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Need help with the final laps of this long ring theory problem
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Ryan Reich
science forum Guru Wannabe


Joined: 21 May 2005
Posts: 120

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

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)(3-2i). 3+2i and 3-2i 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=(3-2i). (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
3-2i 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
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Doug B
science forum beginner


Joined: 03 Feb 2005
Posts: 27

PostPosted: Sun Jun 18, 2006 2:40 pm    Post subject: Need help with the final laps of this long ring theory problem Reply with 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)(3-2i). 3+2i and 3-2i 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=(3-2i). (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
3-2i is prime. Can anyone help me out with this last sprint (or
correct me if I'm way off course)?
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