Homomorphic image of principal ideal ring

• May 14th 2011, 12:50 PM
AlexP
Homomorphic image of principal ideal ring
I'm trying to show that the homomorphic image of a principal ideal ring is again a principal ideal ring. Let $\displaystyle \phi:R \to S$. If we take a general principal ideal $\displaystyle \langle a \rangle$ and an element of this ideal, $\displaystyle ra$, then we have $\displaystyle \phi(ra) = \phi(r)\phi(a)$, but this doesn't prove it unless $\displaystyle \phi$ is surjective, correct? Because if it's not surjective then we don't know that every element in S is of the form $\displaystyle \phi(r)$, right? This is what I'm stuck on. Any hints (not answers)?
• May 14th 2011, 12:56 PM
Tinyboss
You want to show that the homomorphic image is a PID. Every function is a surjection onto its image.
• May 14th 2011, 01:12 PM
Deveno
φ:R-->φ(R) is always onto.
• May 14th 2011, 05:59 PM
AlexP
Yes, true. So do we then have the desired result since subrings are preserved, and $\displaystyle \phi(ra)=\phi(r)\phi(a)$ so that the elements of the image are of the form $\displaystyle s\phi(a)$ with $\displaystyle s \in S$?