Rings

Maths and statistics discussion, revision, exam and homework help.

Announcements Posted on
Enter our travel-writing competition for the chance to win a Nikon 1 J3 camera 20-05-2013
IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
Search Thread
Advanced Search
Sign in to Reply
  1. JBKProductions's Avatar
    1 13-05-2012  21:59
    • Overlord in Training
    • Posts: 2,105
    Rings
    A bit of a proof that I don't understand again.
    First of all I'll state the theorem
    If R= \mathbb{Z} / p^k \mathbb{Z} where k \geq 1 and p is prime then |R^*|= p^k-p^{k-1} .
    Starting about halfway through it says this, "Suppose p divides a ( a \in R ). Then a = p\hat{a} and p^{k-1}a \cong 0 \ mod \ p^k ." Why is p^{k-1}a \cong 0 \ mod \ p^k if p divides a?
    EDIT: IF more information is needed let me know, I'm just too lazy to type it all out if it is not needed.
    Last edited by JBKProductions; 13-05-2012 at 22:10.
  2. mathz's Avatar
    2 13-05-2012  22:19
    • Exalted Member
    Re: Rings
    maybe im missing something here but if a=pa' then p^k-1a=p^ka'
  3. ztibor's Avatar
    3 13-05-2012  22:22
    • Peer Of The TSR Realm
    • Location: Hungary
    • Posts: 1,540
    Re: Rings
    (Original post by JBKProductions)
    A bit of a proof that I don't understand again.
    First of all I'll state the theorem
    If R= \mathbb{Z} / p^k \mathbb{Z} where k \geq 1 and p is prime then |R^*|= p^k-p^{k-1} .
    Starting about halfway through it says this, "Suppose p divides a ( a \in R ). Then a = p\hat{a} and p^{k-1}a \cong 0 \ mod \ p^k ." Why is p^{k-1}a \cong 0 \ mod \ p^k if p divides a?
    I think because p^{k-1}a=p^{k-1}(p\hat{a})=p^k \hat{a}
  4. JBKProductions's Avatar
    4 13-05-2012  22:27
    • Overlord in Training
    • Posts: 2,105
    Rings
    Ok thanks, I understand it now.
  5. JBKProductions's Avatar
    5 13-05-2012  22:49
    • Overlord in Training
    • Posts: 2,105
    Re: Rings
    I have attached the proof here, can someone explain the last part to me, I'm not sure exactly why there are exactly p^{k-1} elements divisible by p.
    Last edited by JBKProductions; 14-05-2012 at 20:50.
  6. nuodai's Avatar
    6 13-05-2012  23:38
    • PS Helper
    • TSR Legend
    Re: Rings
    (Original post by JBKProductions)
    I have attached the proof here, can someone explain the last part to me, I'm not sure exactly why there are exactly p^{k-1} elements divisible by p.
    Think about what \mathbb{Z} / p^k\mathbb{Z} is: it's (essentially) the remainder that you get when you divide an integer by p^k; that is \mathbb{Z} / p^k \mathbb{Z} = \{ 0, 1, 2, \dots, p^k-2, p^k-1 \}. How many of these are divisible by p? Well you have 0, p, 2p, 3p, \dots, p^{k-2}p, (p^{k-2}+1)p, \dots, (p^{k-1}-1)p. (Note that p^{k-1}.p=p^k=0.)

    How many of these are there?
    Last edited by nuodai; 13-05-2012 at 23:46.
  7. JBKProductions's Avatar
    7 14-05-2012  00:10
    • Overlord in Training
    • Posts: 2,105
    Re: Rings
    (Original post by nuodai)
    Think about what \mathbb{Z} / p^k\mathbb{Z} is: it's (essentially) the remainder that you get when you divide an integer by p^k; that is \mathbb{Z} / p^k \mathbb{Z} = \{ 0, 1, 2, \dots, p^k-2, p^k-1 \}. How many of these are divisible by p? Well you have 0, p, 2p, 3p, \dots, p^{k-2}p, (p^{k-2}+1)p, \dots, (p^{k-1}-1)p. (Note that p^{k-1}.p=p^k=0.)

    How many of these are there?
    EDIT: I understand it now, thanks.
    Last edited by JBKProductions; 14-05-2012 at 00:21.
Sign in to Reply
Search Thread
Advanced Search
Share this discussion:  
Useful resources

Articles:

Study Help rules and posting guidelinesStudy Help Member of the Month nominationsGuide to MathematicsMathematics resourcesMathematics websitesMathematics revision notes in the TSR wikiCambridge Assessment admissions tests (including STEP)How to use LaTeXCareers in Mathematics

Useful Dates:

Quick Link:

Unanswered Maths Threads

Groups associated with this forum:

View associated groups
Article updates
Moderators

We have a brilliant team of more than 60 volunteers looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is moderated by:

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22

Registered Office: International House, Queens Road, Brighton, BN1 3XE

Shortcuts

Get Started

TSR Group

Using TSR

Info

Connect with TSR

Reputation gems:
The Reputation gems seen here indicate how well reputed the user is, red gem indicate negative reputation and green indicates a good rep.
Post rating score:
These scores show if a post has been positively or negatively rated by our members.