# Primes in set of rational numbers

1. Jan 21, 2010

### 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.

2. Jan 22, 2010

### 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 $X=(2^n)$.

3. Jan 22, 2010

### ice109

doesn't 1 = 1 mod 2?

4. Jan 22, 2010

### rasmhop

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

5. Jan 22, 2010

### ice109

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. Jan 22, 2010

### rasmhop

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. Jan 23, 2010

### HallsofIvy

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

8. Jan 23, 2010

### rasmhop

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. Jan 25, 2010

### math_grl

Last edited by a moderator: Apr 24, 2017