Mixing
For a general ring $R$, if $R$ has a unique right-unity, then it has a unity.
4
$begingroup$
Problem: For a general ring $R$ I want to show that:
if $R$ has a unique right unity, then it has an overall unity.
Attempt:
Suppose $e$ is the unique right unity of $R$. Then for any $r in R$, we have that
$re=r$ and so $er=ere$ and $er-ere=0$. From here I feel the answer should be obvious but I'm not seeing it right away. Help appreciated.
abstract-algebra ring-theory
edited Jan 19 at 20:33
Matt Samuel
38.7k63769
asked Jan 19 at 20:20
IntegrateThisIntegrateThis
1,9111718
$endgroup$
add a comment |
4
$begingroup$
Problem: For a general ring $R$ I want to show that:
if $R$ has a unique right unity, then it has an overall unity.
Attempt:
Suppose $e$ is the unique right unity of $R$. Then for any $r in R$, we have that
$re=r$ and so $er=ere$ and $er-ere=0$. From here I feel the answer should be obvious but I'm not seeing it right away. Help appreciated.
abstract-algebra ring-theory
edited Jan 19 at 20:33
Matt Samuel
38.7k63769
asked Jan 19 at 20:20
IntegrateThisIntegrateThis
1,9111718
$endgroup$
$begingroup$
Do you mean $R$ has an identity or $R$ has a unit?
$endgroup$
– John Douma
Jan 19 at 20:27
$begingroup$
@JohnDouma unity=identity I believe.
$endgroup$
– IntegrateThis
Jan 19 at 20:30
$begingroup$
I have never seen that term.
$endgroup$
– John Douma
Jan 19 at 20:30
$begingroup$
@JohnDouma ok. well I got it from Nicholson's Abstract Algebra.
$endgroup$
– IntegrateThis
Jan 19 at 20:31
$begingroup$
I didn't say it doesn't exist. I just have never heard of it.
$endgroup$
– John Douma
Jan 19 at 20:31
add a comment |
4
4
4
$begingroup$
Problem: For a general ring $R$ I want to show that:
if $R$ has a unique right unity, then it has an overall unity.
Attempt:
Suppose $e$ is the unique right unity of $R$. Then for any $r in R$, we have that
$re=r$ and so $er=ere$ and $er-ere=0$. From here I feel the answer should be obvious but I'm not seeing it right away. Help appreciated.
abstract-algebra ring-theory
edited Jan 19 at 20:33
Matt Samuel
38.7k63769
asked Jan 19 at 20:20
IntegrateThisIntegrateThis
1,9111718
$endgroup$
Problem: For a general ring $R$ I want to show that:
if $R$ has a unique right unity, then it has an overall unity.
Attempt:
Suppose $e$ is the unique right unity of $R$. Then for any $r in R$, we have that
$re=r$ and so $er=ere$ and $er-ere=0$. From here I feel the answer should be obvious but I'm not seeing it right away. Help appreciated.
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Jan 19 at 20:33
Matt Samuel
38.7k63769
asked Jan 19 at 20:20
IntegrateThisIntegrateThis
1,9111718
edited Jan 19 at 20:33
Matt Samuel
38.7k63769
asked Jan 19 at 20:20
IntegrateThisIntegrateThis
1,9111718
edited Jan 19 at 20:33
Matt Samuel
38.7k63769
edited Jan 19 at 20:33
Matt Samuel
38.7k63769
edited Jan 19 at 20:33
Matt Samuel
38.7k63769
38.7k63769
asked Jan 19 at 20:20
IntegrateThisIntegrateThis
1,9111718
asked Jan 19 at 20:20
IntegrateThisIntegrateThis
1,9111718
asked Jan 19 at 20:20
IntegrateThisIntegrateThis
1,9111718
1,9111718
$begingroup$
Do you mean $R$ has an identity or $R$ has a unit?
$endgroup$
– John Douma
Jan 19 at 20:27
$begingroup$
@JohnDouma unity=identity I believe.
$endgroup$
– IntegrateThis
Jan 19 at 20:30
$begingroup$
I have never seen that term.
$endgroup$
– John Douma
Jan 19 at 20:30
$begingroup$
@JohnDouma ok. well I got it from Nicholson's Abstract Algebra.
$endgroup$
– IntegrateThis
Jan 19 at 20:31
$begingroup$
I didn't say it doesn't exist. I just have never heard of it.
$endgroup$
– John Douma
Jan 19 at 20:31
add a comment |
$begingroup$
Do you mean $R$ has an identity or $R$ has a unit?
$endgroup$
– John Douma
Jan 19 at 20:27
$begingroup$
@JohnDouma unity=identity I believe.
$endgroup$
– IntegrateThis
Jan 19 at 20:30
$begingroup$
I have never seen that term.
$endgroup$
– John Douma
Jan 19 at 20:30
$begingroup$
@JohnDouma ok. well I got it from Nicholson's Abstract Algebra.
$endgroup$
– IntegrateThis
Jan 19 at 20:31
$begingroup$
I didn't say it doesn't exist. I just have never heard of it.
$endgroup$
– John Douma
Jan 19 at 20:31
$begingroup$
Do you mean $R$ has an identity or $R$ has a unit?
$endgroup$
– John Douma
Jan 19 at 20:27
$begingroup$
Do you mean $R$ has an identity or $R$ has a unit?
$endgroup$
– John Douma
Jan 19 at 20:27
$begingroup$
@JohnDouma unity=identity I believe.
$endgroup$
– IntegrateThis
Jan 19 at 20:30
$begingroup$
@JohnDouma unity=identity I believe.
$endgroup$
– IntegrateThis
Jan 19 at 20:30
$begingroup$
I have never seen that term.
$endgroup$
– John Douma
Jan 19 at 20:30
$begingroup$
I have never seen that term.
$endgroup$
– John Douma
Jan 19 at 20:30
$begingroup$
@JohnDouma ok. well I got it from Nicholson's Abstract Algebra.
$endgroup$
– IntegrateThis
Jan 19 at 20:31
$begingroup$
@JohnDouma ok. well I got it from Nicholson's Abstract Algebra.
$endgroup$
– IntegrateThis
Jan 19 at 20:31
$begingroup$
I didn't say it doesn't exist. I just have never heard of it.
$endgroup$
– John Douma
Jan 19 at 20:31
$begingroup$
I didn't say it doesn't exist. I just have never heard of it.
$endgroup$
– John Douma
Jan 19 at 20:31
add a comment |
1 Answer
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$begingroup$
The key here is that it is unique. Suppose there is no element $s$ such that $sr=rs=r$ for all $r$. Then there exists $r$ such that $erneq r$. Then $e'=e+er-r$ is a distinct right identity, which is a contradiction.
answered Jan 19 at 20:32
Matt SamuelMatt Samuel
38.7k63769
$endgroup$
$begingroup$
This seems correct to me. Just having difficulty seeing why $e'$ is an identity.
$endgroup$
– IntegrateThis
Jan 19 at 20:34
$begingroup$
@Integrate It's a proof by contradiction. It shows that a two sided identity must exist. Since this would also be a right identity, it is in fact equal to $e$. It's not required to show that though.
$endgroup$
– Matt Samuel
Jan 19 at 20:38
$begingroup$
Nevermind. All good thanks!
$endgroup$
– IntegrateThis
Jan 19 at 20:40
$begingroup$
@Integrate That wouldn't be true necessarily. $e'$ is a fabricated element that doesn't exist. You could phrase it as a direct proof by saying that if the ring has a right identity that is not a left identity, then the right identity is not unique. There's no mention of a left identity there.
$endgroup$
– Matt Samuel
Jan 19 at 20:41
$begingroup$
@Integrate Sure, no problem.
$endgroup$
– Matt Samuel
Jan 19 at 20:41
add a comment |
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1 Answer
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1 Answer
1
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votes
5
$begingroup$
The key here is that it is unique. Suppose there is no element $s$ such that $sr=rs=r$ for all $r$. Then there exists $r$ such that $erneq r$. Then $e'=e+er-r$ is a distinct right identity, which is a contradiction.
answered Jan 19 at 20:32
Matt SamuelMatt Samuel
38.7k63769
$endgroup$
$begingroup$
This seems correct to me. Just having difficulty seeing why $e'$ is an identity.
$endgroup$
– IntegrateThis
Jan 19 at 20:34
$begingroup$
@Integrate It's a proof by contradiction. It shows that a two sided identity must exist. Since this would also be a right identity, it is in fact equal to $e$. It's not required to show that though.
$endgroup$
– Matt Samuel
Jan 19 at 20:38
$begingroup$
Nevermind. All good thanks!
$endgroup$
– IntegrateThis
Jan 19 at 20:40
$begingroup$
@Integrate That wouldn't be true necessarily. $e'$ is a fabricated element that doesn't exist. You could phrase it as a direct proof by saying that if the ring has a right identity that is not a left identity, then the right identity is not unique. There's no mention of a left identity there.
$endgroup$
– Matt Samuel
Jan 19 at 20:41
$begingroup$
@Integrate Sure, no problem.
$endgroup$
– Matt Samuel
Jan 19 at 20:41
add a comment |
5
$begingroup$
The key here is that it is unique. Suppose there is no element $s$ such that $sr=rs=r$ for all $r$. Then there exists $r$ such that $erneq r$. Then $e'=e+er-r$ is a distinct right identity, which is a contradiction.
answered Jan 19 at 20:32
Matt SamuelMatt Samuel
38.7k63769
$endgroup$
$begingroup$
This seems correct to me. Just having difficulty seeing why $e'$ is an identity.
$endgroup$
– IntegrateThis
Jan 19 at 20:34
$begingroup$
@Integrate It's a proof by contradiction. It shows that a two sided identity must exist. Since this would also be a right identity, it is in fact equal to $e$. It's not required to show that though.
$endgroup$
– Matt Samuel
Jan 19 at 20:38
$begingroup$
Nevermind. All good thanks!
$endgroup$
– IntegrateThis
Jan 19 at 20:40
$begingroup$
@Integrate That wouldn't be true necessarily. $e'$ is a fabricated element that doesn't exist. You could phrase it as a direct proof by saying that if the ring has a right identity that is not a left identity, then the right identity is not unique. There's no mention of a left identity there.
$endgroup$
– Matt Samuel
Jan 19 at 20:41
$begingroup$
@Integrate Sure, no problem.
$endgroup$
– Matt Samuel
Jan 19 at 20:41
add a comment |
5
5
5
$begingroup$
The key here is that it is unique. Suppose there is no element $s$ such that $sr=rs=r$ for all $r$. Then there exists $r$ such that $erneq r$. Then $e'=e+er-r$ is a distinct right identity, which is a contradiction.
answered Jan 19 at 20:32
Matt SamuelMatt Samuel
38.7k63769
$endgroup$
The key here is that it is unique. Suppose there is no element $s$ such that $sr=rs=r$ for all $r$. Then there exists $r$ such that $erneq r$. Then $e'=e+er-r$ is a distinct right identity, which is a contradiction.
answered Jan 19 at 20:32
Matt SamuelMatt Samuel
38.7k63769
answered Jan 19 at 20:32
Matt SamuelMatt Samuel
38.7k63769
answered Jan 19 at 20:32
Matt SamuelMatt Samuel
38.7k63769
answered Jan 19 at 20:32
Matt SamuelMatt Samuel
38.7k63769
38.7k63769
$begingroup$
This seems correct to me. Just having difficulty seeing why $e'$ is an identity.
$endgroup$
– IntegrateThis
Jan 19 at 20:34
$begingroup$
@Integrate It's a proof by contradiction. It shows that a two sided identity must exist. Since this would also be a right identity, it is in fact equal to $e$. It's not required to show that though.
$endgroup$
– Matt Samuel
Jan 19 at 20:38
$begingroup$
Nevermind. All good thanks!
$endgroup$
– IntegrateThis
Jan 19 at 20:40
$begingroup$
@Integrate That wouldn't be true necessarily. $e'$ is a fabricated element that doesn't exist. You could phrase it as a direct proof by saying that if the ring has a right identity that is not a left identity, then the right identity is not unique. There's no mention of a left identity there.
$endgroup$
– Matt Samuel
Jan 19 at 20:41
$begingroup$
@Integrate Sure, no problem.
$endgroup$
– Matt Samuel
Jan 19 at 20:41
add a comment |
$begingroup$
This seems correct to me. Just having difficulty seeing why $e'$ is an identity.
$endgroup$
– IntegrateThis
Jan 19 at 20:34
$begingroup$
@Integrate It's a proof by contradiction. It shows that a two sided identity must exist. Since this would also be a right identity, it is in fact equal to $e$. It's not required to show that though.
$endgroup$
– Matt Samuel
Jan 19 at 20:38
$begingroup$
Nevermind. All good thanks!
$endgroup$
– IntegrateThis
Jan 19 at 20:40
$begingroup$
@Integrate That wouldn't be true necessarily. $e'$ is a fabricated element that doesn't exist. You could phrase it as a direct proof by saying that if the ring has a right identity that is not a left identity, then the right identity is not unique. There's no mention of a left identity there.
$endgroup$
– Matt Samuel
Jan 19 at 20:41
$begingroup$
@Integrate Sure, no problem.
$endgroup$
– Matt Samuel
Jan 19 at 20:41
$begingroup$
This seems correct to me. Just having difficulty seeing why $e'$ is an identity.
$endgroup$
– IntegrateThis
Jan 19 at 20:34
$begingroup$
This seems correct to me. Just having difficulty seeing why $e'$ is an identity.
$endgroup$
– IntegrateThis
Jan 19 at 20:34
$begingroup$
@Integrate It's a proof by contradiction. It shows that a two sided identity must exist. Since this would also be a right identity, it is in fact equal to $e$. It's not required to show that though.
$endgroup$
– Matt Samuel
Jan 19 at 20:38
$begingroup$
@Integrate It's a proof by contradiction. It shows that a two sided identity must exist. Since this would also be a right identity, it is in fact equal to $e$. It's not required to show that though.
$endgroup$
– Matt Samuel
Jan 19 at 20:38
$begingroup$
Nevermind. All good thanks!
$endgroup$
– IntegrateThis
Jan 19 at 20:40
$begingroup$
Nevermind. All good thanks!
$endgroup$
– IntegrateThis
Jan 19 at 20:40
$begingroup$
@Integrate That wouldn't be true necessarily. $e'$ is a fabricated element that doesn't exist. You could phrase it as a direct proof by saying that if the ring has a right identity that is not a left identity, then the right identity is not unique. There's no mention of a left identity there.
$endgroup$
– Matt Samuel
Jan 19 at 20:41
$begingroup$
@Integrate That wouldn't be true necessarily. $e'$ is a fabricated element that doesn't exist. You could phrase it as a direct proof by saying that if the ring has a right identity that is not a left identity, then the right identity is not unique. There's no mention of a left identity there.
$endgroup$
– Matt Samuel
Jan 19 at 20:41
$begingroup$
@Integrate Sure, no problem.
$endgroup$
– Matt Samuel
Jan 19 at 20:41
$begingroup$
@Integrate Sure, no problem.
$endgroup$
– Matt Samuel
Jan 19 at 20:41
add a comment |
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$begingroup$
Do you mean $R$ has an identity or $R$ has a unit?
$endgroup$
– John Douma
Jan 19 at 20:27
$begingroup$
@JohnDouma unity=identity I believe.
$endgroup$
– IntegrateThis
Jan 19 at 20:30
$begingroup$
I have never seen that term.
$endgroup$
– John Douma
Jan 19 at 20:30
$begingroup$
@JohnDouma ok. well I got it from Nicholson's Abstract Algebra.
$endgroup$
– IntegrateThis
Jan 19 at 20:31
$begingroup$
I didn't say it doesn't exist. I just have never heard of it.
$endgroup$
– John Douma
Jan 19 at 20:31