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March 31st, 2018, 09:59 AM  #1 
Senior Member Joined: Jan 2015 From: usa Posts: 103 Thanks: 1  Short exact sequence
We consider $$\left\{0\right\}\to \mathbb{Z}/3 \mathbb{Z}\overset{\alpha}{\to} \mathbb{Z}/3\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z}\overset{\beta}{\to} \mathbb{Z}/3 \to \left\{0\right\}$$ I want to find all morphisms $\alpha$ and $\beta$ that make the sequence a short exact one. This is what i wrote  we need $\alpha$ to be injective. This happens when $\alpha(1)\neq (0,0)$, so if $\alpha(1)=(a,b)$ then there are $3^21=8$ possible nonzero pairs $(a,b)$.  we need $\ker(\beta)=\text{im}(\alpha)$ and $\beta$ surjective, from which we get $$\left (\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}\right )/\left <(a,b)\right >\cong\mathbb{Z}/3\mathbb{Z}$$ The possible pairs $(a,b)$ are $(0,1),(1,0),(0,2),(2,0), (1,1) or (1,2) or (2,1) or (2,2) $ But i can not conclude to answer the question completly. Please help me to do so. Thanks in advance 
March 31st, 2018, 11:05 AM  #2 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 284 Thanks: 86 Math Focus: Algebraic Number Theory, Arithmetic Geometry 
Suppose the sequence is exact. Notice that $\beta$ is determined by where it maps $(0,1)$ and $(1,0)$, and it can't map both of them to $0$ (else $\beta$ wouldn't be surjective). Let's first deal with the case that $\beta$ maps $(1,0)$ or $(0,1)$ to $0$. If $\beta(1,0) = 0$ and $\beta(0,1) = 1$, then $Ker(\beta) = \{(0,0), (1,0), (2,0) \}$. So we must have $\alpha(1) = (1,0)$ or $\alpha(1) = (2,0)$. It's easy to see that either option for $\alpha$ then makes the sequence exact. The possibilities
We're now left with the case that $\beta(1,0)$ and $\beta(0,1)$ are both nonzero. If $\beta(1,0) = \beta(0,1)$, then $Ker(\beta) = \{(0,0),(1,2),(2,1) \}$ and so $\alpha(1) = (1,2)$ or $\alpha(2) = (2,1)$. It's easy to see that all possibilities here ($\beta(1,0) = 1$ or $2$; $\alpha(1) = (1,2)$ or $(2,1)$) make the sequence exact. If $\beta(1,0) \neq \beta(0,1)$, then a similar argument works. So, all in all, there should be (unless I've screwed up somewhere!) 16 pairs $(\alpha, \beta)$ that make the sequence exact. 
March 31st, 2018, 11:17 AM  #3 
Senior Member Joined: Jan 2015 From: usa Posts: 103 Thanks: 1 
so why here http://palmer.wellesley.edu/~ivolic/...1Solutions.pdf, they said in the answer of question a) that ''Notice that i and p are the only maps that make this sequence exact''? are all these 16 couple of morphisms make 16 equivalent short exact sequences?
Last edited by mona123; March 31st, 2018 at 11:31 AM. 
March 31st, 2018, 11:31 AM  #4 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 284 Thanks: 86 Math Focus: Algebraic Number Theory, Arithmetic Geometry 
No idea. That's definitely incorrect. Also, for what it's worth, in part (b), condition (iii) is not equivalent to (i) or (ii). To make that statement correct, you could replace (iii) by (iii*) There exists a chain isomorphism from $$\left\{0\right\}\to A \overset{f}{\to} B \overset{g}{\to} C \to \left\{0\right\}$$ to $$\left\{0\right\}\to A \overset{i}{\to} A \oplus C \overset{p}{\to} C \to \left\{0\right\}$$ where $i$ and $p$ are the natural inclusion and projection maps, respectively. 
March 31st, 2018, 11:33 AM  #5 
Senior Member Joined: Jan 2015 From: usa Posts: 103 Thanks: 1 
Are all these 16 couple of morphisms make 16 equivalent short exact sequences?

March 31st, 2018, 11:48 AM  #6 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 284 Thanks: 86 Math Focus: Algebraic Number Theory, Arithmetic Geometry  
March 31st, 2018, 11:53 AM  #7 
Senior Member Joined: Jan 2015 From: usa Posts: 103 Thanks: 1 
Can you please help me with case $Z/9Z$ instead of $Z/3Z\oplus Z/3Z$? thanks in advance
Last edited by mona123; March 31st, 2018 at 11:58 AM. 
March 31st, 2018, 11:57 AM  #8  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 284 Thanks: 86 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
Edit: Obviously not all split exact sequences are chain isomorphic, but I mean that if $$\left\{0\right\}\to A \overset{f}{\to} B \overset{g}{\to} C \to \left\{0\right\}$$ and $$\left\{0\right\}\to A \overset{f'}{\to} B \overset{g'}{\to} C \to \left\{0\right\}$$ are both split, then they are chain isomorphic. Last edited by cjem; March 31st, 2018 at 12:00 PM.  
March 31st, 2018, 12:05 PM  #9 
Senior Member Joined: Jan 2015 From: usa Posts: 103 Thanks: 1 
Ah, i see. Can you please help me with case $Z/9Z$ instead of $Z/3Z⊕Z/3Z$? thanks in advance


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