Timeline for What are the rational functions on a noetherian affine scheme?
Current License: CC BY-SA 4.0
11 events
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| Mar 29, 2022 at 15:58 | comment | added | Leo Alonso | @BrianShin you are completely right, see my second comment. I have to refine the idea, it is somehow valid, but the constant function with value $\infty$ does not live in $K(X)$ in any reasonable sense. I meant some subring of the set of functions with values in $\mathbf{P}^1$. | |
| Mar 29, 2022 at 15:41 | comment | added | Brian Shin | @LeoAlonso I believe the complaint is that maps to $\mathbf{P}^1$ does not form a ring. For example, there is no additive inverse for the constant map valued at $\infty$. | |
| Mar 29, 2022 at 12:19 | comment | added | Georges Elencwajg | @red_trumpet: yes, the map $u$ is indeed injective. | |
| Mar 29, 2022 at 11:50 | comment | added | Leo Alonso | For integral schemes it agrees with the usual definition. A function with values in $\mathbf{P}^1$ is defined by a pair of homogenous polynomials of the same degree. I was wondering if this idea might be generalized. | |
| Mar 29, 2022 at 11:40 | comment | added | Matthieu Romagny | @Leo is this a ring? | |
| Mar 29, 2022 at 11:37 | comment | added | Leo Alonso | Has someone considered the ring of functions with values in $\mathbf{P}^1$? | |
| Mar 29, 2022 at 11:28 | history | edited | Georges Elencwajg | CC BY-SA 4.0 |
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| Mar 29, 2022 at 9:29 | comment | added | red_trumpet | I think the map $u$ is injective: For this it is sufficient to show that $A\to\mathcal A(X)$ is injective, so suppose $a \in A$ is in the kernel. This means for each associated prime $\mathfrak p$ there exists $s_{\mathfrak p}\in A \setminus \mathfrak p$ with $a s_{\mathfrak p}=0$. Hence $s = \prod s_{\mathfrak p}$ is not contained in any associated prime, so that $s \neq 0$ and $s$ is not a zero divisor. However $as=0$ and so $a=0$. I always was under the impression that, at least for reduced $X$, the map $u$ is an isomorphism. But I'm not that sure right now. | |
| Mar 28, 2022 at 23:47 | comment | added | Will Sawin | The sheaf $\mathcal K$ is used in studying line bundles. It is needed to extend the useful characterization "line bundles are (Cartier) divisors modulo rational functions" to a reducible scheme. | |
| Mar 28, 2022 at 23:37 | history | edited | Georges Elencwajg | CC BY-SA 4.0 |
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| Mar 28, 2022 at 23:29 | history | answered | Georges Elencwajg | CC BY-SA 4.0 |