Timeline for Bézout ring with non-trivial Picard group?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 27, 2020 at 10:11 | comment | added | Jesse Elliott | lefuneste, did you verify what I claimed, that the definition of Cl(X) in Hartshorne is not the same as my Cl(R)? I still stand by my answer. | |
| Oct 23, 2020 at 2:22 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
added 22 characters in body
|
| Oct 23, 2020 at 2:18 | comment | added | Jesse Elliott | This isn't correct. "Cl X" in Hartshorne refers to the Weil divisor class group. As I noted, I was referring to the ideal class group of invertible fractional ideals modulo regular principal fractional ideals. | |
| Oct 22, 2020 at 18:39 | comment | added | lefuneste | My example above shows that the assertion in your following sentence according to which $\text{Cl}(R)$ and $\text{Pic}(R)$ are isomorphic for integral domains or noetherian rings is false too. | |
| Oct 22, 2020 at 18:30 | comment | added | lefuneste | You write "there is a canonical inclusion $\text{Cl}(R) \longrightarrow \text{Pic}(R)$...". This is not true: there is no canonical map $\text{Cl}(R) \longrightarrow \text{Pic}(R)$, let alone an injective one. Hartshorne shows on page 134 of his Algebraic Geometry that for the ring $R$ of regular functions on the the affine cone $xy-z^2=0$ in affine $3$-space $\mathbb A^3_k$ over a field $k$ we have $\text{Cl}(R)=\mathbb Z/2 \mathbb Z$ but $\text{Pic}(R)=0$. | |
| Nov 20, 2019 at 19:20 | comment | added | Badam Baplan | Thanks for the comments, Jesse! Another straightforward way to see that (regular) Bezout rings with few zero divisors have trivial picard groups is by noting that $T(R)$ will be semi-local (hence $Pic(T(R)) = 0$), and so again every invertible module is isomorphic to a dense finitely generated ideal of $R$ which will in turn be regular (since the zero-divisors of $R$ are contained in finitely many prime ideals), and thus principal by assumption. | |
| Nov 14, 2019 at 5:29 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
added 285 characters in body
|
| Nov 14, 2019 at 5:22 | history | answered | Jesse Elliott | CC BY-SA 4.0 |