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 March 31st, 2018, 08:59 AM #1 Senior Member   Joined: Jan 2015 From: usa Posts: 104 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 $3^2-1=8$ possible non-zero 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, 10:05 AM #2 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 313 Thanks: 112 Math Focus: Number Theory, Algebraic 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$\beta(1,0) = 0$ and $\beta(0,1) = 2$ $\beta(1,0) = 1$ and $\beta(0,1) = 0$ $\beta(1,0) = 2$ and $\beta(0,1) = 0$ can be dealt with in a similar manner. We're now left with the case that $\beta(1,0)$ and $\beta(0,1)$ are both non-zero. 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, 10:17 AM #3 Senior Member   Joined: Jan 2015 From: usa Posts: 104 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 10:31 AM.
 March 31st, 2018, 10:31 AM #4 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 313 Thanks: 112 Math Focus: Number Theory, Algebraic 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, 10:33 AM #5 Senior Member   Joined: Jan 2015 From: usa Posts: 104 Thanks: 1 Are all these 16 couple of morphisms make 16 equivalent short exact sequences?
March 31st, 2018, 10:48 AM   #6
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Quote:
 Originally Posted by mona123 are all these 16 couple of morphisms make 16 equivalent short exact sequences?
Yes, I suspect these are all split exact sequences (and thus are all isomorphic). You might like to verify this.

 March 31st, 2018, 10:53 AM #7 Senior Member   Joined: Jan 2015 From: usa Posts: 104 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 10:58 AM.
March 31st, 2018, 10:57 AM   #8
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Quote:
 Originally Posted by mona123 Is there a theorem that says ''all split exact sequences are equivalent''?
It depends on what you mean by equivalent. If you mean that they are chain isomorphic, then yes, and it follows from what I said in my second post.

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 11:00 AM.

 March 31st, 2018, 11:05 AM #9 Senior Member   Joined: Jan 2015 From: usa Posts: 104 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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