Timeline for How much does Ext tell me about isomorphisms?
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2020 at 21:46 | comment | added | diracdeltafunk | Yes, thanks, that was a typo! But that's exactly right. | |
| Jun 15, 2020 at 14:50 | comment | added | DJWilliams | @Diracdeltafunk thank you so much. This all seems so much clearer ($Ext$ really is powerfu!l). So just to clarify, if $Ext(Y, X)$ is $\mathbb{Z}/2$ or $mathbb{Z}/3$ then any indecomposable modules that occur in between $X, Y$ must be isomorphic (if the latter, then the exact sequences don't have to be equivalent, though) since they cant belong to the trivial class. Did you mean $|Ext^1|>3$ above, by the way? | |
| Jun 14, 2020 at 21:32 | comment | added | diracdeltafunk | (cont.) However, for $\lvert \text{Ext}^1 \rvert > 2$ we can't use this argument anymore, because even after taking the quotient of $\text{Ext}$ by $a \sim -a$ we will have more than one nontrivial class. In general, though, I agree that $\text{Ext}$ is a spookily powerful tool: that's why it's so popular and deeply studied! Anyway, the implication I was talking about was "if $M$ and $N$ are indecomposable then neither extension corresponds to the identity element of $\text{Ext}^1$". This is true, but only because we had also assumed $\text{Ext}^1 \neq 0$. Still very simple, but not immediate. | |
| Jun 14, 2020 at 21:20 | comment | added | diracdeltafunk | Ah, yes, the assumption that $\text{Ext} \cong \mathbb{Z}/2$ is very strong and it is what allows us to conclude that any two non-split extensions are equivalent. If $\text{Ext} \cong \mathbb{Z}/3$ this will not be the case, however, any two non-split extensions will still be isomorphic, as you suspected! This is because if $0 \to X \xrightarrow{f} Z \xrightarrow{g} Y \to 0$ represents an element $a \in \text{Ext}^1(Y,X)$, then $-a$ is represented by $0 \to X \xrightarrow{-f} Z \xrightarrow{g} Y \to 0$ (check this by computing the Baer sum of these two extensions and verifying it splits) | |
| Jun 14, 2020 at 10:45 | comment | added | DJWilliams | @diracdeltafunk sorry, I was unclear. I meant if ext was $\mathbb{Z}/3$, for example. Then would that mean that any two indecomposable modules in an exact sequence $0\to X\to ?\to Y\to 0$ are still isomorphic? This just seems too good to be true. I suppose I am still unhappy with exactly how powerful ext is at classifying extensions (and how that relates to the modules). Could you also explain your small note further, please. Which implication do you mean? | |
| Jun 14, 2020 at 6:24 | comment | added | diracdeltafunk | Well, $x \mapsto -x$ is the identity map on $\mathbb{Z}/2\mathbb{Z}$, so being in inverse classes is the same as being in the same class! Perhaps I'm misunderstanding what you mean? The important point is that $M$ and $N$ being indecomposable implies that neither extension corresponds to the identity element in $\text{Ext}^1$, and there is only one non-identity element in $\mathbb{Z}/2\mathbb{Z}$. Small note: this implication isn't entirely trivial, you need to use the fact that if $X = 0$ or $Y = 0$ then we would have $\text{Ext}^1(Y,X) = 0 \neq \mathbb{Z}/2\mathbb{Z}$. | |
| Jun 13, 2020 at 21:59 | comment | added | DJWilliams | @spin that was exactly my logic, however I wasnt sure since I am not wholly comfortable with ext. I need to read more on it. It also seemed to me that they would be isomorphic if they belonged to inverse classes. However once again I was not sure of this. | |
| Jun 13, 2020 at 20:06 | review | Close votes | |||
| Jun 14, 2020 at 23:11 | |||||
| Jun 13, 2020 at 19:51 | history | edited | Mikhail Borovoi | CC BY-SA 4.0 |
A typo, formatting, style improved (I hope)
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| Jun 13, 2020 at 19:36 | comment | added | spin | So $M$ and $N$ are indecomposable. In particular these short exact sequences are nonsplit, so they correspond to non-zero elements of $\operatorname{Ext}^1(Y,X)$. If $\operatorname{Ext}^1(Y,X) \cong \mathbb{Z}/2\mathbb{Z}$, this means that these two sequences are equivalent, and in particular $M$ and $N$ are isomorphic. | |
| Jun 13, 2020 at 19:20 | comment | added | DJWilliams | @LSpice yes, sorry. I should have thought more on the wording of this question. First time posting, so I didnt think. My own stupid fault. | |
| Jun 13, 2020 at 19:19 | history | edited | DJWilliams | CC BY-SA 4.0 |
Updated for clarity, and cleared up some typos.
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| Jun 13, 2020 at 18:14 | comment | added | LSpice | Shouldn't your question be how much Ext tells you about isomorphism (as an adjective) of modules, not about isomorphic modules, since the point is to deduce, not to assume, the modules are isomorphic? | |
| Jun 13, 2020 at 17:15 | comment | added | DJWilliams | Sorry, that was a typo. My bad. | |
| Jun 13, 2020 at 17:15 | history | edited | DJWilliams | CC BY-SA 4.0 |
added 8 characters in body
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| Jun 13, 2020 at 17:09 | comment | added | Benjamin Steinberg | ext(X,Y) classifies sequences 0->Y->M->X->0. If it was the other way I think you are OK. | |
| Jun 13, 2020 at 17:08 | comment | added | Maxime Ramzi | Don't you mean $Ext^1(Y,X)$ ? | |
| Jun 13, 2020 at 17:04 | history | edited | Gabe Conant | CC BY-SA 4.0 |
fixed typo
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| Jun 13, 2020 at 16:51 | review | First posts | |||
| Jun 13, 2020 at 17:05 | |||||
| Jun 13, 2020 at 16:51 | history | asked | DJWilliams | CC BY-SA 4.0 |