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## 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!