Timeline for Height of ideal in graded ring

6 events

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May 24, 2012 at 20:54	comment	added	Thomas Kahle		Hi variety, that's not a stupid question. The case only $ht(I) = ht(I')$ occurs when $I$ is already homogenous. I excluded that case in the beginning: I assumed that $p$ is non graded (and I'll also assume that $I$ is not graded, otherwise everything is trivial). We get $\text{ht}(IR_{(0)}=1$ since only the zero ideal has height zero and $IR_{(0)}$ is not the zero ideal since it contains a non-homogeneous element.
May 24, 2012 at 18:35	comment	added	variete		Dear Thomas Kahle, in the last line of your answer, you concluded that $\text{ht}(I)=\text{ht}(I')+1$, but we have to prove that $\text{ht}(I)\le\text{ht}(I')+1$ and why do we get $\text{ht}(IR_{(0)})=1$ ? I am so sorry if you feel my question is stupid. Thank you very much for your answer.
May 24, 2012 at 10:25	comment	added	Thomas Kahle		I've updated my answer.
May 24, 2012 at 10:24	history	edited	Thomas Kahle	CC BY-SA 3.0	More detail on the reduction to the prime case.
May 23, 2012 at 15:31	comment	added	variete		Sorry, but in your answer you has just dealt with only prime ideal. My question is about a general ideal.
May 23, 2012 at 12:36	history	answered	Thomas Kahle	CC BY-SA 3.0