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I am having trouble thinking about a practice problem that I feel should be pretty easy because it is part 1 of a 4 part problem. I am probably overthinking it because I keep trying to construct maps of induced sequences of Hom since this comes from a section where we have introduced the notion of projectivity and injectivity for modules.

Let $R$ be a commutative ring with identity

Suppose $f \in End_{R}(M)$ is an $R$-module endomorphism. If $f$ is surjective then $f$ is not a right divisor of zero in $End_{R}(M)$. Conversely if or every submodule $N \neq M$ with $N \subset M$ there exits a linear form $x^{*} \in E^{*}$ which is zero on $N$ and surjective, every element of $End_{R}(M)$ which is not a right divisor of zero is a surjective endomorphism.

Does the first half of the problem follow simply from the fact that $f$ surjective implies $f^*$ injective? There is a similar statement for injective endomorphsisms in the problem set but I cannot seem to come up with the ideas for either I was also wondering if there where any good texts or lecture notes where they cover module endomorphisims in a way that expalins these type of problems clearly.

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For the first part if an endomorphism is injective then it is not a left divisor of zero. – Frank Murphy Sep 21 '11 at 14:08
up vote 3 down vote accepted

Suppose $f$ is surjective and $gf=0$. To prove $f$ is not a right divisor of zero, we need to show that $g=0$, i.e. that $g(m)=0$ for all $m$. So let $m$ be in $M$. Since $f$ is surjective, $m=f(n)$ for some $n$ in $M$. Thus $g(m)=g(f(n))=(gf)(n)=0(n)=0$.

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In the category of modules you can prove that a morphism is an epimorphism iff is surjective, where epimorphism means right cancellable. So consider another morphism $g$ that $gf=0$ and the fact that $0f=0$ gives you $gf=0f$ now it is right cancellable so $g=0$.So it isn't a right divisor of zero.

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Also always surjective implies epi. – Frank Murphy Sep 21 '11 at 14:06

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