Onto Homomorphisms of Rings

[missavvy]

Homework Statement

f: R -> R1 is an onto ring homomorphism.
Show that f[Z(R)] ⊆ Z(R1)

Homework Equations

The Attempt at a Solution

I'm a little confused. So f is onto, then for all r' belonging to R1, we have f(r)=r' for some r in R.
But if f is onto couldn't R1 has less elements than R? ie. R would have some elements which don't get mapped to those in R1?

Anyways if I were to prove this, I'd show that for some x in f(Z(R)), x also is in Z(R1). But I just don't get why it has to be in Z(R1) if there could be some elements not mapped to R1...
:S
thanks!

 

[mighty2000]

Hi there,

Your confusion about the size of R and R1 is valid, but it doesn't affect the proof. Remember, an onto function means that every element in the codomain is mapped to by at least one element in the domain. This means that even if R1 has less elements than R, every element in R1 is still being mapped to by at least one element in R. So we can still prove that f[Z(R)] ⊆ Z(R1) without worrying about the size of the sets.

To prove this, we need to show that for any x in f[Z(R)], x is also in Z(R1). So let x = f(r) for some r in Z(R). Then for any element r' in R1, we have f(r') = r'' for some r'' in R1 (since f is onto). Since r is in Z(R), we know that r*r' = r'*r for all r' in R. Now, we can use the homomorphism property of f to show that f(r*r') = f(r'*r). This gives us f(r)*f(r') = f(r')*f(r), which means that x*r' = r'*x for all r' in R1. Since this holds for all elements in R1, we can conclude that x is in Z(R1). Therefore, f[Z(R)] ⊆ Z(R1).

I hope this helps! Let me know if you have any further questions.