Timeline for Extending a module structure in a family
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 17, 2017 at 9:17 | history | edited | Wille Liu | CC BY-SA 3.0 |
removal of the first hypothesis
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| Aug 16, 2017 at 23:12 | history | edited | Wille Liu | CC BY-SA 3.0 |
typo
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| Aug 16, 2017 at 22:16 | history | edited | Wille Liu | CC BY-SA 3.0 |
typo
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| Aug 16, 2017 at 22:05 | history | edited | Wille Liu | CC BY-SA 3.0 |
more explanations
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| Aug 16, 2017 at 21:57 | history | edited | Wille Liu | CC BY-SA 3.0 |
typo
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| Aug 16, 2017 at 21:46 | comment | added | Wille Liu | I corrected the wrong perception that $\mathrm{End}_{k}(M_0) = k$ implies irreducibility. I give another try. | |
| Aug 16, 2017 at 21:43 | history | edited | Wille Liu | CC BY-SA 3.0 |
Correction of the error of irreducibility and new proof under stronger assumptions.
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| Aug 15, 2017 at 19:03 | comment | added | Alexander Braverman | No, in my example $End_{A_0}(M_0)=k$ (only scalar matrices commute with all upper triangular matrices). Yes, you can assume that $End_{A_x}(M_x)=k$ for all $x$. | |
| Aug 15, 2017 at 18:27 | comment | added | Wille Liu | Maybe the condition that you have on $M$ was actually $\mathrm{End}_{A_x}M_x = k$ for $x$ generic? | |
| Aug 15, 2017 at 18:09 | comment | added | Wille Liu | But your example doesn't give $\mathrm{End}_{A_0}(M_0) = k$, it is $k[\epsilon] / (\epsilon^2)$ instead, where $\epsilon$ is the matrix $\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$. | |
| Aug 15, 2017 at 16:40 | comment | added | Alexander Braverman | The point is that the map from $(A/I_A)_0$ to $End_k(M_0)$ doesn't have to be injective. | |
| Aug 15, 2017 at 16:39 | comment | added | Alexander Braverman | For example, consider the case when $M=(k[t]^2)$, $B=2\times 2$-matrices over $k[t]$ and $A$ is the subalgebra of of $B$ consisting of matrices which are upper-triangular at $t=0$. Then $I=0$ and all the assumptions are satisfied but the map $A_0\to B_0$ is clearly not injective. | |
| Aug 15, 2017 at 16:36 | comment | added | Alexander Braverman | I think the new argument isn't quite right - the point is that the map from $(A/I_A)_0$ to $(B/I_B)_0$ is not necessarily injective (this doesn't even have to be the case when $I=0$) | |
| Aug 15, 2017 at 10:02 | history | edited | Wille Liu | CC BY-SA 3.0 |
added 26 characters in body
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| Aug 15, 2017 at 9:28 | history | edited | Wille Liu | CC BY-SA 3.0 |
deleted 1 character in body
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| Aug 15, 2017 at 9:23 | comment | added | Wille Liu | Yes, with this new hypothesis. I have added it. | |
| Aug 15, 2017 at 9:22 | history | edited | Wille Liu | CC BY-SA 3.0 |
added 2300 characters in body
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| Aug 14, 2017 at 22:10 | comment | added | Alexander Braverman | Thanks. In my case I also know that the only endomorphisms of $M_0$ over $A_0$ are scalars. Do you think this might help? | |
| Aug 14, 2017 at 18:20 | history | edited | Wille Liu | CC BY-SA 3.0 |
added 92 characters in body
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| Aug 14, 2017 at 18:05 | history | answered | Wille Liu | CC BY-SA 3.0 |