1. ## Modules and Ideals

I need some help trying to prove an exercise in Introduction to Algebra by Cameron
Let R be a commutative ring with identity. Suppose I and J are ideals of R. Then there is a surjective R-module homomorphism from R/I to R/J if and only if I is a subset of J.
I'm able to prove if-part, but I'm having trouble with the 'only if'-part. I assume a surjective homomorphism exists, but I don't know how to use it. I guess I can't assume the construction I used in the if-part (that is, f(I+r)=J+r)?
I know the R-modules R/I and R/J are cyclic but I don't think a homomorphism maps the generator of one to the other (f(I+1)=J+1)? Could you point me in the right direction?

2. Originally Posted by bleys
I need some help trying to prove an exercise in Introduction to Algebra by Cameron
Let R be a commutative ring with identity. Suppose I and J are ideals of R. Then there is a surjective R-module homomorphism from R/I to R/J if and only if I is a subset of J.
I'm able to prove if-part, but I'm having trouble with the 'only if'-part. I assume a surjective homomorphism exists, but I don't know how to use it.
well, it's very easy: let $\displaystyle f:R/I \longrightarrow R/J$ be a surjective R-module homomorphism. then $\displaystyle f(r+I)=1+J,$ for some $\displaystyle r \in R.$ now if $\displaystyle s \in I,$ then

$\displaystyle s+J=s(1+J)=sf(r+I)=srf(1+I)=rsf(1+I)=rf(s+I)=0$

and thus $\displaystyle s \in J,$ i.e. $\displaystyle I \subseteq J.$

3. Oh, so commutativity of R is essential for the proof. Is there some analogous result for a general ring R, possibly non-commutative?
Thanks for the help, NonCommAlg!

4. Originally Posted by bleys
Oh, so commutativity of R is essential for the proof. Is there some analogous result for a general ring R, possibly non-commutative?
good question! no, it's not true for noncommutative rings. for example let $\displaystyle R=M_2(\mathbb{R}),$ the ring of $\displaystyle 2 \times 2$ matrices with real entries.

let $\displaystyle I=\left \{\begin{pmatrix} a & 0 \\ b & 0 \end{pmatrix}: \ a,b \in \mathbb{R} \right \}$ and $\displaystyle J=\left \{\begin{pmatrix} a & a \\ b & b \end{pmatrix}: \ a,b \in \mathbb{R} \right \}.$ see that $\displaystyle I,J$ are left ideals of $\displaystyle R$. obviously $\displaystyle I$ is not contained in $\displaystyle J.$

now define $\displaystyle f: R/I \longrightarrow R/J$ in this way: for any $\displaystyle r=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in R$ we define $\displaystyle f(r+I)= \begin{pmatrix} 0 & b \\ 0 & d \end{pmatrix}+J.$ it is easy to see that $\displaystyle f$ is a

well-defined $\displaystyle R$-homomorphism. $\displaystyle f$ is surjective because if $\displaystyle r=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in R$ and $\displaystyle s=\begin{pmatrix} 0 & b-a \\ 0 & d-c \end{pmatrix} \in R,$ then $\displaystyle f(s+I)=s+J=r+J$

because $\displaystyle r-s \in J.$