If $R$ is a noncommutative ring such that $exists a,b in R$ s.t. $ab=1$ and $ba neq 1$ then $R$ is infinite...
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This question already has an answer here:
If a ring element is right-invertible, but not left-invertible, then it has infinitely many right-inverses. [duplicate]
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If $R$ is a noncommutative ring and there exist $a,bin R$ such that $ab=1$ and $baneq 1$, theN $R$ is infinite.
I need some help with this one, been playing around with elementary algebra but am getting nowhere. Thanks in advance
abstract-algebra ring-theory
edited Dec 8 '18 at 0:39
Arturo Magidin
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Dec 8 '18 at 14:41
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This question already has an answer here:
If a ring element is right-invertible, but not left-invertible, then it has infinitely many right-inverses. [duplicate]
1 answer
If $R$ is a noncommutative ring and there exist $a,bin R$ such that $ab=1$ and $baneq 1$, theN $R$ is infinite.
I need some help with this one, been playing around with elementary algebra but am getting nowhere. Thanks in advance
abstract-algebra ring-theory
edited Dec 8 '18 at 0:39
Arturo Magidin
261k34586906
asked Dec 8 '18 at 0:19
Jesus SavesJesus Saves
112
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marked as duplicate by amWhy, Saad, Brahadeesh, rschwieb abstract-algebra
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Dec 8 '18 at 14:41
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This question already has an answer here:
If a ring element is right-invertible, but not left-invertible, then it has infinitely many right-inverses. [duplicate]
1 answer
If $R$ is a noncommutative ring and there exist $a,bin R$ such that $ab=1$ and $baneq 1$, theN $R$ is infinite.
I need some help with this one, been playing around with elementary algebra but am getting nowhere. Thanks in advance
abstract-algebra ring-theory
edited Dec 8 '18 at 0:39
Arturo Magidin
261k34586906
asked Dec 8 '18 at 0:19
Jesus SavesJesus Saves
112
$endgroup$
This question already has an answer here:
If a ring element is right-invertible, but not left-invertible, then it has infinitely many right-inverses. [duplicate]
1 answer
If $R$ is a noncommutative ring and there exist $a,bin R$ such that $ab=1$ and $baneq 1$, theN $R$ is infinite.
I need some help with this one, been playing around with elementary algebra but am getting nowhere. Thanks in advance
This question already has an answer here:
If a ring element is right-invertible, but not left-invertible, then it has infinitely many right-inverses. [duplicate]
1 answer
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Dec 8 '18 at 0:39
Arturo Magidin
261k34586906
asked Dec 8 '18 at 0:19
Jesus SavesJesus Saves
112
edited Dec 8 '18 at 0:39
Arturo Magidin
261k34586906
asked Dec 8 '18 at 0:19
Jesus SavesJesus Saves
112
edited Dec 8 '18 at 0:39
Arturo Magidin
261k34586906
edited Dec 8 '18 at 0:39
Arturo Magidin
261k34586906
edited Dec 8 '18 at 0:39
Arturo Magidin
261k34586906
261k34586906
asked Dec 8 '18 at 0:19
Jesus SavesJesus Saves
112
asked Dec 8 '18 at 0:19
Jesus SavesJesus Saves
112
asked Dec 8 '18 at 0:19
Jesus SavesJesus Saves
112
112
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Dec 8 '18 at 14:41
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Welcome to MathematicsStackExchange! Thanks for using MathJax. You will enhance your experience when you are able to articulate what avenues of approach have been tried.
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Dec 8 '18 at 0:27
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and please search before asking. Quite often you may find an answer already there.
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Welcome to MathematicsStackExchange! Thanks for using MathJax. You will enhance your experience when you are able to articulate what avenues of approach have been tried.
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Dec 8 '18 at 0:27
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Welcome to MathematicsStackExchange! Thanks for using MathJax. You will enhance your experience when you are able to articulate what avenues of approach have been tried.
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– dantopa
Dec 8 '18 at 0:27
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Welcome to MathematicsStackExchange! Thanks for using MathJax. You will enhance your experience when you are able to articulate what avenues of approach have been tried.
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– dantopa
Dec 8 '18 at 0:27
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and please search before asking. Quite often you may find an answer already there.
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2 Answers
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Consider $f_a(c)=ca$, it is injective $f_a(c)=f_a(d)$ implies that $ca=da$, we deduce that $(ca)b=(da)b=c(ab)=d(ab)=c=d$. If $R$ is finite, $f$ is bijective, there exists $x$ such that $f(x)=xa=1$, you have $xab=x(ab)=x=(xa)b=b$. Contradiction.
answered Dec 8 '18 at 0:36
Tsemo AristideTsemo Aristide
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Suppose $R$ is finite and $ab = 1$. Then in particular $a^n = 1$ for some $n$ or $a^m = 0$ for some $m$ minimal. In the second case, $0 = a^m b = a^{m-1}$, a contradiction by minimality of $m$ so we must be in the first case. In the first case, $b = a^n b = a^{n-1}$ and hence $ba = a^n = 1$.
By the contrapositive, if $ab = 1$ and $ba neq 1$ then $R$ must be infinite.
answered Dec 8 '18 at 0:32
ODFODF
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2 Answers
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2 Answers
2
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oldest
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active
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active
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$begingroup$
Consider $f_a(c)=ca$, it is injective $f_a(c)=f_a(d)$ implies that $ca=da$, we deduce that $(ca)b=(da)b=c(ab)=d(ab)=c=d$. If $R$ is finite, $f$ is bijective, there exists $x$ such that $f(x)=xa=1$, you have $xab=x(ab)=x=(xa)b=b$. Contradiction.
answered Dec 8 '18 at 0:36
Tsemo AristideTsemo Aristide
56.9k11444
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add a comment |
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Consider $f_a(c)=ca$, it is injective $f_a(c)=f_a(d)$ implies that $ca=da$, we deduce that $(ca)b=(da)b=c(ab)=d(ab)=c=d$. If $R$ is finite, $f$ is bijective, there exists $x$ such that $f(x)=xa=1$, you have $xab=x(ab)=x=(xa)b=b$. Contradiction.
answered Dec 8 '18 at 0:36
Tsemo AristideTsemo Aristide
56.9k11444
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add a comment |
$begingroup$
Consider $f_a(c)=ca$, it is injective $f_a(c)=f_a(d)$ implies that $ca=da$, we deduce that $(ca)b=(da)b=c(ab)=d(ab)=c=d$. If $R$ is finite, $f$ is bijective, there exists $x$ such that $f(x)=xa=1$, you have $xab=x(ab)=x=(xa)b=b$. Contradiction.
answered Dec 8 '18 at 0:36
Tsemo AristideTsemo Aristide
56.9k11444
$endgroup$
Consider $f_a(c)=ca$, it is injective $f_a(c)=f_a(d)$ implies that $ca=da$, we deduce that $(ca)b=(da)b=c(ab)=d(ab)=c=d$. If $R$ is finite, $f$ is bijective, there exists $x$ such that $f(x)=xa=1$, you have $xab=x(ab)=x=(xa)b=b$. Contradiction.
answered Dec 8 '18 at 0:36
Tsemo AristideTsemo Aristide
56.9k11444
edited Dec 8 '18 at 0:41
answered Dec 8 '18 at 0:36
Tsemo AristideTsemo Aristide
56.9k11444
answered Dec 8 '18 at 0:36
Tsemo AristideTsemo Aristide
56.9k11444
answered Dec 8 '18 at 0:36
Tsemo AristideTsemo Aristide
56.9k11444
56.9k11444
add a comment |
add a comment |
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Suppose $R$ is finite and $ab = 1$. Then in particular $a^n = 1$ for some $n$ or $a^m = 0$ for some $m$ minimal. In the second case, $0 = a^m b = a^{m-1}$, a contradiction by minimality of $m$ so we must be in the first case. In the first case, $b = a^n b = a^{n-1}$ and hence $ba = a^n = 1$.
By the contrapositive, if $ab = 1$ and $ba neq 1$ then $R$ must be infinite.
answered Dec 8 '18 at 0:32
ODFODF
1,431510
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add a comment |
$begingroup$
Suppose $R$ is finite and $ab = 1$. Then in particular $a^n = 1$ for some $n$ or $a^m = 0$ for some $m$ minimal. In the second case, $0 = a^m b = a^{m-1}$, a contradiction by minimality of $m$ so we must be in the first case. In the first case, $b = a^n b = a^{n-1}$ and hence $ba = a^n = 1$.
By the contrapositive, if $ab = 1$ and $ba neq 1$ then $R$ must be infinite.
answered Dec 8 '18 at 0:32
ODFODF
1,431510
$endgroup$
add a comment |
$begingroup$
Suppose $R$ is finite and $ab = 1$. Then in particular $a^n = 1$ for some $n$ or $a^m = 0$ for some $m$ minimal. In the second case, $0 = a^m b = a^{m-1}$, a contradiction by minimality of $m$ so we must be in the first case. In the first case, $b = a^n b = a^{n-1}$ and hence $ba = a^n = 1$.
By the contrapositive, if $ab = 1$ and $ba neq 1$ then $R$ must be infinite.
answered Dec 8 '18 at 0:32
ODFODF
1,431510
$endgroup$
Suppose $R$ is finite and $ab = 1$. Then in particular $a^n = 1$ for some $n$ or $a^m = 0$ for some $m$ minimal. In the second case, $0 = a^m b = a^{m-1}$, a contradiction by minimality of $m$ so we must be in the first case. In the first case, $b = a^n b = a^{n-1}$ and hence $ba = a^n = 1$.
By the contrapositive, if $ab = 1$ and $ba neq 1$ then $R$ must be infinite.
answered Dec 8 '18 at 0:32
ODFODF
1,431510
answered Dec 8 '18 at 0:32
ODFODF
1,431510
answered Dec 8 '18 at 0:32
ODFODF
1,431510
answered Dec 8 '18 at 0:32
ODFODF
1,431510
1,431510
add a comment |
add a comment |
2
$begingroup$
Welcome to MathematicsStackExchange! Thanks for using MathJax. You will enhance your experience when you are able to articulate what avenues of approach have been tried.
$endgroup$
– dantopa
Dec 8 '18 at 0:27
$begingroup$
and please search before asking. Quite often you may find an answer already there.
$endgroup$
– rschwieb
Dec 8 '18 at 14:42