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Timeline for Simple Ore extensions

Current License: CC BY-SA 3.0

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Feb 4, 2015 at 4:23 comment added Pace Nielsen It appears that this question is now given as an open problem in the following paper by Johan Öinert and others: worldscientific.com/doi/abs/10.1142/S0219498812501927 Also, Steven's comment is cited in the paper.
Nov 17, 2011 at 16:03 answer added Steven Deprez timeline score: 2
Nov 17, 2011 at 15:48 answer added Steven Deprez timeline score: 1
Jun 23, 2011 at 15:22 comment added Johan Öinert Johan, you obviously intended to write "non-trivial ideal" instead of "ideal". That was perhaps a stupid remark. While I'm at it, I might as well point out that in my previous comment I assumed that $\sigma$ was NOT injective. Otherwise $\ker\sigma = \{0\}$ is of course an ideal in the Ore extension, and the first sentense of my comment would be false.
Jun 23, 2011 at 7:25 comment added Johan It is worth noting that $\ker \sigma$ does not have to be invariant under $\delta$. In fact there does not have to be any ideal that is invariant under both $\sigma$ and $\delta$.
Jun 20, 2011 at 19:32 comment added Mariano Suárez-Álvarez Ups, that's what I meant, actually :)
Jun 19, 2011 at 23:11 comment added Johan Öinert Mariano, $\ker\sigma \subseteq R$ is never an ideal of the Ore extension, but your idea is correct. It can be made more general: It is true if $\delta(\ker\sigma)\subseteq \ker\sigma$, for then (put $A:=R[x;\sigma,\delta]$) the set $(\ker\sigma) A$ is a proper ideal of the Ore extension. (I like to think of $R[x;\sigma,\delta]$ as a free left $R$-module with basis $\{1,x,x^2,...\}$.)
Jun 19, 2011 at 21:26 comment added Mariano Suárez-Álvarez More generally, it is true if $\delta(\ker\sigma)=0$, for then $\ker\sigma$ is an ideal in the Ore extension.
Jun 19, 2011 at 20:35 history asked Johan Öinert CC BY-SA 3.0