first properties of 1-forms

PROPOSITION 9.20   Let $p:X\to S$ and $q:Y\to S$ be separated morphisms. Let $f:X\to Y$ be a separated morphism of schemes such that $q\circ f=p.$ Then we have a following exact sequence.

$$f^*\Omega^1_{Y/S} \to \Omega^1_{X/S} \to \Omega^1_{X/Y} \to 0$$

PROOF.. Since the question is local on $X,Y,S$ , we may assume that these schemes are affine. Then the claim is deduced by the following arguments (Corollary 9.22), and a Yoneda-type argument. $\qedsymbol$

LEMMA 9.21   Let $A,B,C$ be (unital commutative associative) rings. Assume we are given homomorphisms $A\to B \to C$ of rings. Then for any $A$ -module $M$ , we have an exact sequence

$$*_M 0 \to \operatorname{Der}_{A/B}(A,M)\overset{\alpha_M}{\to} \operatorname{Der}_{A/C}(A,M) \overset{\beta_M}{\to}\operatorname{Der}_{B/C}(B,M)$$

The sequence is natural in the sense that if we have another $A$ -module $N$ and an $A$ -module homomorphism $\varphi:M\to N$ , then we have a commutative diagram.

$$\begin{CD}0 @»> \operatorname{Der}_{A/B}(A,M)@>\alpha_M» \opera... ...peratorname{Der}_{A/C}(A,N)@>\beta_N»\operatorname{Der}_{B/C}(B,N) \end{CD}$$

PROOF.. Any $A$ -derivation $D:A\to M$ over $B$ may be regarded as a derivation over $C$ which we denote by $\alpha_M(D)$ .

Any $A$ -derivation $D:A\to M$ over $C$ defines by restriction a $B$ -derivation $B\to M$ over $C$ which we denote by $\beta_M(D)$ .

The rest is easy observation. $\qedsymbol$

With the help of universality of $d$ , we obtain the following corollary.

COROLLARY 9.22   Let $A,B,C$ be (unital commutative associative) rings. Assume we are given homomorphisms $A\to B \to C$ of rings. Then for any $A$ -module $M$ , we have an exact sequence

$$**_M 0 \to \operatorname{Hom}_{A}(\Omega^1_{B/A},M) \overset{\alpha_M}... ...A/C},M) \overset{\beta_M}{\to}\operatorname{Hom}_A (A\otimes_B\Omega^1_{B/C},M)$$

It is natural in a sense similar to the Lemma above. $\qedsymbol$

ARRAY(0x9294884)

PROPOSITION 9.23 (A Yoneda type argument)   Let $A$ be a commutative associative ring.

1. Let $M_1,M_2$ be $A$ -modules. Assume for each $A$ -module $M$ , we are given a homomorphism
   $$\alpha_M: \operatorname{Hom}_A(M_1,M)\to \operatorname{Hom}_A(M_2,M).$$
   We assume that the assignment $M\mapsto \alpha_M$ is natural. That means, for any $A$ -modules $M,N$ and for any $A$ -module homomorphism $\phi:M\to N$ , we have the following commutative diagram.
   $$\begin{CD} \operatorname{Hom}_A(M_1,M)@>\alpha_M» \operatorname{... ...\operatorname{Hom}_A(M_1,N)@>\alpha_N» \operatorname{Hom}_A(M_2,N)\\ \end{CD}$$
   In other words, we have
   $$\phi\circ \alpha_M(\psi) = \alpha_N(\phi\circ\psi)$$
   for any $\psi \in \operatorname{Hom}_A(M_1,M)$ . Then $\alpha$ is ``representable''. That means, there exists a unique $f_\alpha\in \operatorname{Hom}_A(M_2,M_1)$ such that
   $$\alpha_M(\psi)= \psi\circ f_\alpha$$
   holds for all $\psi \in \operatorname{Hom}_A(M_1,M)$ .
2. Let $M_1,M_2,M_3$ be $A$ -modules. Assume for each $A$ -module $M$ , we are given homomorphisms

   $$\alpha_M: \operatorname{Hom}_A(M_1,M)\to \operatorname{Hom}_A(M_2,M),$$

   $$\beta_M: \operatorname{Hom}_A(M_2,M)\to \operatorname{Hom}_A(M_3,M).$$

   
   

   We assume that the assignments $\alpha,\beta$ is natural. Assume furthermore that for any $A$ -module $M$ , a sequence

   $$***_M 0\to \operatorname{Hom}_A(M_1,M) \overset{\alpha_M}{\to} \operatorname{Hom}_A(M_2,M) \overset{\beta_M}{\to} \operatorname{Hom}_A(M_3,M)$$

   
   
   is always exact. Then the corresponding sequence
   $$M_3 \overset{f_\beta}{\to} M_2 \overset{f_\alpha}{\to} M_1 \to 0$$
   (which arises due to the claim above) is also exact.

PROOF.. (1) Put

$$f_\alpha=\alpha_{M_1}(\operatorname{id}_{M_1}).$$

Then for any $\psi \in \operatorname{Hom}_A(M_1,M)$ , we have

$$\alpha_M(\psi) =\alpha_M(\psi\circ \operatorname{id}_{M_1}) =\psi\circ \alpha_{M_1}(\operatorname{id}_{M_1}) =\psi \circ f_\alpha.$$

(2) Since $\beta_M \circ \alpha_M=0$ , we deduce that $f_\alpha \circ f_\beta=0$ using the uniqueness of the homomorphism which represents $\beta\circ \alpha$ .

For surjectivity of $f_\alpha$ , we use the sequence $(***_M)$ for $M=M_1/f_\alpha(M_2)$ . For the exactness at the middle term, we use the sequence $(***_M)$ for $M=M_2/f_\beta(M_3)$ . We leave the detail as an easy exercise. $\qedsymbol$