Modules and Ideals

**bleys** · Aug 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** · Aug 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** · Aug 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** · Aug 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 $r-s \in J.$

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