# Short exact sequences

• Dec 6th 2010, 05:47 PM
math8
Short exact sequences
I am claiming the following is not a short exact sequence
$0\rightarrow Z\stackrel{i}{\rightarrow} Z \oplus Z \oplus Z \stackrel{j}{\rightarrow} Z \rightarrow 0$

I am thinking of it this way: If it was a short exact sequence, then i would be injective and j would be onto. Also, $(Z \oplus Z \oplus Z )/ i(Z)$ would be isomorphic to $Z$ .

Now $i(1)= (a,b,c) \neq 0$ for some $a,b,c \in Z$. Now, is it true that
$i(Z)=im(i)=Z<(a,b,c)>$ ? And that $(Z \oplus Z \oplus Z )/ i(Z) = Z/aZ \times Z/bz \times Z/cZ$?If yes, how do you show this is not isomorphic to $Z$ ? If not, what am I doing wrong and how do you prove my claim?
• Dec 6th 2010, 06:30 PM
Tinyboss
That sequence can't be exact. If you have 0->A->B->C->0, then C is isomorphic to B/A', where A' is the image of A under the map A->B. Because of exactness, that map is injective, so that A' is isomorphic to A. You can't take Z^3, mod out by something isomorphic to Z, and get Z.
• Dec 6th 2010, 06:51 PM
math8
Ok, that makes sense, but what would be a good argument to show $(Z \oplus Z \oplus Z) / A'$ is not isomorphic to $Z$?
• Dec 6th 2010, 07:01 PM
math8
Can we say, $} Z \oplus Z \oplus Z$ has 3 generators, now $( Z \oplus Z \oplus Z)/i(Z)$ is isomorphic to $( Z \oplus Z \oplus Z) / Z$ which has at least 2 generators, but $Z$ only has one generator?
• Dec 6th 2010, 07:03 PM
Tinyboss
You have to be careful when talking about the number of generators, because it's not fixed--for instance, the integers are generated by (1) or by (2,3). I think the free rank of an abelian group might be helpful here.
• Dec 6th 2010, 10:08 PM
topspin1617
Or just say it's not exact by definition. To be exact, we would need $\mathrm{im}(i)=\mathrm{ker}(j)$.

Of course, the other issue is still a good question in and of itself, so don't let this stop that discussion.
• Dec 7th 2010, 07:12 AM
math8
I am not sure I understand how $im(i) = Ker(j)$ helps show $Z^3/i(Z)$ is not isomorphic to $Z$. Can you elaborate on that?

Also @ Tinyboss: How, do I use the free rank of an abelian group to prove that? My algebra is very rusty, and I tried to look it up, but I don't think I understand how to apply the definition here...
• Dec 7th 2010, 04:50 PM
Tinyboss
Sorry, what I thought would work, doesn't. I asked a buddy and he said to tensor the sequence with Q, yielding the same sequence except with Q's in place of Z's. Since Q is a field, there is a well-defined notion of dimension, and so that sequence can't be exact. And because tensoring preserves exact sequences, this implies the original sequence wasn't.

I don't understand it 100%, but I'm putting it here since my first guess was wrong.
• Dec 7th 2010, 06:19 PM
topspin1617
Quote:

Originally Posted by math8
I am not sure I understand how $im(i) = Ker(j)$ helps show $Z^3/i(Z)$ is not isomorphic to $Z$. Can you elaborate on that?

Also @ Tinyboss: How, do I use the free rank of an abelian group to prove that? My algebra is very rusty, and I tried to look it up, but I don't think I understand how to apply the definition here...

I don't know that it does. I was just giving another way of answering the original question, about the sequence not being exact.