Re: Ideals
- From: magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin)
- Date: Sat, 16 Jun 2007 20:12:36 +0000 (UTC)
In article <1182022288.900335.44890@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Subluxian <cbrown@xxxxxxxxxxxxxxxxx> wrote:
On Jun 16, 12:04 pm, bockermann <fstr...@xxxxxx> wrote:
Hello,
let A be a ring and suppose that (f_1,...,f_n)=A with Elements f_i of
A.
Is it true that (f_1^k,...,f_n^k)=A for any k>0?
Counterexample: Let A be addition/multiplication mod 3. Then A =
{0,1,2}; but {0^2, 1^2, 2^2} = {0, 1, 1} = {0, 1}.
This is not a counterexample. The IDEAL generated by 0, 1, and 2 is
all of A, but so is the ideal generated by 0 and 1. The question is
about ideals, not sets.
The result holds for commutative rings with 1, at least: If A is a
ring with 1, let k>=1 and let I = (f_1^k,...,f_n^k); if I is not the
entire ring, then there exists a maximal ideal M such that I is
contained in M. Since M is maximal, it is prime. In particular, since
f_i^k lies in M, then f_i lies in M (it being a prime ideal). Thus, M
contains f1, ..., f_n. However, (f_1,..,f_n)=A by assumption,
contradicting that M is a maximal ideal. Thus, the assumption that I
is not the entre ring cannot hold. So (f_1^k,...,f_n^k)=A, as claimed.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
.
- Follow-Ups:
- Re: Ideals
- From: bockermann
- Re: Ideals
- From: Subluxian
- Re: Ideals
- References:
- Ideals
- From: bockermann
- Re: Ideals
- From: Subluxian
- Ideals
- Prev by Date: Number to Letter Counting algorithim needed
- Next by Date: Re: Operator Theory Problem
- Previous by thread: Re: Ideals
- Next by thread: Re: Ideals
- Index(es):
Relevant Pages
|
