1,2$

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Define $d:=\lambda_2-\lambda_1$. Since $\mu'\circ\lambda_1=\mu'\circ\lambda_2$, ${\rm Im\;}d={\rm Ker\;}\mu'=\Omega_{A/k}$. Therefore, we see that $d$ is map from $A$ to $\Omega_{A/k}$. $\Omega_{A/k}$ can be seen as an $A$-module via $a\cdot m=\lambda_i(a)\cdot m$, and it is easy to see from here that $d$ is $k$-derivation from $A$ to $\Omega_{A/k}$. \

Now we prove that the pair $(\Omega_{A/k},d)$ satisfy the requirements of $(M_0,d)$. Given an $A$-module $M$, and $D\in {\rm Der}_k(A,M)$, define the $A$-module homomorphism $f:\Omega_{A/k}\to M$ with $f(x\oplus y)= xDy$. Notice that for any $a\in A$ $$ f(da)=f(1\otimes a {\rm \;mod\; }I^2 - a\otimes 1 {\rm \;mod\; }I^2)=Da - a\cdot D(1)= Da\;\;\; $$ so $D=f\circ d$. \qed

My main concern is the construction of $f$. It is usually constructed in a longer way but I thought I could make it more concise.

Bernard
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coolpenguin
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  • The map $f$ should satisfy $f \circ d = D$ and hence your formula $f(da):=D(a)$ is correct. The problem is to prove that this is a well defined $A$-linear map $f: \Omega^1_A \rightarrow M$. You must convince yourself that you have given such a proof. – hm2020 Oct 13 '21 at 09:43
  • any element $\omega \in I \subseteq A\otimes_k A$ is on the form $\omega=\sum_i x_id(y_i)$ and you must define $f(\overline{\omega}):=\sum_i x_i D(y_i)$for the diagram to commute. Then you must prove that this is well defined: if $\omega-\omega' \in I^2$ you should get $f(\overline{\omega})=f(\overline{\omega'})$. – hm2020 Oct 15 '21 at 10:55
  • The essence of the exercise is to prove that the map $f$ defined above is well defined. – hm2020 Oct 15 '21 at 10:58

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