• Apr 17th 2011, 09:28 AM
slevvio

Is it possible to have a ring isomorphism from $R$ to F, where F is a field but $R$ is not a field?
• Apr 17th 2011, 09:35 AM
abhishekkgp
edit: deleted post.

hi slevvio!!

f: R----> F
f(x)=0 for all x in R.

this is trivial homomorphism i guess. Is this acceptable to you??
• Apr 17th 2011, 09:50 AM
slevvio
Thanks but that is not a ring isomorphism!
• Apr 17th 2011, 09:57 AM
Deveno
if φ is onto, R has to be a field, since then R = φ^-1(F).

if φ is not onto, and φ(R) is just a subring of F, then sure, consider the identity homomorphism which includes Z in Q.
• Apr 17th 2011, 10:08 AM
abhishekkgp
Quote:

Originally Posted by slevvio
Thanks but that is not a ring isomorphism!

my book says that if R and S are rings then f:R ---> S is a homomorphism if the following hold for all a,b in R

1) f(a+b)=f(a)+f(b)
2)f(ab)=f(a)f(b)

aren't these two satisfied in the example i gave??
am i missing something out?(Doh)
• Apr 17th 2011, 10:14 AM
Deveno
Quote:

Originally Posted by abhishekkgp
my book says that if R and S are rings then f:R ---> S is a homomorphism if the following hold for all a,b in R

1) f(a+b)=f(a)+f(b)
2)f(ab)=f(a)f(b)

aren't these two satisfied in the example i gave??

am i missing something out?(Doh)

unless R is trivial, that is not an isomorphism.
• Apr 17th 2011, 08:22 PM
abhishekkgp
Quote:

Originally Posted by Deveno
unless R is trivial, that is not an isomorphism.