Primes in set of rational numbers

  • Thread starter math_grl
  • Start date
  • #1
49
0
There was a part c and d from a question I couldn't answer.
Let [tex]R = \{ a/b : a, b \in \mathbb{Z}, b \equiv 1 (\mod 2) \}[/tex].

a) was find the units, b) was show that [tex]R\setminus U(R)[/tex] is a maximal ideal. Both I was successful. But

c) is find all primes, which I believe i only found one....the rational number 2.

d) find all ideals and show that [tex]R[/tex] is a PID.

Any help would be appreciated.
 

Answers and Replies

  • #2
430
3
c) Write an element,
[tex]p = 2^n \frac{a}{b}[/tex]
where a and b are odd. If n>1, then let, [itex]x = 2^{n-1}[/itex] and [itex]y =2[/itex]. Then [itex]p | xy[/itex], but [itex]p \not | x[/itex] and [itex]p \not| y[/itex]. Thus p is not a prime if n>1.
If n=1, and, p|xy, then either x or y must have a factor 2. Assume WLOG that 2|x, then p|x.
If n=0, then p is a unit.
Thus only associates of 2 are prime.

d) For some ideal X of R let n be the largest non-negative integer such that [itex]2^n[/itex] divides all elements in X. Then [itex]2^n \in X[/itex] since there exists odd integers a,b such that [itex]2^n \frac{a}{b} \in X[/itex], but then [itex]2^n\frac{a}{b}\frac{b}{a}=2^n[/itex]. We have [itex]X=(2^n)[/itex].
 
  • #3
1,710
5
doesn't 1 = 1 mod 2?
 
  • #4
430
3
doesn't 1 = 1 mod 2?

Yes this is true, but how does it relate to anything in any of the previous posts?
 
  • #5
1,710
5
Yes this is true, but how does it relate to anything in any of the previous posts?

i didn't read your proof but doesn't my comment imply that for every prime p in Z the element r=p/1 is in R
 
  • #6
430
3
i didn't read your proof but doesn't my comment imply that for every prime p in Z the element r=p/1 is in R

I misunderstood your comment. math_grl very likely means prime element in the general sense of prime elements in an integral domain. In an integral domain R we say that a non-unit p is a prime element if p|ab imply p|a or p|b. This is equivalent to the ordinary notion of prime numbers in Z, but it need not be the same for other rings.

In the R given by math_grl 3 for instance is not a prime because it's a unit (i.e. 1|3 and 3 |1). To see this note that 3 = 3*1 so 1|3 and 1 = 3*1/3 so 3|1. On the other hand 6 is prime in R.
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
41,847
964
But ice109's point is, I believe, that all of the usual primes still are primes in this domain.
 
  • #8
430
3
But ice109's point is, I believe, that all of the usual primes still are primes in this domain.

But they aren't. All the usual odd primes in Z are units in this domain and therefore not primes. This is due to the fact that if p is an odd prime in Z, then 1/p is in the domain and since p(1/p)=1 we have p|1. He is of course completely right if prime element is taken to mean prime element in Z, but as we're working in R it would seem natural to consider prime elements in R.
 
  • #9
49
0
Last edited by a moderator:

Related Threads on Primes in set of rational numbers

  • Last Post
Replies
2
Views
2K
Replies
10
Views
8K
Replies
5
Views
3K
Replies
4
Views
2K
Replies
6
Views
5K
Replies
3
Views
2K
Replies
1
Views
4K
Replies
3
Views
2K
  • Last Post
Replies
4
Views
3K
Replies
7
Views
4K
Top