In a ring $(A,+, cdot)$ if $aba = a$ then $bab = b$ and all non zero elements in $A$ are invertible. [closed]
$begingroup$
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
edited Feb 12 '16 at 12:45
user26857
39.5k124284
asked May 5 '15 at 1:35
Roiner Segura CuberoRoiner Segura Cubero
1,9251731
$endgroup$
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 |
$begingroup$
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
edited Feb 12 '16 at 12:45
user26857
39.5k124284
asked May 5 '15 at 1:35
Roiner Segura CuberoRoiner Segura Cubero
1,9251731
$endgroup$
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 |
$begingroup$
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
edited Feb 12 '16 at 12:45
user26857
39.5k124284
asked May 5 '15 at 1:35
Roiner Segura CuberoRoiner Segura Cubero
1,9251731
$endgroup$
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
edited Feb 12 '16 at 12:45
user26857
39.5k124284
asked May 5 '15 at 1:35
Roiner Segura CuberoRoiner Segura Cubero
1,9251731
edited Feb 12 '16 at 12:45
user26857
39.5k124284
asked May 5 '15 at 1:35
Roiner Segura CuberoRoiner Segura Cubero
1,9251731
edited Feb 12 '16 at 12:45
user26857
39.5k124284
edited Feb 12 '16 at 12:45
user26857
39.5k124284
edited Feb 12 '16 at 12:45
user26857
39.5k124284
39.5k124284
asked May 5 '15 at 1:35
Roiner Segura CuberoRoiner Segura Cubero
1,9251731
asked May 5 '15 at 1:35
Roiner Segura CuberoRoiner Segura Cubero
1,9251731
asked May 5 '15 at 1:35
Roiner Segura CuberoRoiner Segura Cubero
1,9251731
1,9251731
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
$begingroup$
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.
answered May 5 '15 at 2:44
rschwiebrschwieb
108k12104253
$endgroup$
add a comment |
$begingroup$
$a(bab)a=(aba)ba=aba=a$. So $bab$ does the thing that $b$ does, but $b$ is the unique element that does that, so $bab=b$.
answered May 5 '15 at 1:41
Jorge Fernández HidalgoJorge Fernández Hidalgo
77.1k1394195
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered May 5 '15 at 2:44
rschwiebrschwieb
108k12104253
$endgroup$
add a comment |
$begingroup$
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.
answered May 5 '15 at 2:44
rschwiebrschwieb
108k12104253
$endgroup$
add a comment |
$begingroup$
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.
answered May 5 '15 at 2:44
rschwiebrschwieb
108k12104253
$endgroup$
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.
answered May 5 '15 at 2:44
rschwiebrschwieb
108k12104253
edited May 30 '18 at 10:44
answered May 5 '15 at 2:44
rschwiebrschwieb
108k12104253
answered May 5 '15 at 2:44
rschwiebrschwieb
108k12104253
answered May 5 '15 at 2:44
rschwiebrschwieb
108k12104253
108k12104253
add a comment |
add a comment |
$begingroup$
$a(bab)a=(aba)ba=aba=a$. So $bab$ does the thing that $b$ does, but $b$ is the unique element that does that, so $bab=b$.
answered May 5 '15 at 1:41
Jorge Fernández HidalgoJorge Fernández Hidalgo
77.1k1394195
$endgroup$
add a comment |
$begingroup$
$a(bab)a=(aba)ba=aba=a$. So $bab$ does the thing that $b$ does, but $b$ is the unique element that does that, so $bab=b$.
answered May 5 '15 at 1:41
Jorge Fernández HidalgoJorge Fernández Hidalgo
77.1k1394195
$endgroup$
add a comment |
$begingroup$
$a(bab)a=(aba)ba=aba=a$. So $bab$ does the thing that $b$ does, but $b$ is the unique element that does that, so $bab=b$.
answered May 5 '15 at 1:41
Jorge Fernández HidalgoJorge Fernández Hidalgo
77.1k1394195
$endgroup$
$a(bab)a=(aba)ba=aba=a$. So $bab$ does the thing that $b$ does, but $b$ is the unique element that does that, so $bab=b$.
answered May 5 '15 at 1:41
Jorge Fernández HidalgoJorge Fernández Hidalgo
77.1k1394195
answered May 5 '15 at 1:41
Jorge Fernández HidalgoJorge Fernández Hidalgo
77.1k1394195
answered May 5 '15 at 1:41
Jorge Fernández HidalgoJorge Fernández Hidalgo
77.1k1394195
answered May 5 '15 at 1:41
Jorge Fernández HidalgoJorge Fernández Hidalgo
77.1k1394195
77.1k1394195
add a comment |
add a comment |
