When is the localization map injective?
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
add a comment |
$begingroup$
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
$endgroup$
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
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
add a comment |
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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