Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I love it when an undergraduate catches me out. I'm lecturing a first course in (not necessarily commutative) rings (with 1) and I've spent the last few weeks doing basic module theory. I defined a short exact sequence of (left) $R$-modules and a homomorphism of short exact sequences (a homomorphism from $0\to A\to B\to C\to 0$ to $0\to A'\to B'\to C'\to 0$ is just $R$-module maps $A\to A'$, $B\to B'$ and $C\to C'$ such that the obvious two squares commute). There's hence an obvious notion of an isomorphism of short exact sequences.

Today one of the students asked me if it was possible to have a ring $R$ and modules $A,A',B,B',C,C'$ sitting in two short exact sequences as above, and such that $A$ was isomorphic to $A'$, $B$ was isomorphic to $B'$ and $C$ was isomorphic to $C'$, but that the sequences weren't isomorphic. I said "sure, I'll email you a counterexample later" (the logic being that if this were a theorem, it would be one I knew about). I thought I'd knock up a counterexample on the tube home -- but I failed :-/

If $R$ is a field then short exact sequences split (yes we're assuming AC) so that's not going anywhere. So I thought that $R=k[X]$ would be a good place to start, $k$ a field. In this case an $R$-module is just a $k$-vector space equipped with an endomorphism and I figured this would give me enough flexibility. I wanted $A,A',C,C'$ to be $R/(x^2)$ and tried some messy matrix calculations to figure out an example, but I couldn't get it to work. I then went for $R=k[x,y]$ but now a 2-dimensional vector space is an $R$-module when we give it two commuting linear maps and somehow this set-up had too many endomorphisms for me to face doing the algebra. I then figured that I might want to try a polynomial ring in two noncommuting variables--but then it was my stop and it was time to start thinking about other things.

I am almost certain that there will even be a counterexample with $R$ commutative (that's why I was thinking about the commutative case). Can anyone tell me the trick I'm missing?

share|improve this question

4 Answers 4

up vote 6 down vote accepted

I think saw a question about this in a different guise recently, but I cannot find it again -- it might be an old one. As far as I remember the answer was something like: $$0 \longrightarrow \mathbb Z/p^2 \times \mathbb Z/p \longrightarrow \mathbb Z/p^3 \times \mathbb Z/p^2 \times \mathbb Z/p \longrightarrow \mathbb Z/p^2 \times \mathbb Z/p \longrightarrow 0$$ in two different ways as abelian groups i.e. $\mathbb Z$-modules.

(Edit: here it was, not exactly as I remembered it. I may have misunderstood something important, because the answer in the old answer had an extra factor of $\mathbb Z/p$ in $A$ and $B$ and was claimed to be minimal, but I don't see exactly why that extra factor is needed).

(Edit 2: Here's an explicit construction, including the $A\to B$ morphisms).

share|improve this answer
You're trying with $R$ a PID, which is what I tried, but you're going deeper into the problem than I did before I gave up: my candidate $B$s had length 4 and you're suggesting that I need to go up to length 6. I think this could well work; I'll see if I can construct a rigorous argument on the tube in to work tomorrow :-) [unless someone wants to help out while I'm asleep...] –  Kevin Buzzard Dec 8 '11 at 22:18
Note that I've now found a description of an explicit solution that is slightly more complex than my suggestion. –  Henning Makholm Dec 8 '11 at 22:36
Thanks a lot Henning. I wonder if one can use $k[X]$ instead with a length 7 example? –  Kevin Buzzard Dec 8 '11 at 23:57
Looking at the example on the page you link to -- don't you think that last $\mathbf{Z}/p\mathbf{Z}$ (which splits off as a direct factor in both $G$, $H$ and $K$) is a little strange? Can I really not cross it out? –  Kevin Buzzard Dec 9 '11 at 0:15
@HenningMakholm: your example is fine (order p^6). I believe there are three Aut(B) orbits when p is odd (sizes p^2, p^2, and p-1), so once it fails, it fails pretty badly. There are non-abelian examples of order 16 (so they should have been mentioned by Vipul, but aren't relevant to Kevin). –  Jack Schmidt Dec 12 '11 at 15:51

One can visualize the example as follows: there are two ways to embed the module

*   *


*   *
|   |
*   *   *

so that the quotient is

*   *

Here dots represent the simple module and vertical lines non-split extensions: one can either embed it so thatit becomes the sum of the summand of length two and the socle of the summand of length 3, or as the sum of the radical of the summand of length 3 and the simple summand. It is clear that those two embeddings are not conjugate.

(These pictures imply that it is enough to have a simple module with a non-split self-extension which lives to a module of length 3, which is more or less what Jack wrote)

share|improve this answer

This is just Henning's answer with a more general hypothesis and more of the information inline:

Let $(R,p)$ be a commutative artinian local ring with length at least 3, so that $M=R/p^3$ is a uniserial module of length 3.

Define the following modules:

$$\begin{array}{rccccl} A & = & p/p^3 & \times & p^2/p^2 & \times & R/p & \cong R/p^2 \times R/p \\ B & = & R/p^3 & \times & R/p^2 & \times & R/p & \\ C & = & R/p & \times & R/p^2 & \times & R/R & \cong R/p^2 \times R/p \\ \\ A' & = & p^2/p^3 & \times & R/p^2 & \times & p/p & \cong R/p^2 \times R/p \\ B' & = & R/p^3 & \times & R/p^2 & \times & R/p & \\ C' & = & R/p^2 & \times & R/R & \times & R/p & \cong R/p^2 \times R/p \end{array}$$

Define the obvious morphisms A→B→C and A′→B′→C′ that take (x,y,z) to (x,y,z). The results are short exact sequences, $C=B/A$ and $C'=B'/A'$, with $A\cong A'$, $B = B'$, $C\cong C'$, but there is no automorphism of B taking A to A′.

Indeed, $A \cap pB = p/p^3 \times p^2/p^2 \times p/p \cong R/p$, while $A' \cap pB' = p^2/p^3 \times p/p^2 \times p/p \cong R/p \times R/p$. Of course $\phi(A \cap pB) = \phi(A) \cap p\phi(B) = \phi(A) \cap pB'$, so $\phi(A) = A'$ is impossible.

share|improve this answer

Another way of picturing Mariano's anwer: $$ \begin{matrix} \bullet&\\ |\\ \bullet&&\circ\\ |&&|\\ \circ&&\circ&&\bullet \end{matrix}\quad\quad\text{ versus }\quad\quad \begin{matrix} \bullet&\\ |\\ \circ&&\bullet\\ |&&|\\ \circ&&\bullet&&\circ \end{matrix} $$ The submodule is black and the quotient is white.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.