# Primes in set of rational numbers

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.

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 $X=(2^n)$.

doesn't 1 = 1 mod 2?

doesn't 1 = 1 mod 2?

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

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

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