Timeline for Height of associated primes in regular rings
Current License: CC BY-SA 3.0
10 events
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| Mar 30, 2018 at 19:21 | comment | added | O-Ren Ishii | @JasonStarr Thank you for a helpful answer. Could you explain briefly why every associated prime $\mathfrak{p}$ of $J$ being of height $2$ implies that every associated prime of $I$ has height $1$ or $2$? | |
| Aug 8, 2017 at 11:51 | comment | added | Jason Starr | "for an ideal $I$ generated by $n$ elements ..." In fact, for every local Cohen-Macaulay ring $(R,\mathfrak{m})$, for every sequence $(a_1,\dots,a_n)$ of elements in $\mathfrak{m}$, the sequence is a regular sequence if and only if every minimal prime over $I=\langle a_1,\dots,a_n \rangle$ has height $\geq n$, and then $R/I$ is also Cohen-Macaulay, cf. Theorem 17.4, p. 135, Matsumura, Commutative ring theory. Then, by the unmixedness theorem, every associated prime of $I$ is minimal of height $n$. Thus, if $I$ has embedded primes, then some minimal prime has height $<n$. | |
| Aug 8, 2017 at 3:02 | comment | added | Pham Hung Quy | Nice! Do you have any example for monomial ideal? Could you explain more the sentence: for an ideal $I$ generated by $n$ elements in a regular local ring $R$, every embedded prime is contained in a minimal prime of height strictly less than $n$. | |
| Aug 8, 2017 at 2:43 | vote | accept | Pham Hung Quy | ||
| Aug 7, 2017 at 23:49 | history | edited | Jason Starr | CC BY-SA 3.0 |
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| Aug 7, 2017 at 23:39 | history | edited | Jason Starr | CC BY-SA 3.0 |
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| Aug 7, 2017 at 22:59 | history | edited | Jason Starr | CC BY-SA 3.0 |
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| Aug 7, 2017 at 21:36 | history | edited | Jason Starr | CC BY-SA 3.0 |
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| S Aug 7, 2017 at 21:06 | history | answered | Jason Starr | CC BY-SA 3.0 | |
| S Aug 7, 2017 at 21:06 | history | made wiki | Post Made Community Wiki by Jason Starr |