# counter example

• Dec 1st 2009, 01:03 PM
ux0
counter example
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 $\displaystyle f: R \to S$ is an isomorphism.

I wanted to say....

if $\displaystyle m \geq 2$, then the map $\displaystyle f:\mathbb{Z} \to \mathbb{Z}_m,$ given by $\displaystyle f(n) = [n]$, is not injective, but it is surjective....... but my two rings don't seem to be isomophic..
• Dec 1st 2009, 07:43 PM
NonCommAlg
Quote:

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 $\displaystyle f: R \to S$ is an isomorphism.

for any ring $\displaystyle R$ define the map $\displaystyle f: R[x] \longrightarrow R[x]$ by $\displaystyle f(a_0 + a_1 x + \cdots + a_n x^n)=a_0.$
• Dec 1st 2009, 11:07 PM
lepton
The trivial map, sending all things to the (respective) identity, is an not an isomorphism between a ring and itself.