Timeline for Quotient rings in formal power series
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 14, 2022 at 7:29 | history | edited | Chris Grossack | CC BY-SA 4.0 | added 42 characters in body |
| May 16, 2014 at 20:49 | history | edited | user26857 | edited tags | |
| May 16, 2014 at 19:13 | vote | accept | Wooster | ||
| May 16, 2014 at 19:08 | answer | added | rschwieb | timeline score: 3 | |
| Feb 9, 2014 at 19:19 | comment | added | zcn | Right, in a PID every ideal is principal, so the primes are precisely of the form $(p)$ for $p$ a prime element. Why do you say that $0$ is not a prime element though (in a domain)? Perhaps a matter of taste, but I see no reason not to include it | |
| Feb 9, 2014 at 19:16 | comment | added | Daniel Fischer | @user115654 Right. I often forget that one when dealing with principal ideal domains, since $0$ is not a prime element, and of course the prime ideals are the ideals of the form $(p)$ for primes $p$ then, aren't they? | |
| Feb 9, 2014 at 19:13 | comment | added | zcn | @DanielFischer: and $0$ (since $k[[x]]$) is a domain). But these are all of them | |
| Feb 9, 2014 at 15:06 | comment | added | Daniel Fischer | And in general $(x+\langle x^n\rangle)^n = \langle x^n\rangle$, so the only prime ideal is $\langle x\rangle$. | |
| Feb 9, 2014 at 15:04 | comment | added | Wooster | Well in that case $f = x$ gives $(x + <x^2>)^2 = x^2 + <x^2> = 0 + <x^2>$ so that is also not an integral domain? Thank you for all your help! | |
| Feb 9, 2014 at 14:57 | comment | added | Daniel Fischer | It looks like you're on the right track. What about $\langle x^2\rangle$? | |
| Feb 9, 2014 at 14:56 | comment | added | Wooster | Ah yes I see that. So there are no elements $f + <x^n>$ s.t $(f+<x^n>)^k = f^k + <x^n> = 0 + <x^n>$ for some k. So for example the $<x^3>$ case is not an integral domain because taking $f=x^2$ gives $f^2 = 0$? Am I thinking of this the right way? I'm not quite sure how we can get a condition on n, thanks | |
| Feb 9, 2014 at 14:45 | comment | added | Daniel Fischer | If $\deg f < n$ and $\deg g < n$, then also $\deg (f-g) < n$. If further $x^n \mid (f-g)$, from degree/order considerations it follows that $f-g = 0$. $k[[x]]/\langle x^n\rangle$ is not isomorphic to $\mathbb{Z}/(n)$. It is an $n$-dimensional $k$-vector space (and $k$-algebra). If it is an integral domain, it does in particular contain no nilpotent elements (except $0$). What does that say about $n$? | |
| Feb 9, 2014 at 14:39 | comment | added | Wooster | Okay thanks! Why does the polynomial have to be unique? Is it correct that this is then isomorphic to the integers modulo n? | |
| Feb 9, 2014 at 14:37 | comment | added | Daniel Fischer | Yes, every coset has one unique element that is a polynomial of degree less than $n$. In general, $k[[x]]/\langle x^n\rangle$ is not an integral domain, only for specific $n$. | |
| Feb 9, 2014 at 14:26 | history | asked | Wooster | CC BY-SA 3.0 |