If $a$ is an element of some ring and ${a^n; n=0,1,dots}$ is finite, then $a$ is invertible or a zero divisor...
This question already has an answer here:
Every nonzero element in a finite ring is either a unit or a zero divisor
6 answers
I'm supposed to prove that, in a ring, if $a$ is an element such that the set ${a^n; n=0,1,...}$ is finite, then $a$ is either invertible or a zero divisor. I don't understand how $a$ can not be invertible, since if the set is finite then there has to be an $i$ such that $a^i=1$.
abstract-algebra ring-theory
edited Nov 21 '18 at 14:13
Asaf Karagila♦
302k32426757
asked Nov 21 '18 at 9:08
InsertNameHereInsertNameHere
163
marked as duplicate by Bill Dubuque abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 21 '18 at 14:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Every nonzero element in a finite ring is either a unit or a zero divisor
6 answers
I'm supposed to prove that, in a ring, if $a$ is an element such that the set ${a^n; n=0,1,...}$ is finite, then $a$ is either invertible or a zero divisor. I don't understand how $a$ can not be invertible, since if the set is finite then there has to be an $i$ such that $a^i=1$.
abstract-algebra ring-theory
edited Nov 21 '18 at 14:13
Asaf Karagila♦
302k32426757
asked Nov 21 '18 at 9:08
InsertNameHereInsertNameHere
163
marked as duplicate by Bill Dubuque abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 21 '18 at 14:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
What happens for $a=0$?
– asdq
Nov 21 '18 at 9:17
Though not an exact dupe, many answers in the dupe apply here too.
– Bill Dubuque
Nov 21 '18 at 15:15
e.g. In $,Bbb Z/8,$ the positive powers of $2$ being $,2,4,0,0,ldots$ are not $equiv 1pmod{!8},$ being even $= 2^n !+ 8k$ $ $
– Bill Dubuque
Nov 21 '18 at 15:24
add a comment |
This question already has an answer here:
Every nonzero element in a finite ring is either a unit or a zero divisor
6 answers
I'm supposed to prove that, in a ring, if $a$ is an element such that the set ${a^n; n=0,1,...}$ is finite, then $a$ is either invertible or a zero divisor. I don't understand how $a$ can not be invertible, since if the set is finite then there has to be an $i$ such that $a^i=1$.
abstract-algebra ring-theory
edited Nov 21 '18 at 14:13
Asaf Karagila♦
302k32426757
asked Nov 21 '18 at 9:08
InsertNameHereInsertNameHere
163
This question already has an answer here:
Every nonzero element in a finite ring is either a unit or a zero divisor
6 answers
I'm supposed to prove that, in a ring, if $a$ is an element such that the set ${a^n; n=0,1,...}$ is finite, then $a$ is either invertible or a zero divisor. I don't understand how $a$ can not be invertible, since if the set is finite then there has to be an $i$ such that $a^i=1$.
This question already has an answer here:
Every nonzero element in a finite ring is either a unit or a zero divisor
6 answers
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Nov 21 '18 at 14:13
Asaf Karagila♦
302k32426757
asked Nov 21 '18 at 9:08
InsertNameHereInsertNameHere
163
edited Nov 21 '18 at 14:13
Asaf Karagila♦
302k32426757
asked Nov 21 '18 at 9:08
InsertNameHereInsertNameHere
163
edited Nov 21 '18 at 14:13
Asaf Karagila♦
302k32426757
edited Nov 21 '18 at 14:13
Asaf Karagila♦
302k32426757
edited Nov 21 '18 at 14:13
Asaf Karagila♦
302k32426757
302k32426757
asked Nov 21 '18 at 9:08
InsertNameHereInsertNameHere
163
asked Nov 21 '18 at 9:08
InsertNameHereInsertNameHere
163
asked Nov 21 '18 at 9:08
InsertNameHereInsertNameHere
163
163
marked as duplicate by Bill Dubuque abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 21 '18 at 14:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Bill Dubuque abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 21 '18 at 14:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
What happens for $a=0$?
– asdq
Nov 21 '18 at 9:17
Though not an exact dupe, many answers in the dupe apply here too.
– Bill Dubuque
Nov 21 '18 at 15:15
e.g. In $,Bbb Z/8,$ the positive powers of $2$ being $,2,4,0,0,ldots$ are not $equiv 1pmod{!8},$ being even $= 2^n !+ 8k$ $ $
– Bill Dubuque
Nov 21 '18 at 15:24
add a comment |
1
What happens for $a=0$?
– asdq
Nov 21 '18 at 9:17
Though not an exact dupe, many answers in the dupe apply here too.
– Bill Dubuque
Nov 21 '18 at 15:15
e.g. In $,Bbb Z/8,$ the positive powers of $2$ being $,2,4,0,0,ldots$ are not $equiv 1pmod{!8},$ being even $= 2^n !+ 8k$ $ $
– Bill Dubuque
Nov 21 '18 at 15:24
1
1
What happens for $a=0$?
– asdq
Nov 21 '18 at 9:17
What happens for $a=0$?
– asdq
Nov 21 '18 at 9:17
Though not an exact dupe, many answers in the dupe apply here too.
– Bill Dubuque
Nov 21 '18 at 15:15
Though not an exact dupe, many answers in the dupe apply here too.
– Bill Dubuque
Nov 21 '18 at 15:15
e.g. In $,Bbb Z/8,$ the positive powers of $2$ being $,2,4,0,0,ldots$ are not $equiv 1pmod{!8},$ being even $= 2^n !+ 8k$ $ $
– Bill Dubuque
Nov 21 '18 at 15:24
e.g. In $,Bbb Z/8,$ the positive powers of $2$ being $,2,4,0,0,ldots$ are not $equiv 1pmod{!8},$ being even $= 2^n !+ 8k$ $ $
– Bill Dubuque
Nov 21 '18 at 15:24
add a comment |
2 Answers
2
active
oldest
votes
Since the set is finite there exist $m,n$ such that $a^m=a^n$. Without loss of generality assume that $m>n$. If $a$ is not a zero divisor then $0=a^m-a^n=a^n(a^{m-n}-1)$ implies that $a^{m-n}=1$ hence $a$ is invertible.
If you consider the ring $k[x]/(x^2)$ for example, the powers of $x$ are finite and $x$ is a zero divisor.
answered Nov 21 '18 at 9:17
LeventLevent
2,700925
add a comment |
It is not true that if that set is finite, then $a^i=1$ for some $iinmathbb N$. Take, for instance, the ring of all $2times2$ rational matrices. Then each matrix $M$ of the form$$M=begin{bmatrix}1-d&frac{d-d^2}c\c&dend{bmatrix},$$($cinmathbb{Q}setminus{0}$, $dinmathbb Q$) is such that $M^2=M$. So, the set ${M^n,|,ninmathbb{Z}^+}$ is surely finite. However, there is no natural $i$ such that $M^i=operatorname{Id}$.
Of course, since $M^2=M$, you have $Mtimes(M-operatorname{Id})=0$ and therefore $M$ is a zero divisor.
answered Nov 21 '18 at 9:17
José Carlos SantosJosé Carlos Santos
152k22123226
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Since the set is finite there exist $m,n$ such that $a^m=a^n$. Without loss of generality assume that $m>n$. If $a$ is not a zero divisor then $0=a^m-a^n=a^n(a^{m-n}-1)$ implies that $a^{m-n}=1$ hence $a$ is invertible.
If you consider the ring $k[x]/(x^2)$ for example, the powers of $x$ are finite and $x$ is a zero divisor.
answered Nov 21 '18 at 9:17
LeventLevent
2,700925
add a comment |
Since the set is finite there exist $m,n$ such that $a^m=a^n$. Without loss of generality assume that $m>n$. If $a$ is not a zero divisor then $0=a^m-a^n=a^n(a^{m-n}-1)$ implies that $a^{m-n}=1$ hence $a$ is invertible.
If you consider the ring $k[x]/(x^2)$ for example, the powers of $x$ are finite and $x$ is a zero divisor.
answered Nov 21 '18 at 9:17
LeventLevent
2,700925
add a comment |
Since the set is finite there exist $m,n$ such that $a^m=a^n$. Without loss of generality assume that $m>n$. If $a$ is not a zero divisor then $0=a^m-a^n=a^n(a^{m-n}-1)$ implies that $a^{m-n}=1$ hence $a$ is invertible.
If you consider the ring $k[x]/(x^2)$ for example, the powers of $x$ are finite and $x$ is a zero divisor.
answered Nov 21 '18 at 9:17
LeventLevent
2,700925
Since the set is finite there exist $m,n$ such that $a^m=a^n$. Without loss of generality assume that $m>n$. If $a$ is not a zero divisor then $0=a^m-a^n=a^n(a^{m-n}-1)$ implies that $a^{m-n}=1$ hence $a$ is invertible.
If you consider the ring $k[x]/(x^2)$ for example, the powers of $x$ are finite and $x$ is a zero divisor.
answered Nov 21 '18 at 9:17
LeventLevent
2,700925
answered Nov 21 '18 at 9:17
LeventLevent
2,700925
answered Nov 21 '18 at 9:17
LeventLevent
2,700925
answered Nov 21 '18 at 9:17
LeventLevent
2,700925
2,700925
add a comment |
add a comment |
It is not true that if that set is finite, then $a^i=1$ for some $iinmathbb N$. Take, for instance, the ring of all $2times2$ rational matrices. Then each matrix $M$ of the form$$M=begin{bmatrix}1-d&frac{d-d^2}c\c&dend{bmatrix},$$($cinmathbb{Q}setminus{0}$, $dinmathbb Q$) is such that $M^2=M$. So, the set ${M^n,|,ninmathbb{Z}^+}$ is surely finite. However, there is no natural $i$ such that $M^i=operatorname{Id}$.
Of course, since $M^2=M$, you have $Mtimes(M-operatorname{Id})=0$ and therefore $M$ is a zero divisor.
answered Nov 21 '18 at 9:17
José Carlos SantosJosé Carlos Santos
152k22123226
add a comment |
It is not true that if that set is finite, then $a^i=1$ for some $iinmathbb N$. Take, for instance, the ring of all $2times2$ rational matrices. Then each matrix $M$ of the form$$M=begin{bmatrix}1-d&frac{d-d^2}c\c&dend{bmatrix},$$($cinmathbb{Q}setminus{0}$, $dinmathbb Q$) is such that $M^2=M$. So, the set ${M^n,|,ninmathbb{Z}^+}$ is surely finite. However, there is no natural $i$ such that $M^i=operatorname{Id}$.
Of course, since $M^2=M$, you have $Mtimes(M-operatorname{Id})=0$ and therefore $M$ is a zero divisor.
answered Nov 21 '18 at 9:17
José Carlos SantosJosé Carlos Santos
152k22123226
add a comment |
It is not true that if that set is finite, then $a^i=1$ for some $iinmathbb N$. Take, for instance, the ring of all $2times2$ rational matrices. Then each matrix $M$ of the form$$M=begin{bmatrix}1-d&frac{d-d^2}c\c&dend{bmatrix},$$($cinmathbb{Q}setminus{0}$, $dinmathbb Q$) is such that $M^2=M$. So, the set ${M^n,|,ninmathbb{Z}^+}$ is surely finite. However, there is no natural $i$ such that $M^i=operatorname{Id}$.
Of course, since $M^2=M$, you have $Mtimes(M-operatorname{Id})=0$ and therefore $M$ is a zero divisor.
answered Nov 21 '18 at 9:17
José Carlos SantosJosé Carlos Santos
152k22123226
It is not true that if that set is finite, then $a^i=1$ for some $iinmathbb N$. Take, for instance, the ring of all $2times2$ rational matrices. Then each matrix $M$ of the form$$M=begin{bmatrix}1-d&frac{d-d^2}c\c&dend{bmatrix},$$($cinmathbb{Q}setminus{0}$, $dinmathbb Q$) is such that $M^2=M$. So, the set ${M^n,|,ninmathbb{Z}^+}$ is surely finite. However, there is no natural $i$ such that $M^i=operatorname{Id}$.
Of course, since $M^2=M$, you have $Mtimes(M-operatorname{Id})=0$ and therefore $M$ is a zero divisor.
answered Nov 21 '18 at 9:17
José Carlos SantosJosé Carlos Santos
152k22123226
answered Nov 21 '18 at 9:17
José Carlos SantosJosé Carlos Santos
152k22123226
answered Nov 21 '18 at 9:17
José Carlos SantosJosé Carlos Santos
152k22123226
answered Nov 21 '18 at 9:17
José Carlos SantosJosé Carlos Santos
152k22123226
152k22123226
add a comment |
add a comment |
1
What happens for $a=0$?
– asdq
Nov 21 '18 at 9:17
Though not an exact dupe, many answers in the dupe apply here too.
– Bill Dubuque
Nov 21 '18 at 15:15
e.g. In $,Bbb Z/8,$ the positive powers of $2$ being $,2,4,0,0,ldots$ are not $equiv 1pmod{!8},$ being even $= 2^n !+ 8k$ $ $
– Bill Dubuque
Nov 21 '18 at 15:24