The number of divisors of $0$ in a finite ring
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Let $(A,+,•)$ be a ring with unity, which is not a division ring, and $|A|=25.$ Let $a$ be the number of divisors of $0$ in this ring. Then
a) $a=1;$
b) $a=2;$
c) $a=3;$
d) $a geq 4.$
I guess that the answer is d), since this would be the case if we let $A=mathbbZ/25 mathbbZ.$ However, I couldn't prove it for general $A.$
I know that a finite ring with $0 neq 1$ which doesn't have divisors of $0$ is a division ring. Hence, our $A$ must have divisors of 0. I tried to use the characteristic and I got that it can only be $5$ or $25,$ but this is where I got stuck.
abstract-algebra group-theory ring-theory finite-groups
asked Sep 9 at 9:21
AndrewC
923213
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up vote
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Let $(A,+,•)$ be a ring with unity, which is not a division ring, and $|A|=25.$ Let $a$ be the number of divisors of $0$ in this ring. Then
a) $a=1;$
b) $a=2;$
c) $a=3;$
d) $a geq 4.$
I guess that the answer is d), since this would be the case if we let $A=mathbbZ/25 mathbbZ.$ However, I couldn't prove it for general $A.$
I know that a finite ring with $0 neq 1$ which doesn't have divisors of $0$ is a division ring. Hence, our $A$ must have divisors of 0. I tried to use the characteristic and I got that it can only be $5$ or $25,$ but this is where I got stuck.
abstract-algebra group-theory ring-theory finite-groups
asked Sep 9 at 9:21
AndrewC
923213
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(A,+,•)$ be a ring with unity, which is not a division ring, and $|A|=25.$ Let $a$ be the number of divisors of $0$ in this ring. Then
a) $a=1;$
b) $a=2;$
c) $a=3;$
d) $a geq 4.$
I guess that the answer is d), since this would be the case if we let $A=mathbbZ/25 mathbbZ.$ However, I couldn't prove it for general $A.$
I know that a finite ring with $0 neq 1$ which doesn't have divisors of $0$ is a division ring. Hence, our $A$ must have divisors of 0. I tried to use the characteristic and I got that it can only be $5$ or $25,$ but this is where I got stuck.
abstract-algebra group-theory ring-theory finite-groups
asked Sep 9 at 9:21
AndrewC
923213
Let $(A,+,•)$ be a ring with unity, which is not a division ring, and $|A|=25.$ Let $a$ be the number of divisors of $0$ in this ring. Then
a) $a=1;$
b) $a=2;$
c) $a=3;$
d) $a geq 4.$
I guess that the answer is d), since this would be the case if we let $A=mathbbZ/25 mathbbZ.$ However, I couldn't prove it for general $A.$
I know that a finite ring with $0 neq 1$ which doesn't have divisors of $0$ is a division ring. Hence, our $A$ must have divisors of 0. I tried to use the characteristic and I got that it can only be $5$ or $25,$ but this is where I got stuck.
abstract-algebra group-theory ring-theory finite-groups
abstract-algebra group-theory ring-theory finite-groups
asked Sep 9 at 9:21
AndrewC
923213
asked Sep 9 at 9:21
AndrewC
923213
asked Sep 9 at 9:21
AndrewC
923213
asked Sep 9 at 9:21
AndrewC
923213
asked Sep 9 at 9:21
AndrewC
923213
923213
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
votes
up vote
4
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If $eneq 0$ is a left zero divisor in $A$, then the set of zero divisors contains the left ideal $Ae$. Since this ideal is non-trivial (it contains $e$) and it is not equal to $A$ (it doesn't contain $1$) it is of size $5$. Thus you have at least $4$ zero divisors.
answered Sep 9 at 9:33
SMM
2,03049
4
Great answer. I think it suffices to consider the additive subgroup generated by $e$ by the way. :)
– awllower
Sep 9 at 9:36
1
Why would the ideal only have zero divisors? Consider the ring of 2x2 matrices over a field. The diagonal matrix $e$ with entries 1 and 0 is a zero divisor. But the ideal it generates is the whole ring because the matrix ring is simple.
– Jyrki Lahtonen
Sep 9 at 9:38
True, there are no such matrix rings of 25 elements, but... you need to dig deeper to prove that. I will take @awllower's comment!
– Jyrki Lahtonen
Sep 9 at 9:39
@JyrkiLahtonen I considered the ring to be commutative, but you are right, I don't have this information. So we assume that $e$ is a left zero divisor and $langle erangle$ is the left ideal. Thanks!
– SMM
Sep 9 at 9:46
A nice fix. Sorry about jumping the gun a bit.
– Jyrki Lahtonen
Sep 9 at 10:05
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
If $eneq 0$ is a left zero divisor in $A$, then the set of zero divisors contains the left ideal $Ae$. Since this ideal is non-trivial (it contains $e$) and it is not equal to $A$ (it doesn't contain $1$) it is of size $5$. Thus you have at least $4$ zero divisors.
answered Sep 9 at 9:33
SMM
2,03049
4
Great answer. I think it suffices to consider the additive subgroup generated by $e$ by the way. :)
– awllower
Sep 9 at 9:36
1
Why would the ideal only have zero divisors? Consider the ring of 2x2 matrices over a field. The diagonal matrix $e$ with entries 1 and 0 is a zero divisor. But the ideal it generates is the whole ring because the matrix ring is simple.
– Jyrki Lahtonen
Sep 9 at 9:38
True, there are no such matrix rings of 25 elements, but... you need to dig deeper to prove that. I will take @awllower's comment!
– Jyrki Lahtonen
Sep 9 at 9:39
@JyrkiLahtonen I considered the ring to be commutative, but you are right, I don't have this information. So we assume that $e$ is a left zero divisor and $langle erangle$ is the left ideal. Thanks!
– SMM
Sep 9 at 9:46
A nice fix. Sorry about jumping the gun a bit.
– Jyrki Lahtonen
Sep 9 at 10:05
add a comment |Â
up vote
4
down vote
accepted
If $eneq 0$ is a left zero divisor in $A$, then the set of zero divisors contains the left ideal $Ae$. Since this ideal is non-trivial (it contains $e$) and it is not equal to $A$ (it doesn't contain $1$) it is of size $5$. Thus you have at least $4$ zero divisors.
answered Sep 9 at 9:33
SMM
2,03049
4
Great answer. I think it suffices to consider the additive subgroup generated by $e$ by the way. :)
– awllower
Sep 9 at 9:36
1
Why would the ideal only have zero divisors? Consider the ring of 2x2 matrices over a field. The diagonal matrix $e$ with entries 1 and 0 is a zero divisor. But the ideal it generates is the whole ring because the matrix ring is simple.
– Jyrki Lahtonen
Sep 9 at 9:38
True, there are no such matrix rings of 25 elements, but... you need to dig deeper to prove that. I will take @awllower's comment!
– Jyrki Lahtonen
Sep 9 at 9:39
@JyrkiLahtonen I considered the ring to be commutative, but you are right, I don't have this information. So we assume that $e$ is a left zero divisor and $langle erangle$ is the left ideal. Thanks!
– SMM
Sep 9 at 9:46
A nice fix. Sorry about jumping the gun a bit.
– Jyrki Lahtonen
Sep 9 at 10:05
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
If $eneq 0$ is a left zero divisor in $A$, then the set of zero divisors contains the left ideal $Ae$. Since this ideal is non-trivial (it contains $e$) and it is not equal to $A$ (it doesn't contain $1$) it is of size $5$. Thus you have at least $4$ zero divisors.
answered Sep 9 at 9:33
SMM
2,03049
If $eneq 0$ is a left zero divisor in $A$, then the set of zero divisors contains the left ideal $Ae$. Since this ideal is non-trivial (it contains $e$) and it is not equal to $A$ (it doesn't contain $1$) it is of size $5$. Thus you have at least $4$ zero divisors.
answered Sep 9 at 9:33
SMM
2,03049
edited Sep 9 at 9:47
answered Sep 9 at 9:33
SMM
2,03049
answered Sep 9 at 9:33
SMM
2,03049
answered Sep 9 at 9:33
SMM
2,03049
2,03049
4
Great answer. I think it suffices to consider the additive subgroup generated by $e$ by the way. :)
– awllower
Sep 9 at 9:36
1
Why would the ideal only have zero divisors? Consider the ring of 2x2 matrices over a field. The diagonal matrix $e$ with entries 1 and 0 is a zero divisor. But the ideal it generates is the whole ring because the matrix ring is simple.
– Jyrki Lahtonen
Sep 9 at 9:38
True, there are no such matrix rings of 25 elements, but... you need to dig deeper to prove that. I will take @awllower's comment!
– Jyrki Lahtonen
Sep 9 at 9:39
@JyrkiLahtonen I considered the ring to be commutative, but you are right, I don't have this information. So we assume that $e$ is a left zero divisor and $langle erangle$ is the left ideal. Thanks!
– SMM
Sep 9 at 9:46
A nice fix. Sorry about jumping the gun a bit.
– Jyrki Lahtonen
Sep 9 at 10:05
add a comment |Â
4
Great answer. I think it suffices to consider the additive subgroup generated by $e$ by the way. :)
– awllower
Sep 9 at 9:36
1
Why would the ideal only have zero divisors? Consider the ring of 2x2 matrices over a field. The diagonal matrix $e$ with entries 1 and 0 is a zero divisor. But the ideal it generates is the whole ring because the matrix ring is simple.
– Jyrki Lahtonen
Sep 9 at 9:38
True, there are no such matrix rings of 25 elements, but... you need to dig deeper to prove that. I will take @awllower's comment!
– Jyrki Lahtonen
Sep 9 at 9:39
@JyrkiLahtonen I considered the ring to be commutative, but you are right, I don't have this information. So we assume that $e$ is a left zero divisor and $langle erangle$ is the left ideal. Thanks!
– SMM
Sep 9 at 9:46
A nice fix. Sorry about jumping the gun a bit.
– Jyrki Lahtonen
Sep 9 at 10:05
4
4
Great answer. I think it suffices to consider the additive subgroup generated by $e$ by the way. :)
– awllower
Sep 9 at 9:36
Great answer. I think it suffices to consider the additive subgroup generated by $e$ by the way. :)
– awllower
Sep 9 at 9:36
1
1
Why would the ideal only have zero divisors? Consider the ring of 2x2 matrices over a field. The diagonal matrix $e$ with entries 1 and 0 is a zero divisor. But the ideal it generates is the whole ring because the matrix ring is simple.
– Jyrki Lahtonen
Sep 9 at 9:38
Why would the ideal only have zero divisors? Consider the ring of 2x2 matrices over a field. The diagonal matrix $e$ with entries 1 and 0 is a zero divisor. But the ideal it generates is the whole ring because the matrix ring is simple.
– Jyrki Lahtonen
Sep 9 at 9:38
True, there are no such matrix rings of 25 elements, but... you need to dig deeper to prove that. I will take @awllower's comment!
– Jyrki Lahtonen
Sep 9 at 9:39
True, there are no such matrix rings of 25 elements, but... you need to dig deeper to prove that. I will take @awllower's comment!
– Jyrki Lahtonen
Sep 9 at 9:39
@JyrkiLahtonen I considered the ring to be commutative, but you are right, I don't have this information. So we assume that $e$ is a left zero divisor and $langle erangle$ is the left ideal. Thanks!
– SMM
Sep 9 at 9:46
@JyrkiLahtonen I considered the ring to be commutative, but you are right, I don't have this information. So we assume that $e$ is a left zero divisor and $langle erangle$ is the left ideal. Thanks!
– SMM
Sep 9 at 9:46
A nice fix. Sorry about jumping the gun a bit.
– Jyrki Lahtonen
Sep 9 at 10:05
A nice fix. Sorry about jumping the gun a bit.
– Jyrki Lahtonen
Sep 9 at 10:05
add a comment |Â
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