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