Example of PID and its maximal ideal

  • Context: Graduate 
  • Thread starter Thread starter tsang
  • Start date Start date
  • Tags Tags
    Example Pid
Click For Summary
SUMMARY

The discussion centers on the properties of the principal ideal domain (PID) \(\mathbb{Z}_{(p)}\), defined as the subring of \(\mathbb{Q}\) where the denominator is not divisible by a prime \(p\). It is established that \(\mathbb{Z}_{(p)}\) is a PID with exactly one maximal ideal, which can be understood through the structure of its ideals. Specifically, all ideals in \(\mathbb{Z}_{(p)}\) can be expressed in the form \(\{\frac{m}{n} \in \mathbb{Z}_{(p)} \mid m \in I\}\) for some ideal \(I\) in \(\mathbb{Z}\), illustrating that the addition of non-divisible denominators reduces the number of ideals.

PREREQUISITES
  • Understanding of principal ideal domains (PIDs)
  • Familiarity with the structure of the integers \(\mathbb{Z}\)
  • Knowledge of subrings and their properties in ring theory
  • Basic concepts of ideals in ring theory
NEXT STEPS
  • Study the properties of principal ideal domains (PIDs) in greater detail
  • Explore the concept of maximal ideals in ring theory
  • Learn about the structure and properties of the ring \(\mathbb{Z}_{(p)}\)
  • Investigate the relationship between ideals in \(\mathbb{Z}\) and their extensions in subrings
USEFUL FOR

Mathematicians, algebraists, and students studying abstract algebra, particularly those interested in ring theory and the properties of principal ideal domains.

tsang
Messages
15
Reaction score
0
I got one example on my notes about PID and maximal ideal. I feel it is a strange example as it doesn't make sense to me, and there are no explanations. It says:

For a prime p\in\mathbb{N}, denote by \mathbb{Z}_{(p)} the subring of \mathbb{Q} given by
\mathbb{Z}_{(p)}={\frac{m}{n} \in\mathbb{Q}|p does not divide n}.
Then \mathbb{Z}_{(p)} is a PID, and it has exactly one maximal ideal.


I can't see the reason of this example at all, and I'm not able to imagine what are the ideals like in Z_(p), can anyone please explain to me why it is a PID and only has one maximal ideal? Thanks a lot.
 
Physics news on Phys.org
Show that all ideals are of the form

\{\frac{m}{n}\in \mathbb{Z}_{(p)}~\vert~m\in I\}

for I an ideal in \mathbb{Z}.
 
intuitively, the more units there are, i.e. the more denominators are allowed, the fewer ideals there are, since any ideal containing a unit is the whole ring.

so you started from a pid, namely Z, and added in a lot of denominators, i.e. anything not divisible by p, so you lost a lot of ideals, and the remaining ones are probably still generated by the same principal generators as before.

i.e. try intersecting an ideal of your ring with Z, and see if the generator of that ideal also generates your original ideal.

anytime the set of non units itself forms an ideal, that ideal contains all others. what are the non units here?
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
2K
Replies
3
Views
1K
  • · Replies 31 ·
2
Replies
31
Views
3K
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K