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Let's say we have $f:A \to B$ where $A$ and $B$ are R-modules and $f$ is a R-module homomorphism.

If A has a torsion, then $\{a \in A \mid ar=0 \text{ for some nonzero } r \in R\}\neq \varnothing $. Let's say $a \in A$ and $0 \neq r \in R$ are such that $ar=0$. Then $rf(a)=f(ra)=f(0)=0$. so $f(A)$ has a torsion.

Therefore, can we conclude that

"If $f(A)$ is torsion free, then $A$ is torsion free."?

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1 Answer 1

up vote 2 down vote accepted

Certainly not. Take $\text{Whatever}\to0$. Or, perhaps, more egregiously $R\oplus\text{Whatever}\to R$.

EDIT: Of course, if $f$ was injective then this is certainly true because $A$ sits inside $B$ and subthings of torsion-free things are torsion-free.

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Thank you! Torsion submodule being nonempty was always ture, and all in all my question was wrong. – Emily Apr 11 '12 at 16:41

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