Modules and Ideals

**bleys** · August 28th 2010, 12:13 AM

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?

**NonCommAlg** · August 28th 2010, 04:40 PM

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 $f:R/I \longrightarrow R/J$ be a surjective R-module homomorphism. then $f(r+I)=1+J,$ for some $r \in R.$ now if $s \in I,$ then

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

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

**bleys** · August 29th 2010, 12:10 AM

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!

**NonCommAlg** · August 29th 2010, 01:43 AM

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 $R=M_2(\mathbb{R}),$ the ring of $2 \times 2$ matrices with real entries.

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

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

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

because $r-s \in J.$

Math Help - Modules and Ideals

LinkBack

Thread Tools

Display

Modules and Ideals

Similar Math Help Forum Discussions

Product ideals vs. products of ideals

Prime Ideals, Maximal Ideals

Proof Help! Modules and left ideals

Modules

When are principal ideals prime ideals?

Search Tags