Well-defined, surjective ring homomorphism.

Hey guys.

I have a commutative ring R, with two ideals, I and J, in R, where I $\displaystyle \subseteq$ J.

Now I have to show that $\displaystyle \varphi : R / I \rightarrow R / J $ given by $\displaystyle \varphi ( x + I ) = ( x + J ) $ is a well defined, surjective ring homomorphism.

Right now I cant even figure out what i need to do, to prove that it is well-defined. I'm missing parts of what I should have learned, but as far as I've been able to figure out, a homomorphism is well-defined when for any $\displaystyle a,b \in R/I$, $\displaystyle a = b \Leftrightarrow \varphi (a) = \varphi (b)$.

But I can't figure out how to prove that. If someone could give me a hint, I would really appreciate it. Perhaps I need a more stringent definition of "well-defined" or if you could give me an example of a proof kind of like mine?

All help is appreciated.

Morten

Re: Well-defined, surjective ring homomorphism.

Quote:

Originally Posted by

**m112358** Right now I cant even figure out what i need to do, to prove that it is well-defined.

We have to prove that $\displaystyle \varphi (x+I)$ does not depend on the representative element $\displaystyle x$ . If $\displaystyle x+I=y+I$ then $\displaystyle x-y\in I\subseteq J$ that is , $\displaystyle x+J=y+J$ and as a consequence $\displaystyle \varphi (x+I)=\varphi (y+I)$ .

Re: Well-defined, surjective ring homomorphism.

Arh ok, now I understand. Alright, that was simple, guess I had been staring too blindly to notice. Thanks for the help :)

Morten