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Let $f:M\rightarrow N$ be an $R-$ homomorphism. Prove that if $f$ is monic, then $l_{R}\left(M\right)\supseteq l_{R}(N)$ , whereas if $f$ is epic, then $l_{R}(M)\subseteq l_{R}(N)$ . This is exercise 3 page 51, Rings and Categories of modules - Frank W. Anderson & Kent R. Fuller, Second edition. Help me some hints to prove it. Thank you in advance.

EDIT: Let $M$ be a left $R-$ module. Then for $X\subseteq M$ , the left annihilator of $X$ in R is

$l_{R}(X)=\left\{ r\in R:rx=0\left(x\in X\right)\right\} $

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Hint for the monic case: If $r\not\in l_R(M)$ then $rm\neq 0$ for some $m\in M$ ($m\neq 0$). Thus $f(rm)\neq 0$ because $f$ is injective.

Hint for the epic case: If $r\in l_R(M)$ then $rm = 0$ for all $m\in M$. Thus $f(rm) = 0$ for all $m$, but $f$ is surjective.

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Suppose $f$ is monic and let $r\in l_R(N)$. Take $x\in M$; then $$ f(rx)=rf(x)=0 $$ so $rx=0$. Since $x\in M$ is arbitrary, we get $r\in l_R(M)$.

Suppose $f$ is epic and let $r\in l_R(M)$. You have to show that $r\in l_R(N)$, that is, $ry=0$ for all $y\in N$.

If $y\in N$, we have $y=f(x)$, so $ry=rf(x)=f(rx)=f(0)=0$

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  • $\begingroup$ +1 vote: Actually the proof you give me is more details than those of @Myseft. But he notice me write the definition of left annihilator so I can understand everything more precise. $\endgroup$ – chuyenvien94 Dec 5 '14 at 13:48

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