Timeline for Annihilators of sum of two ideals
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 10, 2022 at 9:36 | comment | added | Zach Teitler | Yes, for a monomial element and monomial ideals it is true. | |
| Mar 10, 2022 at 8:57 | comment | added | Amir Mafi | I think the question is true when $I$ and $J$ are monomial ideals and $x$ is monomial element of a polynomial ring. | |
| Mar 10, 2022 at 8:55 | comment | added | Amir Mafi | Dear Zach, Thank you so much for your nice explanations. | |
| Mar 10, 2022 at 8:04 | comment | added | Zach Teitler | Sorry, my last comment was unclear. I was taking the ideals $I=(x)$ and $J=(y)$, and the element $x+y$. The original question uses "$x$" for the element, and for the value of "$x$" I am taking $x+y$. Perhaps it would be better to restate the question as: is $(I+J):a = (I:a)+(J:a)$? And then I am taking $a=x+y$. Or perhaps I should have used a polynomial ring with different variable names... Anyway, I apologize for any confusion. | |
| Mar 10, 2022 at 7:30 | comment | added | Zach Teitler | Sorry, but no, it is not sufficient for $I$ and $J$ to be monomial ideals. In $R = k[x,y]$ with $I=(x)$, $J=(y)$, the left hand side is $(I+J):(x+y) = (x,y):(x+y) = (1)$. Since $I:(x+y) = I$ and $J:(x+y) = J$, the right hand side is $I+J=(x,y)$. It's true that if $R = S[x_1,\dotsc,x_n]$ is a polynomial ring, $I$ and $J$ are monomial ideals, and $a \in R$ is a monomial, then $(I+J):a = (I:a)+(J:a)$. | |
| Mar 10, 2022 at 6:18 | comment | added | Amir Mafi | For monomial ideals is true | |
| Mar 10, 2022 at 6:11 | history | edited | Amir Mafi | CC BY-SA 4.0 |
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| Mar 7, 2022 at 6:42 | history | answered | Amir Mafi | CC BY-SA 4.0 |