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...
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Every nonzero element in a finite ring is either a unit or a zero divisor
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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 yesterday
Asaf Karagila♦
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InsertNameHere
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marked as duplicate by Bill Dubuque abstract-algebra
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This question already has an answer here:
Every nonzero element in a finite ring is either a unit or a zero divisor
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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 yesterday
Asaf Karagila♦
299k32420750
asked yesterday
InsertNameHere
92
marked as duplicate by Bill Dubuque abstract-algebra
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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.
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What happens for $a=0$?
– asdq
yesterday
Though not an exact dupe, many answers in the dupe apply here too.
– Bill Dubuque
yesterday
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
yesterday
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up vote
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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 yesterday
Asaf Karagila♦
299k32420750
asked yesterday
InsertNameHere
92
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 yesterday
Asaf Karagila♦
299k32420750
asked yesterday
InsertNameHere
92
edited yesterday
Asaf Karagila♦
299k32420750
asked yesterday
InsertNameHere
92
edited yesterday
Asaf Karagila♦
299k32420750
edited yesterday
Asaf Karagila♦
299k32420750
edited yesterday
Asaf Karagila♦
299k32420750
299k32420750
asked yesterday
InsertNameHere
92
asked yesterday
InsertNameHere
92
asked yesterday
InsertNameHere
92
92
marked as duplicate by Bill Dubuque abstract-algebra
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yesterday
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
yesterday
Though not an exact dupe, many answers in the dupe apply here too.
– Bill Dubuque
yesterday
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
yesterday
add a comment |
1
What happens for $a=0$?
– asdq
yesterday
Though not an exact dupe, many answers in the dupe apply here too.
– Bill Dubuque
yesterday
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
yesterday
1
1
What happens for $a=0$?
– asdq
yesterday
What happens for $a=0$?
– asdq
yesterday
Though not an exact dupe, many answers in the dupe apply here too.
– Bill Dubuque
yesterday
Though not an exact dupe, many answers in the dupe apply here too.
– Bill Dubuque
yesterday
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
yesterday
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
yesterday
add a comment |
2 Answers
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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 yesterday
Levent
3,386825
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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 yesterday
José Carlos Santos
140k19111204
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
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 yesterday
Levent
3,386825
add a comment |
up vote
4
down vote
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 yesterday
Levent
3,386825
add a comment |
up vote
4
down vote
up vote
4
down vote
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 yesterday
Levent
3,386825
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 yesterday
Levent
3,386825
answered yesterday
Levent
3,386825
answered yesterday
Levent
3,386825
answered yesterday
Levent
3,386825
3,386825
add a comment |
add a comment |
up vote
1
down vote
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 yesterday
José Carlos Santos
140k19111204
add a comment |
up vote
1
down vote
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 yesterday
José Carlos Santos
140k19111204
add a comment |
up vote
1
down vote
up vote
1
down vote
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 yesterday
José Carlos Santos
140k19111204
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 yesterday
José Carlos Santos
140k19111204
answered yesterday
José Carlos Santos
140k19111204
answered yesterday
José Carlos Santos
140k19111204
answered yesterday
José Carlos Santos
140k19111204
140k19111204
add a comment |
add a comment |
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1
What happens for $a=0$?
– asdq
yesterday
Though not an exact dupe, many answers in the dupe apply here too.
– Bill Dubuque
yesterday
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
yesterday