Abstract Algebra: Homomorphism f Determined by f(1) in Z

[oddiseas]

Homework Statement

Let R be any ring and f:Z→R a homomorphism.

a)Show that f is completely determined by the single value f(1)
b)Determine all possible homomorphisms f in the case when R = Z.

Homework Equations

The Attempt at a Solution

This question has me totally confused. I have gone through all the properties of homororphisms in the book but i am still confused.How is the homomorphism completely determined by one value anyway?

 

[losiu99]

The definition of homomorphism says that
[tex]f(n)=n f(1)[/tex]
and so knowledge of f(1) is enough to compute f for any integer.

 

[Tedjn]

To summarize the above: f(1) determines f for every integer because the integers are generated by 1. Have you made any progress on part b?

 

[oddiseas]

i can't find this property anywhere in my notes. Is this a property of homomorphismsm in general or only in the cas on the integers?

 

[Tedjn]

Apply the defining property of a homomorphism with the domain being the group of integers. Does that make sense?

 

[Hurkyl]

If figuring out the entire homomorphism from the value of f(1) is too hard, then try something simpler. What is f(0)? f(2)?

 

[oddiseas]

but for a homomorphism we have:

f(n)=f(1*n)=f(1)*f(n)
thus f(1) =1
so how is it completeley determined by f(1), which is always one anyway, i thought it would therefore be totally determined by the domain, depending on whether in is integers or rational numbers etc

 

[yossell]

A ring has + as well as *, and the homomorphism must respect both. Use the fact that + is preserved as well to extract more information - such as f(2)

 

[oddiseas]

ok thanks, using the additive property i see now that we get f(n)=nf(1) and it is determuned by this value. Is it sufficient for part b to use the multiplicative property, and the fact that the identity in z is 1 and thus f(1) maps to 1 to show that the only homomorphism is the identity map.Or is there something more in depth i could do?