Show that a f: Z -> R , n -> n*1(subr) is a homomorphism of rings

  • Thread starter General_Sax
  • Start date
  • Tags
  • #1

Homework Statement

Show that a f: Z → R , n → n*1R is a homomorphism of rings

Homework Equations

The Attempt at a Solution

I'm not sure how to exactly go about answering this question, but I'm going to try to start with the definition:

f(a+b) = f(a) + f(b)
f(a*b) = f(a) * f(b)
f(1Z) = 1R

Ok, so if I actually have a clue what the problem statement means -- which is just slightly possible -- then this function maps values 'n' to 'n' times the identity unit of R. Which is equal to the identity unit of Z, so I think that part of the definition is more-or-less covered.

How do I go about using the first two parts of the definition?

The function is just f(n) = n right?
so do I just sub in (a+b) for n?
f(a+b) = f(a) + f(b) ?
Is there anyway I can actually show that this is the case? I mean isn't this intuitive/properly defined upon the integers?

same for the other part
f(ab) = f(a)*f(b) ... I'm just not sure how to actually show this...

thanks for any help.
  • #2
no, f(n) = n is't true.

for example, if R = Z6, f(23) = 5.

realize that n*1R isn't "n", it's:

1R+1R+...+1R (n summands).

Suggested for: Show that a f: Z -> R , n -> n*1(subr) is a homomorphism of rings