Short exact sequences

Feb 2009
98
0
I am claiming the following is not a short exact sequence
\(\displaystyle 0\rightarrow Z\stackrel{i}{\rightarrow} Z \oplus Z \oplus Z \stackrel{j}{\rightarrow} Z \rightarrow 0\)

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, \(\displaystyle (Z \oplus Z \oplus Z )/ i(Z) \) would be isomorphic to \(\displaystyle Z\) .

Now \(\displaystyle i(1)= (a,b,c) \neq 0\) for some \(\displaystyle a,b,c \in Z\). Now, is it true that
\(\displaystyle i(Z)=im(i)=Z<(a,b,c)>\) ? And that \(\displaystyle (Z \oplus Z \oplus Z )/ i(Z) = Z/aZ \times Z/bz \times Z/cZ \)?If yes, how do you show this is not isomorphic to \(\displaystyle Z\) ? If not, what am I doing wrong and how do you prove my claim?
 
Jul 2008
433
159
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.
 
Feb 2009
98
0
Ok, that makes sense, but what would be a good argument to show \(\displaystyle (Z \oplus Z \oplus Z) / A' \) is not isomorphic to \(\displaystyle Z\)?
 
Feb 2009
98
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Can we say, \(\displaystyle } Z \oplus Z \oplus Z\) has 3 generators, now \(\displaystyle ( Z \oplus Z \oplus Z)/i(Z) \) is isomorphic to \(\displaystyle ( Z \oplus Z \oplus Z) / Z\) which has at least 2 generators, but \(\displaystyle Z\) only has one generator?
 
Jul 2008
433
159
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.
 
Nov 2010
193
55
Or just say it's not exact by definition. To be exact, we would need \(\displaystyle \mathrm{im}(i)=\mathrm{ker}(j)\).

Of course, the other issue is still a good question in and of itself, so don't let this stop that discussion.
 
Feb 2009
98
0
I am not sure I understand how \(\displaystyle im(i) = Ker(j)\) helps show \(\displaystyle Z^3/i(Z)\) is not isomorphic to \(\displaystyle Z\). 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...
 
Jul 2008
433
159
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.
 
Nov 2010
193
55
I am not sure I understand how \(\displaystyle im(i) = Ker(j)\) helps show \(\displaystyle Z^3/i(Z)\) is not isomorphic to \(\displaystyle Z\). 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.