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Let $A,B$ be two $R$-modules over a commutative ring $R$ (restrict to $R = \mathbb{Z}$ or $R= \mathbb{K}$ a field where appropriate). Suppose $A$ and $B$ fit into an exact sequence

$0 \to R^p \to A \to R^r \to R^q \to B \to 0$

for integers $p,q,r > 0$, where $R^{n} = R \oplus R ... \oplus R$ ($n$-times). Then what can be said about $A$ and $B$? Specifically,

1) Are there any cases where $A$ and $B$ be computed (most generally in terms of $p,q,r$)?

If not

2) Given an allowed $B$ what can be said about $A$?

3) Given an allowed $A$ what can be said about $B$?

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  • $\begingroup$ Can you be a bit more specific about why you need this? When $R$ is a vector space, for instance, all the answers are essentially trivial. When $R$ is a PID (e.g. $\mathbb{Z}$), it is also quite easy to give very precise answer. But it all depends on the generality you care for. Also, it would help if you explain what "computed" means in question 1). $\endgroup$
    – Vladimir Dotsenko
    Commented Jan 7, 2016 at 17:15
  • $\begingroup$ For example, I want to know what can be said about relative homology groups $H_1(X,A)$ and $H_0(X,A)$ when it is known that $H_1(A) = 0$, $H_1(X) \cong R^p$, and the connectivity of $A$ and $X$ is $r$ and $q$, respectively. This is just an analogy, but should suffice. Computed means compute $H_1(X,A)$ and $H_0(X,A)$ in the usual sense (or at least determine rank). $\endgroup$
    – Chan
    Commented Jan 7, 2016 at 17:26
  • $\begingroup$ Note I have removed the condition $r \geq q$ (it was a mistake!). $\endgroup$
    – Chan
    Commented Jan 7, 2016 at 17:43
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    $\begingroup$ It follows that $[A] = [B]$ in the quotient of the Grothendieck group $K_0(R)$ by the subgroup generated by $[R]$. $\endgroup$
    – Qiaochu Yuan
    Commented Jan 7, 2016 at 17:48
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    $\begingroup$ If the projective dimension of $B$ is at most $2$, then $A$ is projective. If the projective dimension of $B$ is $3$, then the projective dimension of $A$ is at most $1$. If the projective dimension of $B$ is greater than $3$, then the projective dimension of $A$ is exactly two less than the projective dimension of $B$. $\endgroup$
    – Steven Landsburg
    Commented Jan 7, 2016 at 18:12

1 Answer 1

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Assume $R$ is a PID.

Clearly, there is a short exact sequence $0\to M_2\to A\to M_1\to 0$, where $M_1\subset R^r$ is the image of the map $A\to R^r$, so a free module of rank $d\le r$, and $M_2\cong R^p$ is the kernel of that map. There are no extensions between free modules, so $A$ is free of rank $p+d$.

Furthermore, there is an exact sequence $0\to M_3\to M_4\to B\to 0$, where $M_3\subset R^q$ is the kernel of the map $R^q\to B$, and $M_4\cong R^q$. Note that due to exactness $M_3$ is the same as the image of the map $R^r\to R^q$, which is the quotient of $R^r$ by $M_1$; thus, it is a free module of rank $r-d$. Therefore, modulo torsion $B$ is free of rank $q-r+d$; however, unlike $A$, $B$ might have torsion. (Consider the case $A=R^p$, the map $R^p\to A$ being the identity, the map $A\to R^r$ being zero, then the rest $0\to R^r\to R^q\to B\to 0$ is just a presentation of the most general finitely generated $B$ by generators and relations.)

If you only fix modules in the exact sequence but don't have information on maps, this probably is as much as you can establish.

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