Author 
Message 
Ryan Reich science forum Guru Wannabe
Joined: 21 May 2005
Posts: 120

Posted: Sun Jun 18, 2006 3:58 pm Post subject:
Re: Need help with the final laps of this long ring theory problem



On 18 Jun 2006 07:40:21 0700, Doug B <doug_protocols@yahoo.com> wrote:
Quote:  Hi, I've been working on this long ring theory problem:
"Let R={a+bi  a,b in Z} be the ring of Gaussian integers.
a) Show that I=(13) is not prime.
b) Show that R/I is not a field.
c) Find an ideal J, not R itself, such that J contains I and R/J is a
field."
For a, we observe 13 = 9 + 4 = (3+2i)(32i). 3+2i and 32i are clearly
not in I so primality is forfeit.
For b, we recall that in any commutative ring with 1, maximal ideals
are prime. So I is not maximal, so R/I is not a field.
Part c is where I'm stuck. I'm guessing the answer is either J=(3+2i)
and/or J=(32i). (R is a PID afterall, so how much more complicated
could J be?) These certainly contain I, no problem. The difficulty is
proving 1. that this J is not R itself, 2. that this J is maximal
(well, technically the former is immediate if we obtain the latter, of
course). I guess it really comes down to showing that one of 3+2i or
32i is prime. Can anyone help me out with this last sprint (or
correct me if I'm way off course)?

You've already observed that 3 + 2i (or 3  2i) have norm 13, which is a prime
integer. Norm is multiplicative.

Ryan Reich
ryan.reich@gmail.com
sci.math 

Back to top 


Doug B science forum beginner
Joined: 03 Feb 2005
Posts: 27

Posted: Sun Jun 18, 2006 2:40 pm Post subject:
Need help with the final laps of this long ring theory problem



Hi, I've been working on this long ring theory problem:
"Let R={a+bi  a,b in Z} be the ring of Gaussian integers.
a) Show that I=(13) is not prime.
b) Show that R/I is not a field.
c) Find an ideal J, not R itself, such that J contains I and R/J is a
field."
For a, we observe 13 = 9 + 4 = (3+2i)(32i). 3+2i and 32i are clearly
not in I so primality is forfeit.
For b, we recall that in any commutative ring with 1, maximal ideals
are prime. So I is not maximal, so R/I is not a field.
Part c is where I'm stuck. I'm guessing the answer is either J=(3+2i)
and/or J=(32i). (R is a PID afterall, so how much more complicated
could J be?) These certainly contain I, no problem. The difficulty is
proving 1. that this J is not R itself, 2. that this J is maximal
(well, technically the former is immediate if we obtain the latter, of
course). I guess it really comes down to showing that one of 3+2i or
32i is prime. Can anyone help me out with this last sprint (or
correct me if I'm way off course)? 

Back to top 


Google


Back to top 



The time now is Tue Oct 23, 2018 12:33 am  All times are GMT

