Onto maps of cyclic modules

Posted: August 29, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: ,

Problem. 1) Let R be a commutative ring with unity and I,J some ideals of R. If there exists a surjective R-module homomorphism f: R/I \longrightarrow R/J, then I \subseteq J.

2) Show that the result in 1) may not be true in noncommutative rings.

Solution. 1) We have f(r+I)=1+J, for some r \in R. Now if s \in I, then sr \in I and thus

s+J=s(1+J)=sf(r+I)=f(sr+I)=0.

So s \in J.

2) Let R be the ring of 2 \times 2 matrices with, say, 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 and that 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-module homomorphism. Also, 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. \ \Box

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