## Onto maps of cyclic modules

Posted: August 29, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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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 $r-s \in J. \ \Box$