If $amid b$ and $bmid a$, $a,bin R$, show that $exists w in R^times$ with $b=wa$ [duplicate]
This question already has an answer here:
If $R$ is a commutative ring with identity, and $a, bin R$ are divisible by each other, is it true that they must be associates?
3 answers
$R$ - commutative Ring, and $a$ non-zero-divisor
$R^x$ - multiplicative Monoid? (it's called Einheitsgruppe in German, maybe unitary group?)
I started by stating, that if $amid b$ and $bmid a rightarrow a=b$.
So for $amid b$, $exists xin R$ with $xa=b$ and for $bmid a,exists yin R$ with $yb=a$
Now: $b=xa=xyb=wb$, with $w=xy$.
$stackrel{a=b}{rightarrow} b=wa$
Now the problem that I think I have with my proof, is that R is not a given Ring. For example $a=b$ wouldn't work in $mathbb{Z}$. Is there any other way to do the proof?
elementary-number-theory arithmetic divisibility
asked Nov 29 '18 at 22:26
Neon XdNeon Xd
183
marked as duplicate by amWhy, Bill Dubuque divisibility
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Dec 1 '18 at 16:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
If $R$ is a commutative ring with identity, and $a, bin R$ are divisible by each other, is it true that they must be associates?
3 answers
$R$ - commutative Ring, and $a$ non-zero-divisor
$R^x$ - multiplicative Monoid? (it's called Einheitsgruppe in German, maybe unitary group?)
I started by stating, that if $amid b$ and $bmid a rightarrow a=b$.
So for $amid b$, $exists xin R$ with $xa=b$ and for $bmid a,exists yin R$ with $yb=a$
Now: $b=xa=xyb=wb$, with $w=xy$.
$stackrel{a=b}{rightarrow} b=wa$
Now the problem that I think I have with my proof, is that R is not a given Ring. For example $a=b$ wouldn't work in $mathbb{Z}$. Is there any other way to do the proof?
elementary-number-theory arithmetic divisibility
asked Nov 29 '18 at 22:26
Neon XdNeon Xd
183
marked as duplicate by amWhy, Bill Dubuque divisibility
Users with the divisibility badge can single-handedly close divisibility questions as duplicates and reopen them as needed.
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Dec 1 '18 at 16:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
It's not necessarily true that $a | b$ and $b | a$ implies $a = b$ (even in $mathbb{Z}$, we have $1 | -1$ and $-1 | 1$). What does the notation $R^x$ denote (in particular, what is $x$)?
– platty
Nov 29 '18 at 22:28
Presumably $R^x$ denotes the group of units of $R$ under multiplication
– Badam Baplan
Nov 29 '18 at 22:30
Indeed, for any $a, b in mathbb Z,$ if $a = -b$, we still have $a mid b$ and $b mid a$.
– amWhy
Nov 29 '18 at 22:31
"Einheitsgruppe" = "Group of units"
– Badam Baplan
Nov 29 '18 at 22:43
add a comment |
This question already has an answer here:
If $R$ is a commutative ring with identity, and $a, bin R$ are divisible by each other, is it true that they must be associates?
3 answers
$R$ - commutative Ring, and $a$ non-zero-divisor
$R^x$ - multiplicative Monoid? (it's called Einheitsgruppe in German, maybe unitary group?)
I started by stating, that if $amid b$ and $bmid a rightarrow a=b$.
So for $amid b$, $exists xin R$ with $xa=b$ and for $bmid a,exists yin R$ with $yb=a$
Now: $b=xa=xyb=wb$, with $w=xy$.
$stackrel{a=b}{rightarrow} b=wa$
Now the problem that I think I have with my proof, is that R is not a given Ring. For example $a=b$ wouldn't work in $mathbb{Z}$. Is there any other way to do the proof?
elementary-number-theory arithmetic divisibility
asked Nov 29 '18 at 22:26
Neon XdNeon Xd
183
This question already has an answer here:
If $R$ is a commutative ring with identity, and $a, bin R$ are divisible by each other, is it true that they must be associates?
3 answers
$R$ - commutative Ring, and $a$ non-zero-divisor
$R^x$ - multiplicative Monoid? (it's called Einheitsgruppe in German, maybe unitary group?)
I started by stating, that if $amid b$ and $bmid a rightarrow a=b$.
So for $amid b$, $exists xin R$ with $xa=b$ and for $bmid a,exists yin R$ with $yb=a$
Now: $b=xa=xyb=wb$, with $w=xy$.
$stackrel{a=b}{rightarrow} b=wa$
Now the problem that I think I have with my proof, is that R is not a given Ring. For example $a=b$ wouldn't work in $mathbb{Z}$. Is there any other way to do the proof?
This question already has an answer here:
If $R$ is a commutative ring with identity, and $a, bin R$ are divisible by each other, is it true that they must be associates?
3 answers
elementary-number-theory arithmetic divisibility
elementary-number-theory arithmetic divisibility
asked Nov 29 '18 at 22:26
Neon XdNeon Xd
183
asked Nov 29 '18 at 22:26
Neon XdNeon Xd
183
edited Nov 29 '18 at 23:00
Neon Xd
asked Nov 29 '18 at 22:26
Neon XdNeon Xd
183
asked Nov 29 '18 at 22:26
Neon XdNeon Xd
183
asked Nov 29 '18 at 22:26
Neon XdNeon Xd
183
183
marked as duplicate by amWhy, Bill Dubuque divisibility
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Dec 1 '18 at 16:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by amWhy, Bill Dubuque divisibility
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Dec 1 '18 at 16:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
It's not necessarily true that $a | b$ and $b | a$ implies $a = b$ (even in $mathbb{Z}$, we have $1 | -1$ and $-1 | 1$). What does the notation $R^x$ denote (in particular, what is $x$)?
– platty
Nov 29 '18 at 22:28
Presumably $R^x$ denotes the group of units of $R$ under multiplication
– Badam Baplan
Nov 29 '18 at 22:30
Indeed, for any $a, b in mathbb Z,$ if $a = -b$, we still have $a mid b$ and $b mid a$.
– amWhy
Nov 29 '18 at 22:31
"Einheitsgruppe" = "Group of units"
– Badam Baplan
Nov 29 '18 at 22:43
add a comment |
1
It's not necessarily true that $a | b$ and $b | a$ implies $a = b$ (even in $mathbb{Z}$, we have $1 | -1$ and $-1 | 1$). What does the notation $R^x$ denote (in particular, what is $x$)?
– platty
Nov 29 '18 at 22:28
Presumably $R^x$ denotes the group of units of $R$ under multiplication
– Badam Baplan
Nov 29 '18 at 22:30
Indeed, for any $a, b in mathbb Z,$ if $a = -b$, we still have $a mid b$ and $b mid a$.
– amWhy
Nov 29 '18 at 22:31
"Einheitsgruppe" = "Group of units"
– Badam Baplan
Nov 29 '18 at 22:43
1
1
It's not necessarily true that $a | b$ and $b | a$ implies $a = b$ (even in $mathbb{Z}$, we have $1 | -1$ and $-1 | 1$). What does the notation $R^x$ denote (in particular, what is $x$)?
– platty
Nov 29 '18 at 22:28
It's not necessarily true that $a | b$ and $b | a$ implies $a = b$ (even in $mathbb{Z}$, we have $1 | -1$ and $-1 | 1$). What does the notation $R^x$ denote (in particular, what is $x$)?
– platty
Nov 29 '18 at 22:28
Presumably $R^x$ denotes the group of units of $R$ under multiplication
– Badam Baplan
Nov 29 '18 at 22:30
Presumably $R^x$ denotes the group of units of $R$ under multiplication
– Badam Baplan
Nov 29 '18 at 22:30
Indeed, for any $a, b in mathbb Z,$ if $a = -b$, we still have $a mid b$ and $b mid a$.
– amWhy
Nov 29 '18 at 22:31
Indeed, for any $a, b in mathbb Z,$ if $a = -b$, we still have $a mid b$ and $b mid a$.
– amWhy
Nov 29 '18 at 22:31
"Einheitsgruppe" = "Group of units"
– Badam Baplan
Nov 29 '18 at 22:43
"Einheitsgruppe" = "Group of units"
– Badam Baplan
Nov 29 '18 at 22:43
add a comment |
1 Answer
1
active
oldest
votes
I assume you are working in a commutative ring with identity.
The problem is to show that if two elements of a ring divide each other, then they are associates, i.e. the same up to a unit multiple.
Actually this is not true in general. For example, in the ring
$mathbb{Z}[x,y,z]/(xy -z, zy - x)$ we have that $x mid z$ and $z mid x$, but $y$ is not a unit. Note that $x$ and $z$ are both zero divisors, as $z(y^2 - 1) = x(y^2-1) = 0$.
In general,
$a mid b$ means there exists $u in R$ with $au = b$.
$b mid a$ means there exists $v in R$ with $bv = a$.
and putting these equations together...
$$buv = b$$
$$auv = a$$
It follows that if either $a$ or $b$ is not a zero divisor, then $uv = 1$ and $u,v$ are units, therefore $a$ and $b$ are associates.
answered Nov 29 '18 at 23:01
Badam BaplanBadam Baplan
4,476722
You mean $uv=1$ right?
– Neon Xd
Nov 29 '18 at 23:04
yes! thank you.
– Badam Baplan
Nov 29 '18 at 23:06
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I assume you are working in a commutative ring with identity.
The problem is to show that if two elements of a ring divide each other, then they are associates, i.e. the same up to a unit multiple.
Actually this is not true in general. For example, in the ring
$mathbb{Z}[x,y,z]/(xy -z, zy - x)$ we have that $x mid z$ and $z mid x$, but $y$ is not a unit. Note that $x$ and $z$ are both zero divisors, as $z(y^2 - 1) = x(y^2-1) = 0$.
In general,
$a mid b$ means there exists $u in R$ with $au = b$.
$b mid a$ means there exists $v in R$ with $bv = a$.
and putting these equations together...
$$buv = b$$
$$auv = a$$
It follows that if either $a$ or $b$ is not a zero divisor, then $uv = 1$ and $u,v$ are units, therefore $a$ and $b$ are associates.
answered Nov 29 '18 at 23:01
Badam BaplanBadam Baplan
4,476722
You mean $uv=1$ right?
– Neon Xd
Nov 29 '18 at 23:04
yes! thank you.
– Badam Baplan
Nov 29 '18 at 23:06
add a comment |
I assume you are working in a commutative ring with identity.
The problem is to show that if two elements of a ring divide each other, then they are associates, i.e. the same up to a unit multiple.
Actually this is not true in general. For example, in the ring
$mathbb{Z}[x,y,z]/(xy -z, zy - x)$ we have that $x mid z$ and $z mid x$, but $y$ is not a unit. Note that $x$ and $z$ are both zero divisors, as $z(y^2 - 1) = x(y^2-1) = 0$.
In general,
$a mid b$ means there exists $u in R$ with $au = b$.
$b mid a$ means there exists $v in R$ with $bv = a$.
and putting these equations together...
$$buv = b$$
$$auv = a$$
It follows that if either $a$ or $b$ is not a zero divisor, then $uv = 1$ and $u,v$ are units, therefore $a$ and $b$ are associates.
answered Nov 29 '18 at 23:01
Badam BaplanBadam Baplan
4,476722
You mean $uv=1$ right?
– Neon Xd
Nov 29 '18 at 23:04
yes! thank you.
– Badam Baplan
Nov 29 '18 at 23:06
add a comment |
I assume you are working in a commutative ring with identity.
The problem is to show that if two elements of a ring divide each other, then they are associates, i.e. the same up to a unit multiple.
Actually this is not true in general. For example, in the ring
$mathbb{Z}[x,y,z]/(xy -z, zy - x)$ we have that $x mid z$ and $z mid x$, but $y$ is not a unit. Note that $x$ and $z$ are both zero divisors, as $z(y^2 - 1) = x(y^2-1) = 0$.
In general,
$a mid b$ means there exists $u in R$ with $au = b$.
$b mid a$ means there exists $v in R$ with $bv = a$.
and putting these equations together...
$$buv = b$$
$$auv = a$$
It follows that if either $a$ or $b$ is not a zero divisor, then $uv = 1$ and $u,v$ are units, therefore $a$ and $b$ are associates.
answered Nov 29 '18 at 23:01
Badam BaplanBadam Baplan
4,476722
I assume you are working in a commutative ring with identity.
The problem is to show that if two elements of a ring divide each other, then they are associates, i.e. the same up to a unit multiple.
Actually this is not true in general. For example, in the ring
$mathbb{Z}[x,y,z]/(xy -z, zy - x)$ we have that $x mid z$ and $z mid x$, but $y$ is not a unit. Note that $x$ and $z$ are both zero divisors, as $z(y^2 - 1) = x(y^2-1) = 0$.
In general,
$a mid b$ means there exists $u in R$ with $au = b$.
$b mid a$ means there exists $v in R$ with $bv = a$.
and putting these equations together...
$$buv = b$$
$$auv = a$$
It follows that if either $a$ or $b$ is not a zero divisor, then $uv = 1$ and $u,v$ are units, therefore $a$ and $b$ are associates.
answered Nov 29 '18 at 23:01
Badam BaplanBadam Baplan
4,476722
edited Nov 29 '18 at 23:04
answered Nov 29 '18 at 23:01
Badam BaplanBadam Baplan
4,476722
answered Nov 29 '18 at 23:01
Badam BaplanBadam Baplan
4,476722
answered Nov 29 '18 at 23:01
Badam BaplanBadam Baplan
4,476722
4,476722
You mean $uv=1$ right?
– Neon Xd
Nov 29 '18 at 23:04
yes! thank you.
– Badam Baplan
Nov 29 '18 at 23:06
add a comment |
You mean $uv=1$ right?
– Neon Xd
Nov 29 '18 at 23:04
yes! thank you.
– Badam Baplan
Nov 29 '18 at 23:06
You mean $uv=1$ right?
– Neon Xd
Nov 29 '18 at 23:04
You mean $uv=1$ right?
– Neon Xd
Nov 29 '18 at 23:04
yes! thank you.
– Badam Baplan
Nov 29 '18 at 23:06
yes! thank you.
– Badam Baplan
Nov 29 '18 at 23:06
add a comment |
1
It's not necessarily true that $a | b$ and $b | a$ implies $a = b$ (even in $mathbb{Z}$, we have $1 | -1$ and $-1 | 1$). What does the notation $R^x$ denote (in particular, what is $x$)?
– platty
Nov 29 '18 at 22:28
Presumably $R^x$ denotes the group of units of $R$ under multiplication
– Badam Baplan
Nov 29 '18 at 22:30
Indeed, for any $a, b in mathbb Z,$ if $a = -b$, we still have $a mid b$ and $b mid a$.
– amWhy
Nov 29 '18 at 22:31
"Einheitsgruppe" = "Group of units"
– Badam Baplan
Nov 29 '18 at 22:43