Prove that if a is prime, then b is prime

[83956]

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

Homework Statement

Suppose that "a" and "b" are associates. Prove that if "a" is prime, then "b" is prime.

Homework Equations

The Attempt at a Solution

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

 

[micromass]

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

 

[83956]

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.

 

[micromass]

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.

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?

 

[83956]

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

 

[micromass]

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

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

 

[83956]

"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."

 

[micromass]

"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."

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?

 

[83956]

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

 

[micromass]

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

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?

 

[83956]

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.

 

[micromass]

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.

OK, this is good:

$$a=ucd$$

or, when introducing brackets:

$$a=(uc)d$$

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

 

[83956]

either uc or d is a unit?

 

[micromass]

either uc or d is a unit?

Yes, and thus...

 

[83956]

b is prime because one of its factors is a unit