Timeline for Can a zerodivisor reduce both the depth and the dimension?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 17, 2013 at 7:22 | comment | added | Mahdi Majidi-Zolbanin | This is very good. Thank you! By the way, here we can take $n=2$. | |
| Jan 17, 2013 at 7:21 | vote | accept | Mahdi Majidi-Zolbanin | ||
| Jan 17, 2013 at 6:51 | comment | added | Pham Hung Quy | I have just repaired my answer. | |
| Jan 17, 2013 at 6:50 | history | edited | Pham Hung Quy | CC BY-SA 3.0 |
added 555 characters in body
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| Jan 17, 2013 at 6:28 | comment | added | Mahdi Majidi-Zolbanin | Yes, I agree with $(a,b,c)^2\cap(c)=(ac,bc,c^2)$ but I think an error has occurred before that. Do you agree $(a,b,c)^2\cap(c)\cap(c,d)^2=(c^2,bcd,acd)$? If you agree with what I just wrote then you will see the problem. | |
| Jan 17, 2013 at 6:27 | history | edited | Pham Hung Quy | CC BY-SA 3.0 |
edited body
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| Jan 17, 2013 at 6:23 | comment | added | Pham Hung Quy | $(a, b, c)^2 \cap (c) = (ac, bc, c^2) = (a, b, c)(c)$ | |
| Jan 17, 2013 at 6:12 | comment | added | Mahdi Majidi-Zolbanin | I don't think this is correct. The reason is every element in $(a,b,c)^2\cap(c)\cap(c,d)^2$ must be a multiple of $c$. Therefore, $a^2,b^2\not\in(a,b,c)^2\cap(c)\cap(c,d)^2$. Therefore the isomorphism you wrote is not correct. I compute $(a,b,c)^2\cap(c)\cap(c,d)^2=(c^2,bcd,acd)$. Therefore, $R/(d)\cong k[[a,b,c,d]]/(c^2,d)$, which has dimension and depth both equal to $2$. | |
| Jan 17, 2013 at 5:56 | comment | added | Pham Hung Quy | @ Mahdi: see my edit. | |
| Jan 17, 2013 at 5:55 | history | edited | Pham Hung Quy | CC BY-SA 3.0 |
added 49 characters in body
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| Jan 17, 2013 at 4:54 | comment | added | Mahdi Majidi-Zolbanin | Dear Pham: If I am not mistaken $\mathrm{depth} R/d=2$. In fact, I think $\{a,b\}$ forms a regular sequence in $R/d$. Can you explain why you think $\mathrm{depth} R/d=0$? | |
| Jan 17, 2013 at 4:30 | history | answered | Pham Hung Quy | CC BY-SA 3.0 |