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Math Help - Modules and Ideals

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by bleys View Post
    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.
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  3. #3
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    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!
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  4. #4
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    Quote Originally Posted by bleys View Post
    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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