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Sum of a nilpotent element and a zero divisor in a commutative ring $A$.
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Let $A$ be a commutative ring with identity. Is it always true that the sum of a nilpotent element and a zero divisor is a zero divisor? I tried to construct a counterexample but found it was true on the ring $mathbbZ/nmathbbZ$. Can anyone please give a specific example to this problem?
Some of my attempt: For a nilpotent element $x$ and a zero divisor $z$ the sum $x+z$ satisfies $z'(x+z)=z'x$ for some $z' neq 0$, and since $x$ is nilpotent $z'^n(x+z)^n = 0$ for some $n$. I thought we can find a ring $A$ and some $n$, $z$ where $z'^n(x+z)^n-1$ becomes $0$ but it doesn't mean there are no other non-zero elements that annihilates $x+z$.
abstract-algebra commutative-algebra
asked Jul 19 at 1:18
user413924
1516
add a comment |Â
up vote
4
down vote
favorite
Let $A$ be a commutative ring with identity. Is it always true that the sum of a nilpotent element and a zero divisor is a zero divisor? I tried to construct a counterexample but found it was true on the ring $mathbbZ/nmathbbZ$. Can anyone please give a specific example to this problem?
Some of my attempt: For a nilpotent element $x$ and a zero divisor $z$ the sum $x+z$ satisfies $z'(x+z)=z'x$ for some $z' neq 0$, and since $x$ is nilpotent $z'^n(x+z)^n = 0$ for some $n$. I thought we can find a ring $A$ and some $n$, $z$ where $z'^n(x+z)^n-1$ becomes $0$ but it doesn't mean there are no other non-zero elements that annihilates $x+z$.
abstract-algebra commutative-algebra
asked Jul 19 at 1:18
user413924
1516
@Suzet The point is not that $x$ can be a zero divisor, it is that in the remaining case, the non-zero factor that kills $x$ is of a very special form $z'x^k$ for some $k$. Then $z'x^k(x+z)=0$. Ah! comment deleted. Let me add what was there.
– user574889
Jul 19 at 1:47
@cactus Yes, I agree. In fact, I just noticed how trivial my comment was so I had to erase it, it wasn't relevant whatsoever.
– Suzet
Jul 19 at 1:48
1
The whole argument: $(x+z)z'x^n-1=0$. Then either $z'x^n-1neq0$ and $(x+z)$ is a zero divisor, or $z'x^n-1=0$. In the latter case $z'x^kneq0$ for some $k$ maximum (take into account that for $k=0$, $z'x^0=z'neq0$). Then $(x+z)z'x^k=0$ showing that $(x+z)$ is a zero divisor.
– user574889
Jul 19 at 1:51
@Suzet What are you saying? It was the beginning to a complete proof.
– user574889
Jul 19 at 1:52
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $A$ be a commutative ring with identity. Is it always true that the sum of a nilpotent element and a zero divisor is a zero divisor? I tried to construct a counterexample but found it was true on the ring $mathbbZ/nmathbbZ$. Can anyone please give a specific example to this problem?
Some of my attempt: For a nilpotent element $x$ and a zero divisor $z$ the sum $x+z$ satisfies $z'(x+z)=z'x$ for some $z' neq 0$, and since $x$ is nilpotent $z'^n(x+z)^n = 0$ for some $n$. I thought we can find a ring $A$ and some $n$, $z$ where $z'^n(x+z)^n-1$ becomes $0$ but it doesn't mean there are no other non-zero elements that annihilates $x+z$.
abstract-algebra commutative-algebra
asked Jul 19 at 1:18
user413924
1516
Let $A$ be a commutative ring with identity. Is it always true that the sum of a nilpotent element and a zero divisor is a zero divisor? I tried to construct a counterexample but found it was true on the ring $mathbbZ/nmathbbZ$. Can anyone please give a specific example to this problem?
Some of my attempt: For a nilpotent element $x$ and a zero divisor $z$ the sum $x+z$ satisfies $z'(x+z)=z'x$ for some $z' neq 0$, and since $x$ is nilpotent $z'^n(x+z)^n = 0$ for some $n$. I thought we can find a ring $A$ and some $n$, $z$ where $z'^n(x+z)^n-1$ becomes $0$ but it doesn't mean there are no other non-zero elements that annihilates $x+z$.
abstract-algebra commutative-algebra
asked Jul 19 at 1:18
user413924
1516
edited Jul 19 at 1:24
asked Jul 19 at 1:18
user413924
1516
asked Jul 19 at 1:18
user413924
1516
asked Jul 19 at 1:18
user413924
1516
1516
@Suzet The point is not that $x$ can be a zero divisor, it is that in the remaining case, the non-zero factor that kills $x$ is of a very special form $z'x^k$ for some $k$. Then $z'x^k(x+z)=0$. Ah! comment deleted. Let me add what was there.
– user574889
Jul 19 at 1:47
@cactus Yes, I agree. In fact, I just noticed how trivial my comment was so I had to erase it, it wasn't relevant whatsoever.
– Suzet
Jul 19 at 1:48
1
The whole argument: $(x+z)z'x^n-1=0$. Then either $z'x^n-1neq0$ and $(x+z)$ is a zero divisor, or $z'x^n-1=0$. In the latter case $z'x^kneq0$ for some $k$ maximum (take into account that for $k=0$, $z'x^0=z'neq0$). Then $(x+z)z'x^k=0$ showing that $(x+z)$ is a zero divisor.
– user574889
Jul 19 at 1:51
@Suzet What are you saying? It was the beginning to a complete proof.
– user574889
Jul 19 at 1:52
add a comment |Â
@Suzet The point is not that $x$ can be a zero divisor, it is that in the remaining case, the non-zero factor that kills $x$ is of a very special form $z'x^k$ for some $k$. Then $z'x^k(x+z)=0$. Ah! comment deleted. Let me add what was there.
– user574889
Jul 19 at 1:47
@cactus Yes, I agree. In fact, I just noticed how trivial my comment was so I had to erase it, it wasn't relevant whatsoever.
– Suzet
Jul 19 at 1:48
1
The whole argument: $(x+z)z'x^n-1=0$. Then either $z'x^n-1neq0$ and $(x+z)$ is a zero divisor, or $z'x^n-1=0$. In the latter case $z'x^kneq0$ for some $k$ maximum (take into account that for $k=0$, $z'x^0=z'neq0$). Then $(x+z)z'x^k=0$ showing that $(x+z)$ is a zero divisor.
– user574889
Jul 19 at 1:51
@Suzet What are you saying? It was the beginning to a complete proof.
– user574889
Jul 19 at 1:52
@Suzet The point is not that $x$ can be a zero divisor, it is that in the remaining case, the non-zero factor that kills $x$ is of a very special form $z'x^k$ for some $k$. Then $z'x^k(x+z)=0$. Ah! comment deleted. Let me add what was there.
– user574889
Jul 19 at 1:47
@Suzet The point is not that $x$ can be a zero divisor, it is that in the remaining case, the non-zero factor that kills $x$ is of a very special form $z'x^k$ for some $k$. Then $z'x^k(x+z)=0$. Ah! comment deleted. Let me add what was there.
– user574889
Jul 19 at 1:47
@cactus Yes, I agree. In fact, I just noticed how trivial my comment was so I had to erase it, it wasn't relevant whatsoever.
– Suzet
Jul 19 at 1:48
@cactus Yes, I agree. In fact, I just noticed how trivial my comment was so I had to erase it, it wasn't relevant whatsoever.
– Suzet
Jul 19 at 1:48
1
1
The whole argument: $(x+z)z'x^n-1=0$. Then either $z'x^n-1neq0$ and $(x+z)$ is a zero divisor, or $z'x^n-1=0$. In the latter case $z'x^kneq0$ for some $k$ maximum (take into account that for $k=0$, $z'x^0=z'neq0$). Then $(x+z)z'x^k=0$ showing that $(x+z)$ is a zero divisor.
– user574889
Jul 19 at 1:51
The whole argument: $(x+z)z'x^n-1=0$. Then either $z'x^n-1neq0$ and $(x+z)$ is a zero divisor, or $z'x^n-1=0$. In the latter case $z'x^kneq0$ for some $k$ maximum (take into account that for $k=0$, $z'x^0=z'neq0$). Then $(x+z)z'x^k=0$ showing that $(x+z)$ is a zero divisor.
– user574889
Jul 19 at 1:51
@Suzet What are you saying? It was the beginning to a complete proof.
– user574889
Jul 19 at 1:52
@Suzet What are you saying? It was the beginning to a complete proof.
– user574889
Jul 19 at 1:52
add a comment |Â
1 Answer
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Suppose $zw = 0$ with $w ne 0$, but $x+z$ is not a zero-divisor.
Then $xw = (x+z)w ne 0$, $x^2 w = (x+z) xw ne 0$, and by induction
$x^n w ne 0$ for all positive integers $n$, implying $x^n ne 0$ and $x$ is not nilpotent.
answered Jul 19 at 1:53
Robert Israel
304k22201442
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Suppose $zw = 0$ with $w ne 0$, but $x+z$ is not a zero-divisor.
Then $xw = (x+z)w ne 0$, $x^2 w = (x+z) xw ne 0$, and by induction
$x^n w ne 0$ for all positive integers $n$, implying $x^n ne 0$ and $x$ is not nilpotent.
answered Jul 19 at 1:53
Robert Israel
304k22201442
add a comment |Â
up vote
6
down vote
accepted
Suppose $zw = 0$ with $w ne 0$, but $x+z$ is not a zero-divisor.
Then $xw = (x+z)w ne 0$, $x^2 w = (x+z) xw ne 0$, and by induction
$x^n w ne 0$ for all positive integers $n$, implying $x^n ne 0$ and $x$ is not nilpotent.
answered Jul 19 at 1:53
Robert Israel
304k22201442
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Suppose $zw = 0$ with $w ne 0$, but $x+z$ is not a zero-divisor.
Then $xw = (x+z)w ne 0$, $x^2 w = (x+z) xw ne 0$, and by induction
$x^n w ne 0$ for all positive integers $n$, implying $x^n ne 0$ and $x$ is not nilpotent.
answered Jul 19 at 1:53
Robert Israel
304k22201442
Suppose $zw = 0$ with $w ne 0$, but $x+z$ is not a zero-divisor.
Then $xw = (x+z)w ne 0$, $x^2 w = (x+z) xw ne 0$, and by induction
$x^n w ne 0$ for all positive integers $n$, implying $x^n ne 0$ and $x$ is not nilpotent.
answered Jul 19 at 1:53
Robert Israel
304k22201442
answered Jul 19 at 1:53
Robert Israel
304k22201442
answered Jul 19 at 1:53
Robert Israel
304k22201442
answered Jul 19 at 1:53
Robert Israel
304k22201442
304k22201442
add a comment |Â
add a comment |Â
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@Suzet The point is not that $x$ can be a zero divisor, it is that in the remaining case, the non-zero factor that kills $x$ is of a very special form $z'x^k$ for some $k$. Then $z'x^k(x+z)=0$. Ah! comment deleted. Let me add what was there.
– user574889
Jul 19 at 1:47
@cactus Yes, I agree. In fact, I just noticed how trivial my comment was so I had to erase it, it wasn't relevant whatsoever.
– Suzet
Jul 19 at 1:48
1
The whole argument: $(x+z)z'x^n-1=0$. Then either $z'x^n-1neq0$ and $(x+z)$ is a zero divisor, or $z'x^n-1=0$. In the latter case $z'x^kneq0$ for some $k$ maximum (take into account that for $k=0$, $z'x^0=z'neq0$). Then $(x+z)z'x^k=0$ showing that $(x+z)$ is a zero divisor.
– user574889
Jul 19 at 1:51
@Suzet What are you saying? It was the beginning to a complete proof.
– user574889
Jul 19 at 1:52