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Cohomologies.

Yoshifumi Tsuchimoto

\fbox{07. Ext as a derived functor} We recommend the book of Lang [1] as a good reference. The treatment here follows the book for the most part.

THEOREM 07.1   Let $ \mathcal{C}_1$ be an abelian category with enough injectives, and let $ F:\mathcal{C}_1\to \mathcal{C}_2$ be a covariant additive left functor to another abelian category $ \mathcal{C}_2$ . Then:
  1. $ F\cong R^0 F$ .
  2. For each short exact sequence

    $\displaystyle 0\to M'\to M \to M''\to 0
$

    and for each % latex2html id marker 721
$ n\geq 0$ there is a natural homomorphism

    $\displaystyle \delta^n: R^n F(M'')\to R^{n+1} F(M)
$

    such that we obtain a long exact sequence

    $\displaystyle \dots\to R^n F(M')\to R^n F(M) \to R^n F(M'')\overset{\delta^n}{\to}
R^{n+1} F(M')\to \dots.
$

  3. $ \delta$ is natural. That means, for a morphism of short exact sequences

    $\displaystyle \begin{CD}
0 @»> M'@»> M@»> M''@»> 0\\
@. @VVV @VVV @VVV \\
0@»> N'@»> N@»> N''@»> 0
\end{CD}$

    the $ \delta$ 's give a commutative diagram:

    $\displaystyle \begin{CD}
R^n F(M'') @> \delta^n » R^{n+1}F(M') \\
@VVV @VVV \\
R^nF(N'') @> \delta^n » R^{n+1} F(N')
\end{CD}$

  4. For each injective objective object $ I$ of $ A$ and for each $ n>0$ we have $ R^n F(I)$ .

The collection $ \{R^jF\}$ of functors $ R^jF$ is a ``universal delta functor''. See [1].

LEMMA 07.2   Under the assumption of the previous theorem, for any exact sequence $ 0 \to M'\to M\to M''\to 0$ of objects in $ \mathcal{C}_1$ , there exists injective resolutions $ I_{M'},I_{M},I_{M''}$ of $ M',M,M''$ respectively and a commutative diagram

$\displaystyle \begin{CD}
@. 0 @. 0 @. 0@.\\
@. @VVV @VVV @VVV \\
0 @»> M'@»...
...@»> 0\\
@. @VVV @VVV @VVV \\
0@»> I_{M'}@»> I_M@»> I_{M''}@»> 0
\end{CD}$

such that the diagram of resolutions is exact. Thus we obtain a diagram

$\displaystyle \begin{CD}
@. 0 @. 0 @. 0@.\\
@. @VVV @VVV @VVV \\
0 @»> F(M')...
...V @VVV \\
0@»> F(I_{M'})@>\alpha» F(I_M)@>\beta » F(I_{M''})@»> 0
\end{CD}$

such that each row in the last line is exact.

Note that $ j$ -th cohomology of the complex $ F(I_M)$ (respectively, $ F(I_{M'}), F(I_{M''})$ ) gives the $ R^jF(M) (respectively, (R^jF(M'), R^jF(M'')$ .) Using the resolution given in the lemma above, we may prove Theorem 7.1. Let us describe the map $ \delta$ in more detail when $ \mathcal{C}_2$ is a category of modules by ``diagram chasing''. Namely, for $ x \in R^n (M'') $ , let us show how to compute $ \delta(x)$ .

  1. $ x \in R^n (M'') $ may be represented as a class $ [c_x]$ of a cocycle $ c_x \in \operatorname{Ker}(d: F(I_{M''}^n) \to F(I_{M''}^{n+1}))$ .
  2. We take a ``lift'' $ \tilde c_x \in F(I^n_{M})$ such that $ \beta^n(\tilde c_x)=c_x$ . Note that $ \tilde c_x$ is no longer a cocycle.
  3. Consider $ e_x=d \tilde c_x \in F(I^{n+1} M)$ . It is a coboundary and we have $ \beta(e_x)=0$ .
  4. There thus exists an element $ a_x \in F(I^n_{M'})$ such that $ \alpha(a_x)= e_x$ . $ a_x$ is no longer a coboundary but it is a cocycle.
  5. The cohomology class $ [a_x]$ of $ a_x$ is the required $ \delta(x)$ .

Such computation appears frequently when we deal with cohomologies.

DEFINITION 07.3   Let $ A$ be a ring. Let $ M,N$ be $ A$ -modules. Then an extension of $ N$ by $ M$ is a module $ L$ with a exact sequence

(E) $\displaystyle 0 \to N \overset{\alpha}\to L \overset{\beta}\to M \to 0.$

of $ A$ -modules. Let

$\displaystyle 0 \to N \overset{\alpha'}\to L' \overset{\beta'}\to M \to 0.
$

be another extension. Then the two extensions are said to be isomorphic if there exists a commutative diagram

$\displaystyle \begin{CD}
0 @»> N @>\alpha» L @>\beta» M @»> 0\\
@. @V=VV @VVV @V=VV @. \\
0 @»> N @>\alpha'» L' @>\beta'» M @»> 0.\\
\end{CD}$

PROPOSITION 07.4   There exists a bijection between the isomorphism classs of the extensions and elements of the $ \operatorname{Ext}_A^1(M,N)$ . The bijection is given by corresponding the extension ($ E$ ) to the class $ \delta (1_N )\in \operatorname{Ext}^1(M,N)$ of the identity map $ 1_N$ by $ \delta$ associated to the exact sequence $ (E)$ .

See [1, XX,Exercise 27]



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2010-06-15