Timeline for A relation between ideals and annihilators
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 19, 2017 at 22:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 20, 2017 at 21:42 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Apr 20, 2017 at 20:45 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 21, 2017 at 20:22 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| S Feb 19, 2017 at 17:47 | history | suggested | user 1 | CC BY-SA 3.0 |
fixed grammar
|
| Feb 19, 2017 at 17:42 | review | Suggested edits | |||
| S Feb 19, 2017 at 17:47 | |||||
| Feb 17, 2017 at 16:48 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 17, 2017 at 16:36 | answer | added | Jason Starr | timeline score: 1 | |
| Jan 17, 2017 at 9:54 | comment | added | Jason Starr | Typo correction: "there exists $f\in E$" --> "there exists $f\in K$". | |
| Jan 17, 2017 at 9:12 | comment | added | Jason Starr | By the way, the ring $R$ in my previous comment does have the property in the OP's question. An ideal $K$ containing $E$ is contained in no $\mathfrak{p}_a$ precisely if, for every $a$, there exists $f\in E$ with a nonzero term $c x_a^n$ for $c\in k^\times$ and $n\geq 0$. If $K=I+J$ has this property, then let $I'$, resp. $J'$, be generated by those $x_a$ such that $I$, resp. $J$, is contained in $\mathfrak{p}_a$. For every $a$, if $x_a\in I'$, then $x_a\not\in J'$. Thus $I'J' \subset E$. Yet $I+I'$ and $J+J'$ are contained in no $\mathfrak{p}_a$. | |
| Jan 17, 2017 at 9:01 | comment | added | Jason Starr | @KevinBuzzard. Here is an example. Begin with the polynomial ring $S=k[x_1,x_2,x_3,\dots]$ in countably many variables. Let $E$ be the ideal generated by $x_ax_b$ for all $1\leq a < b$. Let $R$ be $S/E$. For every prime $\mathfrak{p}$ of $S$ that contains $E$, for every $a = 1 ,2 ,\dots$, if $x_a\not\in \mathfrak{p}$, then for every $b\neq a$, $x_b\in \mathfrak{p}$. Thus, the minimal primes over $E$ are of the form $\mathfrak{p}_a= \langle x_b | b\neq a \rangle$. For every $f\in \mathfrak{p}_a$, since $f$ has only finitely many terms, there exists $c>a$ such that $f\in \mathfrak{p}_c$. | |
| Jan 16, 2017 at 22:15 | history | edited | Gerry Myerson | CC BY-SA 3.0 |
typo in title
|
| Jan 16, 2017 at 19:50 | comment | added | Kevin Buzzard | Related: is there a (necessarily non-Noetherian) commutative ring (with 1) with the property that some minimal prime is contained in the union of all the others? | |
| Jan 16, 2017 at 19:16 | history | edited | Anderias Beto | CC BY-SA 3.0 |
added 63 characters in body
|
| Jan 16, 2017 at 19:12 | review | First posts | |||
| Jan 16, 2017 at 19:13 | |||||
| Jan 16, 2017 at 19:10 | history | asked | Anderias Beto | CC BY-SA 3.0 |