Timeline for Endomorphism rings of infinitely generated free modules generated by idempotents?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 24, 2020 at 9:06 | comment | added | Cary | @YCor, Shouldn't we replace $ \space 2\begin{pmatrix} I & 0\\ 0 & I\end{pmatrix} $ in your answer? i think we have to subtract $2I$. | |
| Oct 17, 2020 at 15:05 | comment | added | Cary | @YCor yes but how can i find this subring is not a closed subring. in fact I haven't heard about it before | |
| Oct 17, 2020 at 15:03 | comment | added | YCor | You assume stability under infinite sums, which is not automatic for a subring (but is for a closed subring). | |
| Oct 17, 2020 at 15:01 | comment | added | Cary | @YCor yes it doesn't have to be closed under linear combination but in our case since $1$ belongs to $S$ so $\mathbb{Z} \subseteq S$ and hence for every $a_{k}\in \mathbb{Z}$ , $\sum _{k\geq 0}a_{k}p^{n}$ belongs to $S$. | |
| Oct 17, 2020 at 14:47 | comment | added | YCor | Yes the subring generated by idempotents is reduced to $\mathbf{Z}$. A subring doesn't have to be closed under "linear combination". This would be an ideal. In every unital ring the ideal generated by idempotents is the whole ring. | |
| Oct 17, 2020 at 14:38 | comment | added | Cary | @YCor I was thinking about the subring generated bu idempotents of $Zp$, where $Zp$ is the ring of p-adic integers. and I said: let $S$ be the subring generated by idempotents of $Zp$. As 1 belongs to S and S closed under addition so every prime $p$ belongs to $S$. Also since $S$ closed under multiplication for every positive integer$ n$, $p^n$ belongs to $S$. so every linear combination of $p^n$ belongs to $S$ which implies that $S$ equals to $ Zp$. But my friend said $S$ equal to $\mathbb{Z}$. which one is the correct answer? | |
| Oct 5, 2020 at 23:46 | vote | accept | Cary | ||
| Oct 5, 2020 at 9:54 | comment | added | Cary | Thank you so much @Ycor | |
| Oct 5, 2020 at 6:18 | comment | added | YCor | @Cary yes, this is covered by my answer (at least when $\kappa$ is infinite, assuming ZFC, or finite even) | |
| Oct 4, 2020 at 23:13 | comment | added | Cary | i think it must be true for any cardinal $\kappa$ such that $M= R^{\kappa}$ | |
| Oct 3, 2020 at 15:09 | vote | accept | Cary | ||
| Oct 3, 2020 at 15:11 | |||||
| Oct 2, 2020 at 18:44 | comment | added | YCor | Blast, I spent a long time editing, improving this to any module of the form $M^2\oplus N^3$ (which uniformly covers all modules $P^n$ for $n\ge 3$, and hence all ultraproducts of such endomorphism algebras)... and when I refreshed the page after the other answer was deleted, all has disappeared. The argument is not hard but it takes time to write down all these matrices... So well, let's stick to have it as a remark. | |
| Oct 2, 2020 at 18:35 | comment | added | Pace Nielsen | Of course! In other words, when $M\cong N\times N$, then ${\rm End}(M)\cong \mathbb{M}_2({\rm End}(N))$. Great observation. | |
| Oct 2, 2020 at 18:16 | history | answered | YCor | CC BY-SA 4.0 |