# Commuting Homomorphisms

• Apr 5th 2009, 12:09 PM
Coda202
Commuting Homomorphisms
Let phi: R-->R' be a ring homomorphism, I an ideal of R, J an ideal of R' and suppose that phi(I) is a subset of J. Let f and g be the natural homomorphisms, f:R-->R/I and g:R'-->R'/J defined by, f(r) = r+I for all r in R and g(r')=r'+J for all r' in R'.
Now, define a homomorphism h:R/I-->R'/J, prove that h is well-defined, that it is in fact a homomorphism and finally that for all x in R, hf(x)=gphi(x)
• Apr 5th 2009, 01:29 PM
NonCommAlg
Quote:

Originally Posted by Coda202

Let phi: R-->R' be a ring homomorphism, I an ideal of R, J an ideal of R' and suppose that phi(I) is a subset of J. Let f and g be the natural homomorphisms, f:R-->R/I and g:R'-->R'/J defined by, f(r) = r+I for all r in R and g(r')=r'+J for all r' in R'.
Now, define a homomorphism h:R/I-->R'/J, prove that h is well-defined, that it is in fact a homomorphism and finally that for all x in R, hf(x)=gphi(x)

define $h(x+I)=\phi(x) + J, \ \forall x \in R.$ this map is well-defined because $\phi(I) \subseteq J.$ it's a homomorphism because $\phi$ is a homomorphism and finally $hf(x)=h(x+I)=\phi(x) + J=g\phi(x), \ \forall x \in R.$