How to write long exact sequence as short exact sequence of complex?

Question

Let $Ch(A)$ denote the category of complexes in an abelian category $A$.

I was doing Exercise 3.1.1 of Weibel's homological algebra:

Suppose $B$ is a $R$-module,${_r}R=\{s\in R:rs=0\}$,${_r}B=\{x\in B:rx=0\}$,we have a non-projective resolution: $0\rightarrow {_r}R\rightarrow R\overset{r}{\rightarrow}R\rightarrow R/rR\rightarrow 0$ .Show that there is a short exact sequence: $0\rightarrow Tor_2^R(R/rR,B)\rightarrow{_rR\otimes B}\overset{multiply}{\rightarrow}{_rB}\rightarrow Tor_1^R(R/rR,B)\rightarrow 0$.

In fact I solved this Exercise by dimension shifting.but someone gave me the following interesting hints to yield the desired short exact sequence with little explanation:

from $0\rightarrow{_rR}\rightarrow [R\overset{r}{\rightarrow}R]\rightarrow R/{_r}R[-1]\overset{-1}{\rightarrow} 0$ we have: $Tor_1(R/rR[-1],B)\rightarrow {_r}R\otimes B\rightarrow Tor_0([R\overset{r}{\rightarrow}R],B)\rightarrow Tor_0(R/rR[-1],B)\rightarrow 0$.

notice that

$Tor_1(R/rR[-1],B)=Tor_2(R/rR,B),Tor_0([R\overset{r}{\rightarrow}R],B)={_rB},Tor_0(R/rR[-1],B)=Tor_1(R/rR,B)$

from which the conclusion follows.

It seems he write a long exact sequence of $R$-module into a short exact sequence in $Ch(R)$ and directly apply the derived functor $Tor_i(*,B)$.

I have the following questions:

• Here,what actually are those complexes used?I think the three complexes are $0\rightarrow {_rR}\overset{0}{\rightarrow} rR\rightarrow 0$,$0\rightarrow R\overset{r}{\rightarrow} R\rightarrow 0$,$0\rightarrow rR\overset{0}{\rightarrow} R/rR\rightarrow 0$ .Is this correct?

• It seems possible to turn long exact sequence $0\rightarrow A\rightarrow B\rightarrow...\rightarrow C\rightarrow 0$ into a short exact sequence of complex in a similar manner.Is this correct?

• Is applying $Tor_i(*,B)$ or more generally any left derived functor $L_iF$($F$ is a right exact functor on $R$-module) to a complex $C_\cdot$in $Ch(R)$ means applying $L_iF$ term by term to all $C_n$?most importantly,Does any short exact sequence of complexes $0\rightarrow A.\rightarrow B.\rightarrow C.\rightarrow 0$ give rise to a long exact sequence,just like in the case of $R$-module?
• Could you recommend any references about those things?I didn't see any trick like this by now in Weibel's book,but for me it's useful and elegant,I really want to study this.

Any help would be appreciated.

Answer 1

2) I am not sure I understand the statement. If $0\to A_n\to A_{n-1}\to...\to A_0\to 0$ is a long exact sequence, you can always look at (for example) the short exact sequence of complexes $[0\to A_{n-1}\to...\to A_0]\to [A_n\to A_{n-1}\to...\to A_0]\to [A_n\to 0\to...\to 0]$. In this case, this is just a part of the stupid filtration. But this is far from being the only one...

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1) No the complex used are $A_.=[0\rightarrow {_rR}\to 0 \to 0]$ (with ${_rR}$ in degree $0$), $B_.=[0\to R\overset{r}\to R\to 0]$ and $C_.=[0\to 0\to R/rR\to 0]$ with $R/rR$ in degree $1$. This give the diagram :

$$\require{AMScd} \begin{CD} {_r}R@>>>R@>>>0@.\text{degree 0}\\ @VVV@VVV@VVV\\ 0@>>>R@>>>R/rR@.\text{ degree -1} \end{CD} $$

This is not an exact sequence of complexes, but almost : the mapping cone of the map $\varphi:{_rR}\to [R\to R]$ is the complex $[{_rR}\to R\to R]$ and which is quasi-isomorphic to $R/rR[-1]$.

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3) A short exact sequence $A_.\overset\varphi\rightarrow B_.\rightarrow C_.$ such that $C_.$ is quasi-isomorphic to $C(\varphi)_.$ is called a distinguished triangle. They are usually written $$A_.\longrightarrow B_.\longrightarrow C_.\overset{+1}\longrightarrow $$ They have some very nice properties with respect to derived functors. These are best written in the language of derived categories (chapter 10 of Weibel's book, so you may not be there yet...)

These have the property that their long sequence in homology is exact. (Here this is a rather trivial statement). But there is more to it : this statement is preserved by applying a derived functor, though we need to be careful by what this means.

If $F$ is a right exact functor, the total derived functor of $F$ is a "functor" written $LF$ such that $LF(C_.)$ is given by the two step process : (here $C_.$ is a bounded above complex)

• take a quasi-isomorphism $C_.\simeq P_.$ where $P_.$ is a complex of projectives.
• Put $LF(C_.)=F(P_.)$.

You might notice that this is very closed to the definition of $L_iF(M)$. Indeed $L_iF(M)$ is actually the homology of $LF(M)$, in other words $L_iF(M)=H_i(LF(M))$.

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Let us see what this gives with the above complexes and the right exact functor $F=.\otimes B$.

• For $A_.={_rR}$. Take a projective resolution of ${_rR}$ and apply $F$. We get a complex $LF(A_.)$ such that $H_i(LF(A_.))=\operatorname{Tor}_i({_rR},B)$
• For $B_.=[R\overset{r}\to R]$. Take a quasi-isomorphism with a complex of projectives. Wait, this is already the case, so we just have to apply $F$ : $LF(B_.)=[B\overset{r}\to B]$.
• For $C_.=R/rR[-1]$. Take a quasi-isomorphism with a complex of projectives. We might as well take a projective resolution of $R/rR$ and shift it. Then apply $F$. We get a complex $LF(C_.)$ such that $H_i(LF(C_.))=\operatorname{Tor}_{i+1}(R/rR,B)$.

Now we have our sequence of complexes : $$ LF(A_.)\to LF(B_.)\to LF(C_.)$$ the result is that this sequence is again a distinguished triangle, and in particular have a long exact sequence in homology.

$$ H_1(LF(B_.))\to H_1(LF(C_.))\to H_0(LF(A_.))\to H_0(LF(B_.)\to H_0(LF(C_.))\to H_{-1}LF(A_.) $$

Using what we said previously, this exact sequence is exactly $$ 0\to\operatorname{Tor}_2(R/rR,B)\to {_rR}\otimes_R B\to {_rB}\to \operatorname{Tor}_1(R/r,B)\to 0$$

The same long exact sequence in homology but with other degrees give the rest of the exercise.

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4) Weibel book is a good source, as there are plenty of others. Maybe you are not at the relevant chapter yet. The chapter on the derived categories is really important for all this. (And maybe the chapter on spectral sequences).

Answer 2

I am going to give a self-contained solution to Exercise 3.1.1 using only results obtained up to that point in the book. All tensor products are over $R$.

We have three multiplication homomorphisms: \begin{align*} \mathrm{mult}&: R \otimes B \to B, \\ \mathrm{mult}_2&: \phantom{}_rR \otimes B \to B, \\ \mathrm{mult}_3&: rR \otimes B \to B. \end{align*} Only mult is an isomorphism. Consider the following complex of right $R$-modules in which only $R$ is projective \begin{equation*} 0 \xrightarrow{} \phantom{}_rR \xrightarrow{} R \xrightarrow{r\cdot} R \xrightarrow{} R/rR \xrightarrow{} 0 \end{equation*} We apply the functor $\bullet\otimes B$ obtaining the following complex of abelian groups \begin{equation*} 0 \xrightarrow{} \phantom{}_rR \otimes B \xrightarrow{} R \otimes B \xrightarrow{r\cdot\otimes\mathrm{Id}_B} R \otimes B \xrightarrow{} R/rR \otimes B \xrightarrow{} 0 \end{equation*} From the isomorphism $\mathrm{mult}$, this complex becomes \begin{equation*} 0 \xrightarrow{} \phantom{}_rR \otimes B \xrightarrow{\mathrm{mult}_2} B \xrightarrow{r\cdot} B \xrightarrow{} B/rB \xrightarrow{} 0 \end{equation*} By Exercise 2.4.3, $\mathrm{Tor}_2(R/rR, B) = \ker(\mathrm{mult}_2)$. We also note that $\mathrm{Im}(\mathrm{mult}_2) \subseteq \phantom{}_rB$.

We now consider the short exact sequence \begin{equation*} 0 \xrightarrow{} rR \xrightarrow{} R \xrightarrow{} R/rR \xrightarrow{} 0 \end{equation*} and its associated long exact sequence \begin{equation*} 0 \xrightarrow{} {\mathrm{Tor}_1(R/rR, B)} \xrightarrow{} rR \otimes B \xrightarrow{} R \otimes B \xrightarrow{} R/rR \otimes B \xrightarrow{} 0 \end{equation*} Again, from the isomorphism $\mathrm{mult}$, this complex becomes \begin{equation*} 0 \xrightarrow{} {\mathrm{Tor}_1(R/rR, B)} \xrightarrow{} rR \otimes B \xrightarrow{\mathrm{mult}_3} B \xrightarrow{} B/rB \xrightarrow{} 0 \end{equation*} Therefore, $\mathrm{Tor}_1(R/rR, B) = \ker(\mathrm{mult}_3)$. We now define the following map \begin{align*} f: \phantom{}_rB &\to rR \otimes B, \\ f(b) &= r\otimes b. \end{align*} It is clear that $\mathrm{mult}_3\circ f = 0$. If $\mathrm{mult}_3(\sum_{i \in I}rs_i \otimes b_i) = 0$ for some finite set $I$, $s_i \in R$ for $i \in I$ and $b_i \in B$ for $i \in I$, then $\mathrm{mult}_3(\sum_{i \in I}r \otimes s_ib_i) = 0$ and $\sum_{i \in I}rs_i \otimes b_i = f(\sum_{i \in I}s_ib_i)$. This means $\mathrm{Im}(f) = \mathrm{Tor}_1(R/rR, B)$. If for some finite set $I$, $s_i \in \phantom{}_rR$ for $i \in I$ and $b_i \in B$ for $i \in I$, then $f(\mathrm{mult}_2(\sum_{i \in I}s_i\otimes b_i)) = f(\sum_{i \in I}s_ib_i) = r \otimes \sum_{i \in I}s_ib_i = \sum_{i \in I}rs_i \otimes b_i = 0$. Thus, we obtained a complex \begin{equation*} 0 \xrightarrow{} {\mathrm{Tor}_2(R/rR, B)} \xrightarrow{} \phantom{}_rR \otimes B \xrightarrow{\mathrm{mult}_2} \phantom{}_rB \xrightarrow{f} {\mathrm{Tor}_1(R/rR, B)} \xrightarrow{} 0 \end{equation*}

It remains to show exactness at $\phantom{}_rB$. We assume that $b \in \phantom{}_rB$ is such that $0 = f(b) = r\otimes b$. By the solution to this question, there exists a finite set $I$, $r_i \in R$ for $i \in I$ and $b_i \in B$ for $i \in I$ such that: \begin{align*} b &= \sum_{i \in I}r_ib_i, \\ 0 &= rr_i \quad \text{for }i \in I. \end{align*} Therefore, $r_i \in \phantom{}_rR$ and $\mathrm{mult}_2(\sum_{i \in I}r_i\otimes b_i) = b$.

The fact that $\mathrm{Tor}_n(R/rR, B) \cong \mathrm{Tor}_{n - 2}(\phantom{}_rR, B)$ for $n \ge 3$ follows from Exercise 2.4.3.