Find all ring homomorphisms from 3Z to Z?

[mathgirl313]

Homework Statement

Find all ring homomorphisms from 3Z to Z, where 3Z are the integers that are of multiple 3.

Homework Equations

The Attempt at a Solution

So 3Z is cyclic so σ(3) is sufficient to look. Now all of the other examples have finite groups, so |σ(a)| divides the |a| in the domain. Or, the homomorphisms is define with Z as the domain, and then a=a^2. I understand these examples, but then don't know how to start this one...

<3> generates the group, so I know I want to consider it. But if σ(3)=a, but then what? Other examples are a=σ(1)=σ(1*1)=a^2, but that property doesn't hold here, and 1 is not in3Z. I know I'm missing something simple, just can't quite figure out what property I need to put my finger on to make that first step..

Thanks for any help!

 

[Dick]

Homework Statement

Find all ring homomorphisms from 3Z to Z, where 3Z are the integers that are of multiple 3.

Homework Equations

The Attempt at a Solution

So 3Z is cyclic so σ(3) is sufficient to look. Now all of the other examples have finite groups, so |σ(a)| divides the |a| in the domain. Or, the homomorphisms is define with Z as the domain, and then a=a^2. I understand these examples, but then don't know how to start this one...

<3> generates the group, so I know I want to consider it. But if σ(3)=a, but then what? Other examples are a=σ(1)=σ(1*1)=a^2, but that property doesn't hold here, and 1 is not in3Z. I know I'm missing something simple, just can't quite figure out what property I need to put my finger on to make that first step..

Thanks for any help!

You've got the right starting point. So σ(3)+σ(3)+σ(3)=σ(3+3+3)=σ(9)=σ(3*3)=σ(3)*σ(3). So?

 

[mathgirl313]

You've got the right starting point. So σ(3)+σ(3)+σ(3)=σ(3+3+3)=σ(9)=σ(3*3)=σ(3)*σ(3). So?

So 3a=a^2 if a=σ(3). So the possible homomorphisms of a will have that property. And it should them just fail out like the other examples.

 

[Dick]

So 3a=a^2 if a=σ(3). So the possible homomorphisms of a will have that property. And it should them just fail out like the other examples.

If a solves 3a=a^2 then it's a homomorphism. You still need to describe what the homomorphisms are.

 

[mathgirl313]

If a solves 3a=a^2 then it's a homomorphism. You still need to describe what the homomorphisms are.

Then 3a - a^2 = 0, and 3Z has no zero divisors, so factoring the equation gives a=0 or a=3.
Then, if σ(3)=a=0, then σ(x)=0 for all x in 3Z, and σ is the zero homomorphism.
If σ(3)=a=3, then σ(x)=x for all x in 3Z, and σ is the identity homomorphism.

Hence there there two ring homomorphism for 3Z to Z.

 

[Dick]

Then 3a - a^2 = 0, and 3Z has no zero divisors, so factoring the equation gives a=0 or a=3.
Then, if σ(3)=a=0, then σ(x)=0 for all x in 3Z, and σ is the zero homomorphism.
If σ(3)=a=3, then σ(x)=x for all x in 3Z, and σ is the identity homomorphism.

Hence there there two ring homomorphism for 3Z to Z.

Seems fine to me.