Timeline for Generalized height of elements in abelian groups
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 28, 2016 at 14:28 | vote | accept | Ilan Barnea | ||
| Jun 28, 2016 at 8:41 | answer | added | Jeremy Rickard | timeline score: 1 | |
| Jun 27, 2016 at 18:13 | history | edited | Ilan Barnea | CC BY-SA 3.0 |
added 32 characters in body
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| Jun 27, 2016 at 13:54 | comment | added | Jeremy Rickard | I've not checked that this fixes all the problems, hence just a comment. But the idea of the definition of $h^*_p$ seems to be that if there is no $\sigma$ with $a\in p^\sigma A\setminus p^{\sigma+1} A$ then $h^*_p(a)$ should be "big", and $l_p(A)$ is chosen as the smallest ordinal bigger than $h^*_p(b)$ for all those $b\in A$ which already have $h^*_p(b)$ defined. He'd probably have chosen $\infty$ instead of $l_p(A)$ except that then you have to say which infinite ordinals are greater than $\infty$. But if you do choose $\infty$, a notional symbol greater than all ordinals, does that work? | |
| Jun 27, 2016 at 10:50 | comment | added | Ilan Barnea | Exactly, this shows that the statement cannot be true | |
| Jun 27, 2016 at 10:48 | comment | added | Jeremy Rickard | Oh, OK. You meant that's the inequality that would be true if the statement were true, rather than the inequality that is true. | |
| Jun 27, 2016 at 10:46 | comment | added | Ilan Barnea | The statement is "$h^*_p$ does not diminish under homomorphisms" so $h_p^*(f(a))\geq h_p^*(a)$ | |
| Jun 27, 2016 at 10:43 | comment | added | Jeremy Rickard | I think you meant to write $0=h^*_p(f(a))\leq h^*_p(a)$? | |
| Jun 27, 2016 at 10:34 | history | asked | Ilan Barnea | CC BY-SA 3.0 |