In a ring $(A,+, cdot)$ if $aba = a$ then $bab = b$ and all non zero elements in $A$ are invertible. [closed]

7

$begingroup$

Let $left(A,+, cdotright)$ be a ring with $1$ that satisfies the following condition:

For any nonzero $ain A$, there exists a unique $bin A$ such that $aba = a$.

Show that $b$ also satisfies $bab = b$, and that all nonzero elements of $A$ are invertible.

abstract-algebra ring-theory

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edited Feb 12 '16 at 12:45

user26857

39.4k124183

asked May 5 '15 at 1:35

Roiner Segura CuberoRoiner Segura Cubero

1,9051629

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closed as off-topic by user26857, Daniel W. Farlow, Kamil Jarosz, Pragabhava, John B Feb 12 '16 at 15:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, Daniel W. Farlow, Kamil Jarosz, Pragabhava, John B

If this question can be reworded to fit the rules in the help center, please edit the question.

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7

$begingroup$

Let $left(A,+, cdotright)$ be a ring with $1$ that satisfies the following condition:

For any nonzero $ain A$, there exists a unique $bin A$ such that $aba = a$.

Show that $b$ also satisfies $bab = b$, and that all nonzero elements of $A$ are invertible.

abstract-algebra ring-theory

share|cite|improve this question

edited Feb 12 '16 at 12:45

user26857

39.4k124183

asked May 5 '15 at 1:35

Roiner Segura CuberoRoiner Segura Cubero

1,9051629

$endgroup$

closed as off-topic by user26857, Daniel W. Farlow, Kamil Jarosz, Pragabhava, John B Feb 12 '16 at 15:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, Daniel W. Farlow, Kamil Jarosz, Pragabhava, John B

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

7

7

7

2

$begingroup$

Let $left(A,+, cdotright)$ be a ring with $1$ that satisfies the following condition:

For any nonzero $ain A$, there exists a unique $bin A$ such that $aba = a$.

Show that $b$ also satisfies $bab = b$, and that all nonzero elements of $A$ are invertible.

abstract-algebra ring-theory

share|cite|improve this question

edited Feb 12 '16 at 12:45

user26857

39.4k124183

asked May 5 '15 at 1:35

Roiner Segura CuberoRoiner Segura Cubero

1,9051629

$endgroup$

Let $left(A,+, cdotright)$ be a ring with $1$ that satisfies the following condition:

For any nonzero $ain A$, there exists a unique $bin A$ such that $aba = a$.

Show that $b$ also satisfies $bab = b$, and that all nonzero elements of $A$ are invertible.

abstract-algebra ring-theory

abstract-algebra ring-theory

share|cite|improve this question

edited Feb 12 '16 at 12:45

user26857

39.4k124183

asked May 5 '15 at 1:35

Roiner Segura CuberoRoiner Segura Cubero

1,9051629

share|cite|improve this question

edited Feb 12 '16 at 12:45

user26857

39.4k124183

asked May 5 '15 at 1:35

Roiner Segura CuberoRoiner Segura Cubero

1,9051629

share|cite|improve this question

share|cite|improve this question

edited Feb 12 '16 at 12:45

user26857

39.4k124183

edited Feb 12 '16 at 12:45

user26857

39.4k124183

edited Feb 12 '16 at 12:45

user26857

39.4k124183

39.4k124183

asked May 5 '15 at 1:35

Roiner Segura CuberoRoiner Segura Cubero

1,9051629

asked May 5 '15 at 1:35

Roiner Segura CuberoRoiner Segura Cubero

1,9051629

asked May 5 '15 at 1:35

Roiner Segura CuberoRoiner Segura Cubero

1,9051629

1,9051629

closed as off-topic by user26857, Daniel W. Farlow, Kamil Jarosz, Pragabhava, John B Feb 12 '16 at 15:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, Daniel W. Farlow, Kamil Jarosz, Pragabhava, John B

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by user26857, Daniel W. Farlow, Kamil Jarosz, Pragabhava, John B Feb 12 '16 at 15:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, Daniel W. Farlow, Kamil Jarosz, Pragabhava, John B

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

10

$begingroup$

The ring has no nonzero zero divisors: if $ac=0$ for nonzero $a,c$, then you can find a unique $b$ such that $a=aba=a(b+c)a$. But then $b=b+c$ implies $c=0$, a contradiction.

At this point, you can even prove that $A$, as long as it is not equal to zero, has an identity without assuming so in the hypotheses. Given any $a=acaneq 0$, we see that $ac$ is an idempotent, and verify that $ac(acx-x)=0=(xac-x)ac$ implies $ac$ is an identity element for the ring. We write "$1$" for $ac$ hereafter.

Knowing this, you can conclude that $a(ba-1)=0=(ab-1)a$ implies $ba=ab=1$ when a is nonzero. Clearly then $b$ can only be $a^{-1}$ and it follows that $bab=b$ also holds.

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edited May 30 '18 at 10:44

answered May 5 '15 at 2:44

rschwiebrschwieb

106k12102249

$endgroup$

add a comment | 

9

$begingroup$

$a(bab)a=(aba)ba=aba=a$. So $bab$ does the thing that $b$ does, but $b$ is the unique element that does that, so $bab=b$.

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answered May 5 '15 at 1:41

Jorge FernándezJorge Fernández

75.3k1191192

$endgroup$

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

10

$begingroup$

The ring has no nonzero zero divisors: if $ac=0$ for nonzero $a,c$, then you can find a unique $b$ such that $a=aba=a(b+c)a$. But then $b=b+c$ implies $c=0$, a contradiction.

At this point, you can even prove that $A$, as long as it is not equal to zero, has an identity without assuming so in the hypotheses. Given any $a=acaneq 0$, we see that $ac$ is an idempotent, and verify that $ac(acx-x)=0=(xac-x)ac$ implies $ac$ is an identity element for the ring. We write "$1$" for $ac$ hereafter.

Knowing this, you can conclude that $a(ba-1)=0=(ab-1)a$ implies $ba=ab=1$ when a is nonzero. Clearly then $b$ can only be $a^{-1}$ and it follows that $bab=b$ also holds.

share|cite|improve this answer

edited May 30 '18 at 10:44

answered May 5 '15 at 2:44

rschwiebrschwieb

106k12102249

$endgroup$

add a comment | 

10

$begingroup$

The ring has no nonzero zero divisors: if $ac=0$ for nonzero $a,c$, then you can find a unique $b$ such that $a=aba=a(b+c)a$. But then $b=b+c$ implies $c=0$, a contradiction.

At this point, you can even prove that $A$, as long as it is not equal to zero, has an identity without assuming so in the hypotheses. Given any $a=acaneq 0$, we see that $ac$ is an idempotent, and verify that $ac(acx-x)=0=(xac-x)ac$ implies $ac$ is an identity element for the ring. We write "$1$" for $ac$ hereafter.

Knowing this, you can conclude that $a(ba-1)=0=(ab-1)a$ implies $ba=ab=1$ when a is nonzero. Clearly then $b$ can only be $a^{-1}$ and it follows that $bab=b$ also holds.

share|cite|improve this answer

edited May 30 '18 at 10:44

answered May 5 '15 at 2:44

rschwiebrschwieb

106k12102249

$endgroup$

add a comment | 

10

10

10

$begingroup$

The ring has no nonzero zero divisors: if $ac=0$ for nonzero $a,c$, then you can find a unique $b$ such that $a=aba=a(b+c)a$. But then $b=b+c$ implies $c=0$, a contradiction.

At this point, you can even prove that $A$, as long as it is not equal to zero, has an identity without assuming so in the hypotheses. Given any $a=acaneq 0$, we see that $ac$ is an idempotent, and verify that $ac(acx-x)=0=(xac-x)ac$ implies $ac$ is an identity element for the ring. We write "$1$" for $ac$ hereafter.

Knowing this, you can conclude that $a(ba-1)=0=(ab-1)a$ implies $ba=ab=1$ when a is nonzero. Clearly then $b$ can only be $a^{-1}$ and it follows that $bab=b$ also holds.

share|cite|improve this answer

edited May 30 '18 at 10:44

answered May 5 '15 at 2:44

rschwiebrschwieb

106k12102249

$endgroup$

The ring has no nonzero zero divisors: if $ac=0$ for nonzero $a,c$, then you can find a unique $b$ such that $a=aba=a(b+c)a$. But then $b=b+c$ implies $c=0$, a contradiction.

At this point, you can even prove that $A$, as long as it is not equal to zero, has an identity without assuming so in the hypotheses. Given any $a=acaneq 0$, we see that $ac$ is an idempotent, and verify that $ac(acx-x)=0=(xac-x)ac$ implies $ac$ is an identity element for the ring. We write "$1$" for $ac$ hereafter.

Knowing this, you can conclude that $a(ba-1)=0=(ab-1)a$ implies $ba=ab=1$ when a is nonzero. Clearly then $b$ can only be $a^{-1}$ and it follows that $bab=b$ also holds.

share|cite|improve this answer

edited May 30 '18 at 10:44

answered May 5 '15 at 2:44

rschwiebrschwieb

106k12102249

share|cite|improve this answer

share|cite|improve this answer

edited May 30 '18 at 10:44

edited May 30 '18 at 10:44

edited May 30 '18 at 10:44

answered May 5 '15 at 2:44

rschwiebrschwieb

106k12102249

answered May 5 '15 at 2:44

rschwiebrschwieb

106k12102249

answered May 5 '15 at 2:44

rschwiebrschwieb

106k12102249

106k12102249

add a comment | 

add a comment | 

9

$begingroup$

$a(bab)a=(aba)ba=aba=a$. So $bab$ does the thing that $b$ does, but $b$ is the unique element that does that, so $bab=b$.

share|cite|improve this answer

answered May 5 '15 at 1:41

Jorge FernándezJorge Fernández

75.3k1191192

$endgroup$

add a comment | 

9

$begingroup$

$a(bab)a=(aba)ba=aba=a$. So $bab$ does the thing that $b$ does, but $b$ is the unique element that does that, so $bab=b$.

share|cite|improve this answer

answered May 5 '15 at 1:41

Jorge FernándezJorge Fernández

75.3k1191192

$endgroup$

add a comment | 

9

9

9

$begingroup$

$a(bab)a=(aba)ba=aba=a$. So $bab$ does the thing that $b$ does, but $b$ is the unique element that does that, so $bab=b$.

share|cite|improve this answer

answered May 5 '15 at 1:41

Jorge FernándezJorge Fernández

75.3k1191192

$endgroup$

$a(bab)a=(aba)ba=aba=a$. So $bab$ does the thing that $b$ does, but $b$ is the unique element that does that, so $bab=b$.

share|cite|improve this answer

answered May 5 '15 at 1:41

Jorge FernándezJorge Fernández

75.3k1191192

share|cite|improve this answer

share|cite|improve this answer

answered May 5 '15 at 1:41

Jorge FernándezJorge Fernández

75.3k1191192

answered May 5 '15 at 1:41

Jorge FernándezJorge Fernández

75.3k1191192

answered May 5 '15 at 1:41

Jorge FernándezJorge Fernández

75.3k1191192

75.3k1191192

add a comment | 

add a comment | 

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