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Let $R$ be an integral domain and $M$ an $R$-module. An element $m \in M$ is called a torsion element if $\operatorname{Ann}_R(m) \neq 0$, i.e. if $rm = 0$ for some nonzero $r \in R$. Denote the set of torsion elements in $M$ by $T(M)$, called the torsion submodule of $M$. We say that $M$ is torsion-free if $T(M) = 0$.

Let $S$ be a m.c.s. in the integral domain $R$. Show that $$T (S^{−1}M)= S^{−1}(T (M)).$$

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closed as not a real question by Nameless, 5pm, Hagen von Eitzen, Henry T. Horton, Micah Feb 20 '13 at 20:21

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Presumably, you mean "multiplicative closed set," but that is not a standard acronym, and if you want a better chance of getting answers, you should spell out terminology rather than assuming people know what your acronym means. –  Thomas Andrews Feb 19 '13 at 18:52
This looks like the same user as Mathproof P, who earlier today had spammed 6 similarly worded imperative questions with the same idiosyncratic notation and subject matter. –  rschwieb Feb 19 '13 at 18:59
I have deleted all the wrong tags. –  Martin Brandenburg Feb 19 '13 at 19:09

1 Answer 1

up vote 1 down vote accepted

Presumably the inclusion $S^{-1}(T(M)) \subseteq T(S^{-1}M)$ is clear and it's the other direction that needs more explanation.

Hint: Say $\frac{m}{s} \in T(S^{-1}M)$. This means there is an $\frac{a}{s'} \in S^{-1}R$ such that $\frac{am}{ss'} = 0$ in $S^{-1}M$. Using the definition of localization what does $\frac{am}{ss'} = 0$ translate to as a statement about elements of $R$ and $M$?

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