Timeline for Spectrum of a ring (studied by Krull?) of rational functions
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 17, 2020 at 8:54 | answer | added | Anton Mellit | timeline score: 3 | |
| Oct 16, 2020 at 20:05 | vote | accept | lefuneste | ||
| Oct 16, 2020 at 20:04 | comment | added | lefuneste | Dear @Piotr: you write "Maybe I'm mistaken, but what you wrote shows that the maximal ideal is NOT principal" You are not mistaken at all and your fantastic comment explains away all the problems I had with my erroneous belief that the maximal ideal was principal. I can't thank you enough for this brilliant explanation. | |
| Oct 16, 2020 at 19:51 | history | became hot network question | |||
| Oct 16, 2020 at 19:40 | comment | added | Piotr Achinger | So the maximal ideal of $R$ is $xk[x,y]_{(x)}\cap R$, which is not equal to $xR$, in fact it is not finitely generated, because the system of generators $\{fx\,:\, f\in k(y)\}$ does not admit a finite generating subset. See my edit below. | |
| Oct 16, 2020 at 19:15 | comment | added | Piotr Achinger | @lefuneste Maybe I'm mistaken, but what you wrote shows that the maximal ideal is NOT principal. Like in the example above, the maximal ideal is generated by the elements $ax$ with $a\in \mathbf{C}$ and you need infinitely many of those generators... | |
| Oct 16, 2020 at 19:08 | comment | added | Piotr Achinger | Compare with the ring of all $f = \sum_{n\geq 0} a_n x^n$ in $\mathbf{C}[[x]]$ with $a_0\in \mathbf{Q}$. | |
| Oct 16, 2020 at 19:00 | answer | added | Piotr Achinger | timeline score: 6 | |
| Oct 16, 2020 at 19:00 | answer | added | Steven Landsburg | timeline score: 5 | |
| Oct 16, 2020 at 18:58 | comment | added | lefuneste | I think that $R$ is a one-dimensional local domain with principal maximal ideal (and spectrum of cardinality $2$) . However it is not a DVR, because $x$ and $xy$ are in the maximal ideal but neither is divisible by the other. The only explanation for this strange result would be that $R$ is not noetherian. | |
| Oct 16, 2020 at 18:48 | comment | added | Piotr Achinger | It's not obvious to me however whether the underlying space of $\operatorname{Spec} R$ is the pushout of the underlying spaces of the three affine schemes. That would make it homeomorphic to $\operatorname{Spec} A$, i.e. two points, one in the closure of the other. That would agree with my inability to find prime ideals other than $(0)$, $(x)$. | |
| Oct 16, 2020 at 18:46 | comment | added | Piotr Achinger | Now your ring $R$ is seen to be the fiber product $A \times_{k(y)} k$. By Stacks Project, Tag 07RS stacks.math.columbia.edu/tag/07RS , $\operatorname{Spec} R$ is the pushout of $\operatorname{Spec} k \leftarrow \operatorname{Spec} k(y) \to \operatorname{Spec} A$ in the category of schemes. | |
| Oct 16, 2020 at 18:41 | comment | added | Piotr Achinger | In other words, the localization $A$ of $k[x, y]$ at the prime ideal $(x)$ is a discrete valuation ring with residue field $k(y)$. We are looking at the preimage of $k\subseteq k(y)$ in $A$. So you can consider the abstract setting: consider a dvr $A$ with maximal ideal $\mathfrak{m}$ and a subfield $k \subseteq A/\mathfrak{m}$; what does the spectrum of the preimage of $k$ in $A$ look like? | |
| Oct 16, 2020 at 18:29 | comment | added | lefuneste | @Steven Landsburg: Yes, your representation of the elements of $R$ is efficient and absolutely correct. | |
| Oct 16, 2020 at 18:15 | comment | added | Steven Landsburg | So just to clarify: Your ring consists of all $\Big(cp(Y)+Xf(X,Y)\Big)/\Big(p(Y)+Xg(X,Y)\Big)$ such that $p(Y)\neq 0$ and $c\in k$. Yes? | |
| Oct 16, 2020 at 15:06 | comment | added | YCor | OK: so what remains from my previous comment is that this is a local ring with the two given prime ideals (0 and maximal), but I don't immediately see whether there are any others. | |
| Oct 16, 2020 at 14:56 | comment | added | lefuneste | Dear @YCor: Apologies for my ambiguous formulation. I have made an edit making what I meant more explicit, since I want the fraction you mention to belong to $R$ | |
| Oct 16, 2020 at 14:51 | history | edited | lefuneste | CC BY-SA 4.0 |
added 510 characters in body
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| Oct 16, 2020 at 12:55 | comment | added | YCor | If my understanding is correct, it's a local ring, with two obvious prime ideals, 0 and the maximal one (zero on the vertical axis). Further prime ideals are obtained by choosing a non-vertical line $D$, intersecting the vertical axis at $p$ and mapping $f\in R$ to its germ at $p$ on $D$. | |
| Oct 16, 2020 at 12:46 | comment | added | YCor | I understand that $\frac{y+x}{y-x}\notin R$, right? (it's constant on this $x=0$ axis but not "defined" at $(0,0)$). | |
| Oct 16, 2020 at 11:20 | history | asked | lefuneste | CC BY-SA 4.0 |