If $amid b$ and $bmid a$, $a,bin R$, show that $exists w in R^times$ with $b=wa$ [duplicate]












2












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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







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marked as duplicate by amWhy, Bill Dubuque divisibility
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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




    $begingroup$
    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 $x$)?
    $endgroup$
    – platty
    Nov 29 '18 at 22:28












  • $begingroup$
    Presumably $R^x$ denotes the group of units of $R$ under multiplication
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 22:30










  • $begingroup$
    Indeed, for any $a, b in mathbb Z,$ if $a = -b$, we still have $a mid b$ and $b mid a$.
    $endgroup$
    – amWhy
    Nov 29 '18 at 22:31












  • $begingroup$
    "Einheitsgruppe" = "Group of units"
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 22:43
















2












$begingroup$



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







share|cite|improve this question











$endgroup$



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




    $begingroup$
    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 $x$)?
    $endgroup$
    – platty
    Nov 29 '18 at 22:28












  • $begingroup$
    Presumably $R^x$ denotes the group of units of $R$ under multiplication
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 22:30










  • $begingroup$
    Indeed, for any $a, b in mathbb Z,$ if $a = -b$, we still have $a mid b$ and $b mid a$.
    $endgroup$
    – amWhy
    Nov 29 '18 at 22:31












  • $begingroup$
    "Einheitsgruppe" = "Group of units"
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 22:43














2












2








2





$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 23:00







Neon Xd

















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
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




    $begingroup$
    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 $x$)?
    $endgroup$
    – platty
    Nov 29 '18 at 22:28












  • $begingroup$
    Presumably $R^x$ denotes the group of units of $R$ under multiplication
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 22:30










  • $begingroup$
    Indeed, for any $a, b in mathbb Z,$ if $a = -b$, we still have $a mid b$ and $b mid a$.
    $endgroup$
    – amWhy
    Nov 29 '18 at 22:31












  • $begingroup$
    "Einheitsgruppe" = "Group of units"
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 22:43














  • 1




    $begingroup$
    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 $x$)?
    $endgroup$
    – platty
    Nov 29 '18 at 22:28












  • $begingroup$
    Presumably $R^x$ denotes the group of units of $R$ under multiplication
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 22:30










  • $begingroup$
    Indeed, for any $a, b in mathbb Z,$ if $a = -b$, we still have $a mid b$ and $b mid a$.
    $endgroup$
    – amWhy
    Nov 29 '18 at 22:31












  • $begingroup$
    "Einheitsgruppe" = "Group of units"
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 22:43








1




1




$begingroup$
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 $x$)?
$endgroup$
– platty
Nov 29 '18 at 22:28






$begingroup$
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 $x$)?
$endgroup$
– platty
Nov 29 '18 at 22:28














$begingroup$
Presumably $R^x$ denotes the group of units of $R$ under multiplication
$endgroup$
– Badam Baplan
Nov 29 '18 at 22:30




$begingroup$
Presumably $R^x$ denotes the group of units of $R$ under multiplication
$endgroup$
– Badam Baplan
Nov 29 '18 at 22:30












$begingroup$
Indeed, for any $a, b in mathbb Z,$ if $a = -b$, we still have $a mid b$ and $b mid a$.
$endgroup$
– amWhy
Nov 29 '18 at 22:31






$begingroup$
Indeed, for any $a, b in mathbb Z,$ if $a = -b$, we still have $a mid b$ and $b mid a$.
$endgroup$
– amWhy
Nov 29 '18 at 22:31














$begingroup$
"Einheitsgruppe" = "Group of units"
$endgroup$
– Badam Baplan
Nov 29 '18 at 22:43




$begingroup$
"Einheitsgruppe" = "Group of units"
$endgroup$
– Badam Baplan
Nov 29 '18 at 22:43










1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You mean $uv=1$ right?
    $endgroup$
    – Neon Xd
    Nov 29 '18 at 23:04










  • $begingroup$
    yes! thank you.
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 23:06


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You mean $uv=1$ right?
    $endgroup$
    – Neon Xd
    Nov 29 '18 at 23:04










  • $begingroup$
    yes! thank you.
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 23:06
















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You mean $uv=1$ right?
    $endgroup$
    – Neon Xd
    Nov 29 '18 at 23:04










  • $begingroup$
    yes! thank you.
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 23:06














1












1








1





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 '18 at 23:04

























answered Nov 29 '18 at 23:01









Badam BaplanBadam Baplan

4,586722




4,586722












  • $begingroup$
    You mean $uv=1$ right?
    $endgroup$
    – Neon Xd
    Nov 29 '18 at 23:04










  • $begingroup$
    yes! thank you.
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 23:06


















  • $begingroup$
    You mean $uv=1$ right?
    $endgroup$
    – Neon Xd
    Nov 29 '18 at 23:04










  • $begingroup$
    yes! thank you.
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 23:06
















$begingroup$
You mean $uv=1$ right?
$endgroup$
– Neon Xd
Nov 29 '18 at 23:04




$begingroup$
You mean $uv=1$ right?
$endgroup$
– Neon Xd
Nov 29 '18 at 23:04












$begingroup$
yes! thank you.
$endgroup$
– Badam Baplan
Nov 29 '18 at 23:06




$begingroup$
yes! thank you.
$endgroup$
– Badam Baplan
Nov 29 '18 at 23:06



-

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