counter example

**ux0** · December 1st 2009, 01:03 PM

i just need a counter example, this has been bothering me all day...!

If R and S are isomorphic commutative rings, then any ring homomorphism $f: R \to S$ is an isomorphism.

I wanted to say....

if $m \geq 2$ , then the map $f:\mathbb{Z} \to \mathbb{Z}_m,$ given by $f(n) = [n]$ , is not injective, but it is surjective....... but my two rings don't seem to be isomophic..

**NonCommAlg** · December 1st 2009, 07:43 PM

Originally Posted by ux0

i just need a counter example, this has been bothering me all day...!

If R and S are isomorphic commutative rings, then any ring homomorphism $f: R \to S$ is an isomorphism.

for any ring $R$ define the map $f: R[x] \longrightarrow R[x]$ by $f(a_0 + a_1 x + \cdots + a_n x^n)=a_0.$

**lepton** · December 1st 2009, 11:07 PM

The trivial map, sending all things to the (respective) identity, is an not an isomorphism between a ring and itself.

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