Timeline for Can a non-zero non-prime ideal become prime in a smaller ring?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 15, 2023 at 17:30 | answer | added | Sándor Kovács | timeline score: 2 | |
| May 22, 2023 at 22:14 | history | became hot network question | |||
| May 22, 2023 at 16:00 | comment | added | R. van Dobben de Bruyn | @M.G. interesting. I suppose I sometimes think of this example as coming from the free algebras on the monoids $\mathbf N$ and $\mathbf N \setminus \{1\}$, which is kind of similar to what you're describing in that $\mathbf N \setminus \{1\}$ is the monoid obtained from the ideal $\mathbf N_{\geq 2} \subseteq \mathbf N$ (which is only a semigroup) by adjoining a zero element. (The commutative monoid point of view is common in toric and logarithmic algebraic geometry.) | |
| May 22, 2023 at 15:44 | vote | accept | M.G. | ||
| May 22, 2023 at 15:43 | answer | added | R. van Dobben de Bruyn | timeline score: 7 | |
| May 22, 2023 at 15:40 | comment | added | M.G. | @R.vanDobbendeBruyn: One general procedure indicated by your example seems to be to look at ideals $\mathfrak{a}$ as non-unital $k$-algebras and then picking such an $\mathfrak{a}$ whose "unitization" is strictly smaller than the original $k$-algebra. | |
| May 22, 2023 at 15:21 | comment | added | R. van Dobben de Bruyn | I was just working on writing up an answer with some further comments! | |
| May 22, 2023 at 15:18 | comment | added | M.G. | @R.vanDobbendeBruyn: you should probably post your example as an answer so as to not leave the question open-ended. | |
| May 22, 2023 at 15:12 | comment | added | Karl Schwede | I suppose it's worth saying, for a ring and its normalization $R \subseteq S$, the conductor $\mathrm{Ann}_R(S/R)$ is the largest ideal that is simultaneously an ideal in both rings. Normalizations then give lots of such examples (if you view Specs of non-normal rings as gluings of normal rings). | |
| May 22, 2023 at 15:11 | comment | added | Dry Bones | Oh of course, it's the same ideal, not ideals with the same generator! Ha! XD | |
| May 22, 2023 at 15:00 | comment | added | M.G. | @R.vanDobbendeBruyn: thank you for the nice simple example to (Q1)! I was thinking in a very different direction... To (Q2), yeah the point was for $S$ to be a source of non-trivial ideals, but it no longer matters since you provided an answer to (Q1). | |
| May 22, 2023 at 14:57 | comment | added | R. van Dobben de Bruyn | @DryBones I too gave a similar example at first (as did Jason), but the question is about a situation where $I$ is an ideal in both $R$ and $S$ (here you have to take the extension $IS$ which is the ideal generated by $I$ in $S$). | |
| May 22, 2023 at 14:55 | comment | added | Dry Bones | What about the extension $\mathbb R[x]\subset\mathbb C[x]$ and the ideal $(x^2+1)$ ? | |
| May 22, 2023 at 14:53 | comment | added | Jason Starr | @R.vanDobbendeBruyn You are correct: I misread the question. | |
| May 22, 2023 at 14:50 | comment | added | R. van Dobben de Bruyn | (Q1) The ideal $(x^2,x^3) \subseteq k[x]$ is not prime, but its restriction to the subring $k[x^2,x^3]$ is. (Q2) The inclusion $\mathbf Z \subseteq \mathbf Q$ satisfies $\dim R > \dim S$, but there are not a lot of nonzero ideals in $\mathbf Q$. | |
| May 22, 2023 at 14:14 | history | asked | M.G. | CC BY-SA 4.0 |