# Prove that if a is prime, then b is prime

1. Jul 27, 2011

### 83956

Prove that if "a" is prime, then "b" is prime

1. The problem statement, all variables and given/known data
Suppose that "a" and "b" are associates. Prove that if "a" is prime, then "b" is prime.

2. Relevant equations

3. The attempt at a solution

By the definition of an associate a=ub where u is a unit.

2. Jul 27, 2011

### micromass

Staff Emeritus
Re: Associates

What did you try already?? If you show us where you're stuck, then we'll know where to help!!

3. Jul 27, 2011

### 83956

Re: Associates

This is my understanding.

a=ub where u is a unit. By the definition of associates a divides b, and b divides a. So if a is a prime as in the problem statement, then the only divisors of a are 1 and a. Thus, b is either 1 or a. But 1 is a unit so b is a prime equal to a.

But I am being told my reasoning is not right. I understand that the units vary in different rings, but don't quite understand how to factor that into this problem.

4. Jul 27, 2011

### micromass

Staff Emeritus
Re: Associates

What is your definition of a prime number?? I very much doubt that you have defined a prime number as 'the only divisors are 1 and a' in arbitrary rings. Could you look up the definition of a prime number for me?

5. Jul 27, 2011

### 83956

Re: Associates

The units in the ring are also divisors of a prime, right?

6. Jul 27, 2011

### micromass

Staff Emeritus
Re: Associates

Yes, the units divide everything. But can you look in your book to find the exact definition of a prime element in a ring?

7. Jul 27, 2011

### 83956

Re: Associates

"An integer that is not a unit is a prime if it can't be written as a product unless one factor is a unit."

8. Jul 27, 2011

### micromass

Staff Emeritus
Re: Associates

Yes, that's exactly what I'm looking for (although many people call this irreducible instead of prime).

So let a and b be associates, and let a be prime. We must prove b to be prime. So assume that b is written as a product:

$$b=cd$$

can you deduce that either c or d must be a unit?

9. Jul 27, 2011

### 83956

Re: Associates

Since b is nonzero, c or d must be a unit since they are not zero divisors

10. Jul 27, 2011

### micromass

Staff Emeritus
Re: Associates

OK, that actually makes no sense at all What do zero-divisors have to do with this problem??

You'll have to do something with your a. You have right now that

$$b=dc$$

How can you introduce a into this equation?

11. Jul 27, 2011

### 83956

Re: Associates

well since a=ub we can plug in b=a/u so a/u=cd, a=ucd.

I'm sorry - I am taking a class in which I teach myself completely, and needless to say it is very frustrating and stressful. This problem should be so easy, and yet it has me stumped ten times over.

12. Jul 27, 2011

### micromass

Staff Emeritus
Re: Associates

OK, this is good:

$$a=ucd$$

or, when introducing brackets:

$$a=(uc)d$$

but a is prime!! What does the primality of a imply?

13. Jul 27, 2011

### 83956

Re: Associates

either uc or d is a unit?

14. Jul 27, 2011

### micromass

Staff Emeritus
Re: Associates

Yes, and thus...

15. Jul 27, 2011

### 83956

Re: Associates

b is prime because one of its factors is a unit