Question regarding isomorphism of localizations of modules

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Suppose $S$ is a multiplicative subset of a ring $A.$ Let $M$ be an $A$-module. We have the localizations $S^-1A$ and $S^-1M.$ However, my book says that there is a natural isomorphism between $S^-1M cong S^-1A otimes_A M$ defined by the mapping $fracxs mapsto frac1s otimes x.$ I don't exactly see how this is an isomorphism much less a well defined a map. Suppose $fracxs = fracx's'.$ That is, there exists some $s''$ such that $s''(xs' - sx') = 0.$ But why does that imply $frac1s otimes x = frac1s' otimes x'$? I don't know if my following reasoning is correct:



$frac1s' otimes x' = frac1s'ss'' otimes s''sx = frac1s'ss'' otimes s''s'x = frac1s otimes x.$



Injectivity: Suppose $frac1s otimes x = frac1s' otimes x'.$ Then $frac1ss' otimes s'x = frac1ss' otimes sx'.$ Thus, $sx' = s'x$ which implies $fracxs = fracx's'.$



Surjectivity: Consider an arbitrary $fracas otimes x.$ This is simply $frac1s otimes ax$ which is the image of $fracaxs.$



Does this work?




abstract-algebra commutative-algebra




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  • Your arguments are all correct. Alternatively, to justify that your map is an isomorphism, one can also build its inverse.
    – Suzet
    Jul 30 at 0:41















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Suppose $S$ is a multiplicative subset of a ring $A.$ Let $M$ be an $A$-module. We have the localizations $S^-1A$ and $S^-1M.$ However, my book says that there is a natural isomorphism between $S^-1M cong S^-1A otimes_A M$ defined by the mapping $fracxs mapsto frac1s otimes x.$ I don't exactly see how this is an isomorphism much less a well defined a map. Suppose $fracxs = fracx's'.$ That is, there exists some $s''$ such that $s''(xs' - sx') = 0.$ But why does that imply $frac1s otimes x = frac1s' otimes x'$? I don't know if my following reasoning is correct:



$frac1s' otimes x' = frac1s'ss'' otimes s''sx = frac1s'ss'' otimes s''s'x = frac1s otimes x.$



Injectivity: Suppose $frac1s otimes x = frac1s' otimes x'.$ Then $frac1ss' otimes s'x = frac1ss' otimes sx'.$ Thus, $sx' = s'x$ which implies $fracxs = fracx's'.$



Surjectivity: Consider an arbitrary $fracas otimes x.$ This is simply $frac1s otimes ax$ which is the image of $fracaxs.$



Does this work?




abstract-algebra commutative-algebra




share|cite|improve this question



















  • Your arguments are all correct. Alternatively, to justify that your map is an isomorphism, one can also build its inverse.
    – Suzet
    Jul 30 at 0:41













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Suppose $S$ is a multiplicative subset of a ring $A.$ Let $M$ be an $A$-module. We have the localizations $S^-1A$ and $S^-1M.$ However, my book says that there is a natural isomorphism between $S^-1M cong S^-1A otimes_A M$ defined by the mapping $fracxs mapsto frac1s otimes x.$ I don't exactly see how this is an isomorphism much less a well defined a map. Suppose $fracxs = fracx's'.$ That is, there exists some $s''$ such that $s''(xs' - sx') = 0.$ But why does that imply $frac1s otimes x = frac1s' otimes x'$? I don't know if my following reasoning is correct:



$frac1s' otimes x' = frac1s'ss'' otimes s''sx = frac1s'ss'' otimes s''s'x = frac1s otimes x.$



Injectivity: Suppose $frac1s otimes x = frac1s' otimes x'.$ Then $frac1ss' otimes s'x = frac1ss' otimes sx'.$ Thus, $sx' = s'x$ which implies $fracxs = fracx's'.$



Surjectivity: Consider an arbitrary $fracas otimes x.$ This is simply $frac1s otimes ax$ which is the image of $fracaxs.$



Does this work?




abstract-algebra commutative-algebra




share|cite|improve this question











Suppose $S$ is a multiplicative subset of a ring $A.$ Let $M$ be an $A$-module. We have the localizations $S^-1A$ and $S^-1M.$ However, my book says that there is a natural isomorphism between $S^-1M cong S^-1A otimes_A M$ defined by the mapping $fracxs mapsto frac1s otimes x.$ I don't exactly see how this is an isomorphism much less a well defined a map. Suppose $fracxs = fracx's'.$ That is, there exists some $s''$ such that $s''(xs' - sx') = 0.$ But why does that imply $frac1s otimes x = frac1s' otimes x'$? I don't know if my following reasoning is correct:



$frac1s' otimes x' = frac1s'ss'' otimes s''sx = frac1s'ss'' otimes s''s'x = frac1s otimes x.$



Injectivity: Suppose $frac1s otimes x = frac1s' otimes x'.$ Then $frac1ss' otimes s'x = frac1ss' otimes sx'.$ Thus, $sx' = s'x$ which implies $fracxs = fracx's'.$



Surjectivity: Consider an arbitrary $fracas otimes x.$ This is simply $frac1s otimes ax$ which is the image of $fracaxs.$



Does this work?





abstract-algebra commutative-algebra





share|cite|improve this question










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asked Jul 30 at 0:38









伽罗瓦

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  • Your arguments are all correct. Alternatively, to justify that your map is an isomorphism, one can also build its inverse.
    – Suzet
    Jul 30 at 0:41

















  • Your arguments are all correct. Alternatively, to justify that your map is an isomorphism, one can also build its inverse.
    – Suzet
    Jul 30 at 0:41
















Your arguments are all correct. Alternatively, to justify that your map is an isomorphism, one can also build its inverse.
– Suzet
Jul 30 at 0:41





Your arguments are all correct. Alternatively, to justify that your map is an isomorphism, one can also build its inverse.
– Suzet
Jul 30 at 0:41
















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