Commutative Ring with Identity
Multi tool use
$begingroup$
How can I show that $(mathbb Q,oplus,cdot)$ is a commutative ring with identity where $oplus$ and $cdot$ are defined as, $aoplus b=a+b-1$ and $acdot b=a+b$?
According to the book, an algebraic structure $(R,oplus,cdot)$ is called a ring if the following conditions are satisfied:
$(R,oplus)$ is an abelian group.- Associativity of multiplication holds: $acdot(bcdot c) = (acdot b)cdot c$.
- The left distributive law $acdot(boplus c)=(acdot b)oplus(acdot c)$ and the right distributive law $(boplus c)cdot a=(bcdot a)oplus(ccdot a)$ are satisfied by "$oplus$" and "$cdot$".
Though I was somehow able to prove first 2 conditions, the third condition is not getting satisfied. It's an "show that..." question, so the statement is definitely true. Can someone help me?
abstract-algebra ring-theory
edited Nov 29 '18 at 9:11
user1551
72.5k566127
asked Nov 29 '18 at 8:40
P.BendreP.Bendre
164
$endgroup$
|
show 3 more comments
$begingroup$
How can I show that $(mathbb Q,oplus,cdot)$ is a commutative ring with identity where $oplus$ and $cdot$ are defined as, $aoplus b=a+b-1$ and $acdot b=a+b$?
According to the book, an algebraic structure $(R,oplus,cdot)$ is called a ring if the following conditions are satisfied:
$(R,oplus)$ is an abelian group.- Associativity of multiplication holds: $acdot(bcdot c) = (acdot b)cdot c$.
- The left distributive law $acdot(boplus c)=(acdot b)oplus(acdot c)$ and the right distributive law $(boplus c)cdot a=(bcdot a)oplus(ccdot a)$ are satisfied by "$oplus$" and "$cdot$".
Though I was somehow able to prove first 2 conditions, the third condition is not getting satisfied. It's an "show that..." question, so the statement is definitely true. Can someone help me?
abstract-algebra ring-theory
edited Nov 29 '18 at 9:11
user1551
72.5k566127
asked Nov 29 '18 at 8:40
P.BendreP.Bendre
164
$endgroup$
1
$begingroup$
What do you need to show for something to be a ring and commutative? Moreover, what are $I$ and $+$ in this case?
$endgroup$
– Jonas Lenz
Nov 29 '18 at 8:44
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:49
$begingroup$
All I know that 1.it should be abelian group 2.it should follow associative and commutative law 3.it should follow left,right distributive law.Though I was able to prove first two conditions ,I am not able to figure out the 3rd condition.
$endgroup$
– P.Bendre
Nov 29 '18 at 8:50
$begingroup$
Can you please show your computations for distributivity?
$endgroup$
– Stockfish
Nov 29 '18 at 9:05
$begingroup$
for instance yes - you want to expand both sides and rearrange terms such that you get equality
$endgroup$
– Stockfish
Nov 29 '18 at 9:07
|
show 3 more comments
$begingroup$
How can I show that $(mathbb Q,oplus,cdot)$ is a commutative ring with identity where $oplus$ and $cdot$ are defined as, $aoplus b=a+b-1$ and $acdot b=a+b$?
According to the book, an algebraic structure $(R,oplus,cdot)$ is called a ring if the following conditions are satisfied:
$(R,oplus)$ is an abelian group.- Associativity of multiplication holds: $acdot(bcdot c) = (acdot b)cdot c$.
- The left distributive law $acdot(boplus c)=(acdot b)oplus(acdot c)$ and the right distributive law $(boplus c)cdot a=(bcdot a)oplus(ccdot a)$ are satisfied by "$oplus$" and "$cdot$".
Though I was somehow able to prove first 2 conditions, the third condition is not getting satisfied. It's an "show that..." question, so the statement is definitely true. Can someone help me?
abstract-algebra ring-theory
edited Nov 29 '18 at 9:11
user1551
72.5k566127
asked Nov 29 '18 at 8:40
P.BendreP.Bendre
164
$endgroup$
How can I show that $(mathbb Q,oplus,cdot)$ is a commutative ring with identity where $oplus$ and $cdot$ are defined as, $aoplus b=a+b-1$ and $acdot b=a+b$?
According to the book, an algebraic structure $(R,oplus,cdot)$ is called a ring if the following conditions are satisfied:
$(R,oplus)$ is an abelian group.- Associativity of multiplication holds: $acdot(bcdot c) = (acdot b)cdot c$.
- The left distributive law $acdot(boplus c)=(acdot b)oplus(acdot c)$ and the right distributive law $(boplus c)cdot a=(bcdot a)oplus(ccdot a)$ are satisfied by "$oplus$" and "$cdot$".
Though I was somehow able to prove first 2 conditions, the third condition is not getting satisfied. It's an "show that..." question, so the statement is definitely true. Can someone help me?
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Nov 29 '18 at 9:11
user1551
72.5k566127
asked Nov 29 '18 at 8:40
P.BendreP.Bendre
164
edited Nov 29 '18 at 9:11
user1551
72.5k566127
asked Nov 29 '18 at 8:40
P.BendreP.Bendre
164
edited Nov 29 '18 at 9:11
user1551
72.5k566127
edited Nov 29 '18 at 9:11
user1551
72.5k566127
edited Nov 29 '18 at 9:11
user1551
72.5k566127
72.5k566127
asked Nov 29 '18 at 8:40
P.BendreP.Bendre
164
asked Nov 29 '18 at 8:40
P.BendreP.Bendre
164
asked Nov 29 '18 at 8:40
P.BendreP.Bendre
164
164
1
$begingroup$
What do you need to show for something to be a ring and commutative? Moreover, what are $I$ and $+$ in this case?
$endgroup$
– Jonas Lenz
Nov 29 '18 at 8:44
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:49
$begingroup$
All I know that 1.it should be abelian group 2.it should follow associative and commutative law 3.it should follow left,right distributive law.Though I was able to prove first two conditions ,I am not able to figure out the 3rd condition.
$endgroup$
– P.Bendre
Nov 29 '18 at 8:50
$begingroup$
Can you please show your computations for distributivity?
$endgroup$
– Stockfish
Nov 29 '18 at 9:05
$begingroup$
for instance yes - you want to expand both sides and rearrange terms such that you get equality
$endgroup$
– Stockfish
Nov 29 '18 at 9:07
|
show 3 more comments
1
$begingroup$
What do you need to show for something to be a ring and commutative? Moreover, what are $I$ and $+$ in this case?
$endgroup$
– Jonas Lenz
Nov 29 '18 at 8:44
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:49
$begingroup$
All I know that 1.it should be abelian group 2.it should follow associative and commutative law 3.it should follow left,right distributive law.Though I was able to prove first two conditions ,I am not able to figure out the 3rd condition.
$endgroup$
– P.Bendre
Nov 29 '18 at 8:50
$begingroup$
Can you please show your computations for distributivity?
$endgroup$
– Stockfish
Nov 29 '18 at 9:05
$begingroup$
for instance yes - you want to expand both sides and rearrange terms such that you get equality
$endgroup$
– Stockfish
Nov 29 '18 at 9:07
1
1
$begingroup$
What do you need to show for something to be a ring and commutative? Moreover, what are $I$ and $+$ in this case?
$endgroup$
– Jonas Lenz
Nov 29 '18 at 8:44
$begingroup$
What do you need to show for something to be a ring and commutative? Moreover, what are $I$ and $+$ in this case?
$endgroup$
– Jonas Lenz
Nov 29 '18 at 8:44
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:49
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:49
$begingroup$
All I know that 1.it should be abelian group 2.it should follow associative and commutative law 3.it should follow left,right distributive law.Though I was able to prove first two conditions ,I am not able to figure out the 3rd condition.
$endgroup$
– P.Bendre
Nov 29 '18 at 8:50
$begingroup$
All I know that 1.it should be abelian group 2.it should follow associative and commutative law 3.it should follow left,right distributive law.Though I was able to prove first two conditions ,I am not able to figure out the 3rd condition.
$endgroup$
– P.Bendre
Nov 29 '18 at 8:50
$begingroup$
Can you please show your computations for distributivity?
$endgroup$
– Stockfish
Nov 29 '18 at 9:05
$begingroup$
Can you please show your computations for distributivity?
$endgroup$
– Stockfish
Nov 29 '18 at 9:05
$begingroup$
for instance yes - you want to expand both sides and rearrange terms such that you get equality
$endgroup$
– Stockfish
Nov 29 '18 at 9:07
$begingroup$
for instance yes - you want to expand both sides and rearrange terms such that you get equality
$endgroup$
– Stockfish
Nov 29 '18 at 9:07
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
It is obviously not a ring. In what follows, $0$ and $1$ refer to the neutral elements of the original ring addition and multiplication, respectively.
Just note that $1$ is the neutral element for $oplus$, and so it should satisfy $acdot 1=1$ for all $ain R$, if $R$ is to be a ring. (This is because we know the additive identity is multiplicatively absorbing in a ring.)
But it does not: $acdot 1:=a+1neq 1$, for any $aneq 0$.
In case you have radically mistyped your problem, I would encourage you to search for duplicates before asking. This and this and this are all similar, and may explain the answer to you faster than re-asking.
answered Nov 29 '18 at 14:22
rschwiebrschwieb
106k12102249
$endgroup$
add a comment |
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$begingroup$
It is obviously not a ring. In what follows, $0$ and $1$ refer to the neutral elements of the original ring addition and multiplication, respectively.
Just note that $1$ is the neutral element for $oplus$, and so it should satisfy $acdot 1=1$ for all $ain R$, if $R$ is to be a ring. (This is because we know the additive identity is multiplicatively absorbing in a ring.)
But it does not: $acdot 1:=a+1neq 1$, for any $aneq 0$.
In case you have radically mistyped your problem, I would encourage you to search for duplicates before asking. This and this and this are all similar, and may explain the answer to you faster than re-asking.
answered Nov 29 '18 at 14:22
rschwiebrschwieb
106k12102249
$endgroup$
add a comment |
$begingroup$
It is obviously not a ring. In what follows, $0$ and $1$ refer to the neutral elements of the original ring addition and multiplication, respectively.
Just note that $1$ is the neutral element for $oplus$, and so it should satisfy $acdot 1=1$ for all $ain R$, if $R$ is to be a ring. (This is because we know the additive identity is multiplicatively absorbing in a ring.)
But it does not: $acdot 1:=a+1neq 1$, for any $aneq 0$.
In case you have radically mistyped your problem, I would encourage you to search for duplicates before asking. This and this and this are all similar, and may explain the answer to you faster than re-asking.
answered Nov 29 '18 at 14:22
rschwiebrschwieb
106k12102249
$endgroup$
add a comment |
$begingroup$
It is obviously not a ring. In what follows, $0$ and $1$ refer to the neutral elements of the original ring addition and multiplication, respectively.
Just note that $1$ is the neutral element for $oplus$, and so it should satisfy $acdot 1=1$ for all $ain R$, if $R$ is to be a ring. (This is because we know the additive identity is multiplicatively absorbing in a ring.)
But it does not: $acdot 1:=a+1neq 1$, for any $aneq 0$.
In case you have radically mistyped your problem, I would encourage you to search for duplicates before asking. This and this and this are all similar, and may explain the answer to you faster than re-asking.
answered Nov 29 '18 at 14:22
rschwiebrschwieb
106k12102249
$endgroup$
It is obviously not a ring. In what follows, $0$ and $1$ refer to the neutral elements of the original ring addition and multiplication, respectively.
Just note that $1$ is the neutral element for $oplus$, and so it should satisfy $acdot 1=1$ for all $ain R$, if $R$ is to be a ring. (This is because we know the additive identity is multiplicatively absorbing in a ring.)
But it does not: $acdot 1:=a+1neq 1$, for any $aneq 0$.
In case you have radically mistyped your problem, I would encourage you to search for duplicates before asking. This and this and this are all similar, and may explain the answer to you faster than re-asking.
answered Nov 29 '18 at 14:22
rschwiebrschwieb
106k12102249
edited Nov 29 '18 at 14:27
answered Nov 29 '18 at 14:22
rschwiebrschwieb
106k12102249
answered Nov 29 '18 at 14:22
rschwiebrschwieb
106k12102249
answered Nov 29 '18 at 14:22
rschwiebrschwieb
106k12102249
106k12102249
add a comment |
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phx5avmUZ,SS,wwQw0,UErctD,DNo 3wemT9QSvChW1Nc,CnC2UlQYBY4E,OElpSdL7K4k
1
$begingroup$
What do you need to show for something to be a ring and commutative? Moreover, what are $I$ and $+$ in this case?
$endgroup$
– Jonas Lenz
Nov 29 '18 at 8:44
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:49
$begingroup$
All I know that 1.it should be abelian group 2.it should follow associative and commutative law 3.it should follow left,right distributive law.Though I was able to prove first two conditions ,I am not able to figure out the 3rd condition.
$endgroup$
– P.Bendre
Nov 29 '18 at 8:50
$begingroup$
Can you please show your computations for distributivity?
$endgroup$
– Stockfish
Nov 29 '18 at 9:05
$begingroup$
for instance yes - you want to expand both sides and rearrange terms such that you get equality
$endgroup$
– Stockfish
Nov 29 '18 at 9:07