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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.

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

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.

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

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

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

    $\displaystyle \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.

    $\displaystyle \begin{CD}
\operatorname{Hom}_A(M_1,M)@>\alpha_M» \operatorname{...
...\operatorname{Hom}_A(M_1,N)@>\alpha_N» \operatorname{Hom}_A(M_2,N)\\
\end{CD}$

    In other words, we have

    $\displaystyle \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

    $\displaystyle \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

          $\displaystyle \alpha_M: \operatorname{Hom}_A(M_1,M)\to \operatorname{Hom}_A(M_2,M),$
          $\displaystyle \beta_M: \operatorname{Hom}_A(M_2,M)\to \operatorname{Hom}_A(M_3,M).$

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

    ($ ***_M$ ) $\displaystyle 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

    $\displaystyle 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

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

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

$\displaystyle \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$


next up previous
Next: Étale morphism Up: The sheaf of differential Previous: The sheaf of differential
2007-12-11