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Primes in set of rational numbers 
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#1
Jan2110, 04:27 PM

P: 50

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. 


#2
Jan2210, 11:51 AM

P: 424

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^{n1}[/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, pxy, then either x or y must have a factor 2. Assume WLOG that 2x, then px. 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 nonnegative 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
Jan2210, 12:56 PM

P: 1,705

doesn't 1 = 1 mod 2?



#4
Jan2210, 01:06 PM

P: 424

Primes in set of rational numbers



#5
Jan2210, 01:50 PM

P: 1,705




#6
Jan2210, 02:59 PM

P: 424

In the R given by math_grl 3 for instance is not a prime because it's a unit (i.e. 13 and 3 1). To see this note that 3 = 3*1 so 13 and 1 = 3*1/3 so 31. On the other hand 6 is prime in R. 


#7
Jan2310, 09:47 AM

Math
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Thanks
PF Gold
P: 39,363

But ice109's point is, I believe, that all of the usual primes still are primes in this domain.



#8
Jan2310, 10:04 AM

P: 424




#9
Jan2510, 04:03 PM

P: 50

Thanks rasmhop.
For ice: http://en.wikipedia.org/wiki/Prime_n...ments_in_rings I probably could have said explicitly that R was a ring but figured it was implied as the question concerns PID's and units. 


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