Timeline for Adjoining new factors for primes in UFDs
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 18 at 13:59 | comment | added | Pace Nielsen | @LucGuyot I was not familiar with Nagata's criterion. It is very convenient in this setting, and streamlines things. Thanks! | |
| Apr 18 at 13:30 | comment | added | Pace Nielsen | @JesseElliott Thanks for that reference. Lemma 11.1 is quite close to what we want. | |
| Apr 17 at 22:45 | comment | added | Luc Guyot | Not sure if the following is less ad hoc than what you already have: (1) The element $s$ of $S = R[s, p/s]$ is prime because $S/(s) \simeq R/(p)[s']$. (2) A non-zero element of $S$ has only finitely many divisors up to multiplication by a unit of $R$; for this use the $p$-valuation of the coefficient of least degree to reduce to a similar statement in the UFD $R[s]$. Thus $S$ has ACCP. By Nagata's criterion, the ring $S$ is a UFD; indeed its localization at $s$ is the UFD $R[s, s^{-1}]$. | |
| Apr 17 at 1:45 | comment | added | Jesse Elliott | I recall that Fossum's book has examples computing the class group of $k[X_1,X_2,\ldots,X_n]/(F)$ for various quadratic forms $F$, but I don't own a copy of the book and can't check if it has an example that might include yours. | |
| Apr 17 at 1:34 | comment | added | Jesse Elliott | I suppose a third approach, which might generalize, is to try to compute the (divisor) class group of $R[s,s':ss' = p] = R[X,Y]/(XY-p)$ in terms of the class group of $R$, using techniques of Fossum's book, The Class Group of a Krull domain. If $p$ is prime in $R$, then probably the class groups are the same. If $p$ is the square of a prime in $R$, then I'd guess the class group is cyclic of order $2$ if $R$ is a UFD. That's the only type of generalization I can think of to generalize your result. | |
| Apr 16 at 23:25 | history | asked | Pace Nielsen | CC BY-SA 4.0 |