When is the localization map injective?












0












$begingroup$


Suppose that $R$ is a commutative ring and $Ssubset R$ is multiplicatively closed subset, i.e. $1in S$ and if $a,bin S$ then $abin S$.
Consider the natural mapping $phi:Rto S^{-1}R$ defined by $phi(r)=frac{r}{1}$.



If $S$ does not contain zero-divisors then $phi$ is injective.



I am trying to prove it directly but have some troubles.



My approach: Let $rin text{Ker} phi$ then $phi(r)=frac{0}{1}$. It means that $frac{r}{1}=frac{0}{1}$ then by definition it means that $exists uin S$ such that $u(rcdot 1-0cdot 1)=0$, i.e. $ur=0$. How to conclude that $r=0$?



If $uneq 0$ then $ur=0$ leads to $r=0$.



What if $u=0$?



Would be very grateful for help!






abstract-algebra ring-theory commutative-algebra localization







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    I'm sure that $0$ is a zero-divisor.
    $endgroup$
    – Lord Shark the Unknown
    Dec 14 '18 at 18:01










  • $begingroup$
    @LordSharktheUnknown, I thought that the zero-divisor should be always $neq 0$.
    $endgroup$
    – ZFR
    Dec 14 '18 at 18:04






  • 2




    $begingroup$
    If $S$ contains $0$, then the localization immediately collapses to the zero ring, because every pair of elements $a/b, c/d$ satisfies $0(ad-bc)=0$. So... it seems you should be more flexible in the definition and be ready for the alternative use and/or minor slip-ups by an author using this term.
    $endgroup$
    – rschwieb
    Dec 14 '18 at 18:12










  • $begingroup$
    @rschwieb, I don't understand your answer at all. Could you clarify you comment a bit, please?
    $endgroup$
    – ZFR
    Dec 14 '18 at 18:15










  • $begingroup$
    @ZFR Please indicate the unclear part.
    $endgroup$
    – rschwieb
    Dec 14 '18 at 19:19
















0












$begingroup$


Suppose that $R$ is a commutative ring and $Ssubset R$ is multiplicatively closed subset, i.e. $1in S$ and if $a,bin S$ then $abin S$.
Consider the natural mapping $phi:Rto S^{-1}R$ defined by $phi(r)=frac{r}{1}$.



If $S$ does not contain zero-divisors then $phi$ is injective.



I am trying to prove it directly but have some troubles.



My approach: Let $rin text{Ker} phi$ then $phi(r)=frac{0}{1}$. It means that $frac{r}{1}=frac{0}{1}$ then by definition it means that $exists uin S$ such that $u(rcdot 1-0cdot 1)=0$, i.e. $ur=0$. How to conclude that $r=0$?



If $uneq 0$ then $ur=0$ leads to $r=0$.



What if $u=0$?



Would be very grateful for help!






abstract-algebra ring-theory commutative-algebra localization







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    I'm sure that $0$ is a zero-divisor.
    $endgroup$
    – Lord Shark the Unknown
    Dec 14 '18 at 18:01










  • $begingroup$
    @LordSharktheUnknown, I thought that the zero-divisor should be always $neq 0$.
    $endgroup$
    – ZFR
    Dec 14 '18 at 18:04






  • 2




    $begingroup$
    If $S$ contains $0$, then the localization immediately collapses to the zero ring, because every pair of elements $a/b, c/d$ satisfies $0(ad-bc)=0$. So... it seems you should be more flexible in the definition and be ready for the alternative use and/or minor slip-ups by an author using this term.
    $endgroup$
    – rschwieb
    Dec 14 '18 at 18:12










  • $begingroup$
    @rschwieb, I don't understand your answer at all. Could you clarify you comment a bit, please?
    $endgroup$
    – ZFR
    Dec 14 '18 at 18:15










  • $begingroup$
    @ZFR Please indicate the unclear part.
    $endgroup$
    – rschwieb
    Dec 14 '18 at 19:19














0












0








0





$begingroup$


Suppose that $R$ is a commutative ring and $Ssubset R$ is multiplicatively closed subset, i.e. $1in S$ and if $a,bin S$ then $abin S$.
Consider the natural mapping $phi:Rto S^{-1}R$ defined by $phi(r)=frac{r}{1}$.



If $S$ does not contain zero-divisors then $phi$ is injective.



I am trying to prove it directly but have some troubles.



My approach: Let $rin text{Ker} phi$ then $phi(r)=frac{0}{1}$. It means that $frac{r}{1}=frac{0}{1}$ then by definition it means that $exists uin S$ such that $u(rcdot 1-0cdot 1)=0$, i.e. $ur=0$. How to conclude that $r=0$?



If $uneq 0$ then $ur=0$ leads to $r=0$.



What if $u=0$?



Would be very grateful for help!






abstract-algebra ring-theory commutative-algebra localization







share|cite|improve this question











$endgroup$




Suppose that $R$ is a commutative ring and $Ssubset R$ is multiplicatively closed subset, i.e. $1in S$ and if $a,bin S$ then $abin S$.
Consider the natural mapping $phi:Rto S^{-1}R$ defined by $phi(r)=frac{r}{1}$.



If $S$ does not contain zero-divisors then $phi$ is injective.



I am trying to prove it directly but have some troubles.



My approach: Let $rin text{Ker} phi$ then $phi(r)=frac{0}{1}$. It means that $frac{r}{1}=frac{0}{1}$ then by definition it means that $exists uin S$ such that $u(rcdot 1-0cdot 1)=0$, i.e. $ur=0$. How to conclude that $r=0$?



If $uneq 0$ then $ur=0$ leads to $r=0$.



What if $u=0$?



Would be very grateful for help!






abstract-algebra ring-theory commutative-algebra localization




abstract-algebra ring-theory commutative-algebra localization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 7:17









user26857

39.3k124183




39.3k124183










asked Dec 14 '18 at 17:56









ZFRZFR

4,99631338




4,99631338








  • 4




    $begingroup$
    I'm sure that $0$ is a zero-divisor.
    $endgroup$
    – Lord Shark the Unknown
    Dec 14 '18 at 18:01










  • $begingroup$
    @LordSharktheUnknown, I thought that the zero-divisor should be always $neq 0$.
    $endgroup$
    – ZFR
    Dec 14 '18 at 18:04






  • 2




    $begingroup$
    If $S$ contains $0$, then the localization immediately collapses to the zero ring, because every pair of elements $a/b, c/d$ satisfies $0(ad-bc)=0$. So... it seems you should be more flexible in the definition and be ready for the alternative use and/or minor slip-ups by an author using this term.
    $endgroup$
    – rschwieb
    Dec 14 '18 at 18:12










  • $begingroup$
    @rschwieb, I don't understand your answer at all. Could you clarify you comment a bit, please?
    $endgroup$
    – ZFR
    Dec 14 '18 at 18:15










  • $begingroup$
    @ZFR Please indicate the unclear part.
    $endgroup$
    – rschwieb
    Dec 14 '18 at 19:19














  • 4




    $begingroup$
    I'm sure that $0$ is a zero-divisor.
    $endgroup$
    – Lord Shark the Unknown
    Dec 14 '18 at 18:01










  • $begingroup$
    @LordSharktheUnknown, I thought that the zero-divisor should be always $neq 0$.
    $endgroup$
    – ZFR
    Dec 14 '18 at 18:04






  • 2




    $begingroup$
    If $S$ contains $0$, then the localization immediately collapses to the zero ring, because every pair of elements $a/b, c/d$ satisfies $0(ad-bc)=0$. So... it seems you should be more flexible in the definition and be ready for the alternative use and/or minor slip-ups by an author using this term.
    $endgroup$
    – rschwieb
    Dec 14 '18 at 18:12










  • $begingroup$
    @rschwieb, I don't understand your answer at all. Could you clarify you comment a bit, please?
    $endgroup$
    – ZFR
    Dec 14 '18 at 18:15










  • $begingroup$
    @ZFR Please indicate the unclear part.
    $endgroup$
    – rschwieb
    Dec 14 '18 at 19:19








4




4




$begingroup$
I'm sure that $0$ is a zero-divisor.
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 18:01




$begingroup$
I'm sure that $0$ is a zero-divisor.
$endgroup$
– Lord Shark the Unknown
Dec 14 '18 at 18:01












$begingroup$
@LordSharktheUnknown, I thought that the zero-divisor should be always $neq 0$.
$endgroup$
– ZFR
Dec 14 '18 at 18:04




$begingroup$
@LordSharktheUnknown, I thought that the zero-divisor should be always $neq 0$.
$endgroup$
– ZFR
Dec 14 '18 at 18:04




2




2




$begingroup$
If $S$ contains $0$, then the localization immediately collapses to the zero ring, because every pair of elements $a/b, c/d$ satisfies $0(ad-bc)=0$. So... it seems you should be more flexible in the definition and be ready for the alternative use and/or minor slip-ups by an author using this term.
$endgroup$
– rschwieb
Dec 14 '18 at 18:12




$begingroup$
If $S$ contains $0$, then the localization immediately collapses to the zero ring, because every pair of elements $a/b, c/d$ satisfies $0(ad-bc)=0$. So... it seems you should be more flexible in the definition and be ready for the alternative use and/or minor slip-ups by an author using this term.
$endgroup$
– rschwieb
Dec 14 '18 at 18:12












$begingroup$
@rschwieb, I don't understand your answer at all. Could you clarify you comment a bit, please?
$endgroup$
– ZFR
Dec 14 '18 at 18:15




$begingroup$
@rschwieb, I don't understand your answer at all. Could you clarify you comment a bit, please?
$endgroup$
– ZFR
Dec 14 '18 at 18:15












$begingroup$
@ZFR Please indicate the unclear part.
$endgroup$
– rschwieb
Dec 14 '18 at 19:19




$begingroup$
@ZFR Please indicate the unclear part.
$endgroup$
– rschwieb
Dec 14 '18 at 19:19










1 Answer
1






active

oldest

votes


















-1












$begingroup$

Suppose $r,r'in R$ are mapped to $r/1=r'/1$. Then by definition, $s(r-r')=0$ for some $sin S$. If $S$ has no zero divisor, then $r=r'$. Note that the case $s=0in S$ cannot appear by def. Done.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Note that my approach is the same! Is $sneq 0$ then I am OK with that. What if $s=0$? In this case $rneq r'$ can also be the case.
    $endgroup$
    – ZFR
    Dec 14 '18 at 18:06










  • $begingroup$
    The denominators are coming from S.
    $endgroup$
    – Wuestenfux
    Dec 14 '18 at 18:18










  • $begingroup$
    Could you clarify your last comment, please?
    $endgroup$
    – ZFR
    Dec 14 '18 at 18:22










  • $begingroup$
    Division by zero. View the definition of localization.
    $endgroup$
    – Wuestenfux
    Dec 15 '18 at 10:46













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









-1












$begingroup$

Suppose $r,r'in R$ are mapped to $r/1=r'/1$. Then by definition, $s(r-r')=0$ for some $sin S$. If $S$ has no zero divisor, then $r=r'$. Note that the case $s=0in S$ cannot appear by def. Done.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Note that my approach is the same! Is $sneq 0$ then I am OK with that. What if $s=0$? In this case $rneq r'$ can also be the case.
    $endgroup$
    – ZFR
    Dec 14 '18 at 18:06










  • $begingroup$
    The denominators are coming from S.
    $endgroup$
    – Wuestenfux
    Dec 14 '18 at 18:18










  • $begingroup$
    Could you clarify your last comment, please?
    $endgroup$
    – ZFR
    Dec 14 '18 at 18:22










  • $begingroup$
    Division by zero. View the definition of localization.
    $endgroup$
    – Wuestenfux
    Dec 15 '18 at 10:46


















-1












$begingroup$

Suppose $r,r'in R$ are mapped to $r/1=r'/1$. Then by definition, $s(r-r')=0$ for some $sin S$. If $S$ has no zero divisor, then $r=r'$. Note that the case $s=0in S$ cannot appear by def. Done.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Note that my approach is the same! Is $sneq 0$ then I am OK with that. What if $s=0$? In this case $rneq r'$ can also be the case.
    $endgroup$
    – ZFR
    Dec 14 '18 at 18:06










  • $begingroup$
    The denominators are coming from S.
    $endgroup$
    – Wuestenfux
    Dec 14 '18 at 18:18










  • $begingroup$
    Could you clarify your last comment, please?
    $endgroup$
    – ZFR
    Dec 14 '18 at 18:22










  • $begingroup$
    Division by zero. View the definition of localization.
    $endgroup$
    – Wuestenfux
    Dec 15 '18 at 10:46
















-1












-1








-1





$begingroup$

Suppose $r,r'in R$ are mapped to $r/1=r'/1$. Then by definition, $s(r-r')=0$ for some $sin S$. If $S$ has no zero divisor, then $r=r'$. Note that the case $s=0in S$ cannot appear by def. Done.






share|cite|improve this answer











$endgroup$



Suppose $r,r'in R$ are mapped to $r/1=r'/1$. Then by definition, $s(r-r')=0$ for some $sin S$. If $S$ has no zero divisor, then $r=r'$. Note that the case $s=0in S$ cannot appear by def. Done.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 17 '18 at 9:38

























answered Dec 14 '18 at 18:04









WuestenfuxWuestenfux

3,9701411




3,9701411












  • $begingroup$
    Note that my approach is the same! Is $sneq 0$ then I am OK with that. What if $s=0$? In this case $rneq r'$ can also be the case.
    $endgroup$
    – ZFR
    Dec 14 '18 at 18:06










  • $begingroup$
    The denominators are coming from S.
    $endgroup$
    – Wuestenfux
    Dec 14 '18 at 18:18










  • $begingroup$
    Could you clarify your last comment, please?
    $endgroup$
    – ZFR
    Dec 14 '18 at 18:22










  • $begingroup$
    Division by zero. View the definition of localization.
    $endgroup$
    – Wuestenfux
    Dec 15 '18 at 10:46




















  • $begingroup$
    Note that my approach is the same! Is $sneq 0$ then I am OK with that. What if $s=0$? In this case $rneq r'$ can also be the case.
    $endgroup$
    – ZFR
    Dec 14 '18 at 18:06










  • $begingroup$
    The denominators are coming from S.
    $endgroup$
    – Wuestenfux
    Dec 14 '18 at 18:18










  • $begingroup$
    Could you clarify your last comment, please?
    $endgroup$
    – ZFR
    Dec 14 '18 at 18:22










  • $begingroup$
    Division by zero. View the definition of localization.
    $endgroup$
    – Wuestenfux
    Dec 15 '18 at 10:46


















$begingroup$
Note that my approach is the same! Is $sneq 0$ then I am OK with that. What if $s=0$? In this case $rneq r'$ can also be the case.
$endgroup$
– ZFR
Dec 14 '18 at 18:06




$begingroup$
Note that my approach is the same! Is $sneq 0$ then I am OK with that. What if $s=0$? In this case $rneq r'$ can also be the case.
$endgroup$
– ZFR
Dec 14 '18 at 18:06












$begingroup$
The denominators are coming from S.
$endgroup$
– Wuestenfux
Dec 14 '18 at 18:18




$begingroup$
The denominators are coming from S.
$endgroup$
– Wuestenfux
Dec 14 '18 at 18:18












$begingroup$
Could you clarify your last comment, please?
$endgroup$
– ZFR
Dec 14 '18 at 18:22




$begingroup$
Could you clarify your last comment, please?
$endgroup$
– ZFR
Dec 14 '18 at 18:22












$begingroup$
Division by zero. View the definition of localization.
$endgroup$
– Wuestenfux
Dec 15 '18 at 10:46






$begingroup$
Division by zero. View the definition of localization.
$endgroup$
– Wuestenfux
Dec 15 '18 at 10:46




















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