Primes in set of rational numbers

[math_grl]

There was a part c and d from a question I couldn't answer.
Let R = \{ a/b : a, b \in \mathbb{Z}, b \equiv 1 (\mod 2) \}.

a) was find the units, b) was show that R\setminus U(R) 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 R is a PID.

Any help would be appreciated.

 

[rasmhop]

c) Write an element,
p = 2^n \frac{a}{b}
where a and b are odd. If n>1, then let, x = 2^{n-1} and y =2. Then p | xy, but p \not | x and p \not| y. 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 2^n divides all elements in X. Then 2^n \in X since there exists odd integers a,b such that 2^n \frac{a}{b} \in X, but then 2^n\frac{a}{b}\frac{b}{a}=2^n. We have X=(2^n).

 

[ice109]

doesn't 1 = 1 mod 2?

 

[rasmhop]

doesn't 1 = 1 mod 2?

Yes this is true, but how does it relate to anything in any of the previous posts?

 

[ice109]

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

 

[rasmhop]

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.

 

[HallsofIvy]

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

 

[rasmhop]

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.

 

[math_grl]

Thanks rasmhop.

For ice: http://en.wikipedia.org/wiki/Prime_number#Prime_elements_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.