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

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In summary, a function f from the integers to the real numbers, defined as f(n) = n*1R, is a homomorphism of rings. This function maps values 'n' to 'n' times the identity unit of R, which is equal to the identity unit of Z. To show that f is a homomorphism, we need to use the definition of homomorphism, which states that f(a+b) = f(a) + f(b) and f(ab) = f(a)*f(b). We can see that f(n) = n is not true, as for example, f(23) = 5 in the case of f: Z → Z6. Therefore, we need to use the
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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.
 
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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).
 

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

1. What is a homomorphism of rings?

A homomorphism of rings is a function between two rings that preserves the ring structure. This means that it maps the addition and multiplication operations of one ring to the corresponding operations in the other ring.

2. What is the domain and codomain of the given function?

The given function has a domain of the integers (Z) and a codomain of the real numbers (R).

3. How do you show that the given function is a homomorphism of rings?

To show that the given function is a homomorphism of rings, we need to prove that it preserves the ring structure. This can be done by showing that for any two elements a and b in the domain, the function satisfies the following properties:

  • f(a+b) = f(a) + f(b)
  • f(ab) = f(a)f(b)
  • f(1) = 1

4. What is the significance of the multiplication operation in a ring homomorphism?

The multiplication operation in a ring homomorphism is significant because it ensures that the function preserves the algebraic structure of the ring. In other words, it ensures that the function maps the multiplication of elements in one ring to the multiplication of their corresponding images in the other ring.

5. Can you give an example of a ring homomorphism?

One example of a ring homomorphism is the function f: Z -> Z, n -> n*2, which maps every integer to its double in the same ring. This function preserves the ring structure as:

  • f(a+b) = (a+b)*2 = a*2 + b*2 = f(a) + f(b)
  • f(ab) = (ab)*2 = a*2 * b*2 = f(a)f(b)
  • f(1) = 1*2 = 2 = 1

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