zero-divisors of a ring constitute an ideal
up vote
1
down vote
favorite
I want to know if
"zero-divisors of a ring constitute an ideal iff each pair of zero-divisors of the ring has a nonzero annihilator?"
the crucial point for zero-divisors of a ring to constitute an ideal is to check if the sum of each pair of zero-divisors is again a zero-divisor. so one direction is trivial:
if each pair of zero-divisors of the ring has a nonzero annihilator then
zero-divisors constitute an ideal.
what about the converse?
thanks.
Update
by ''each pair of zero-divisors has a nonzero annihilator'' i mean "for any pair of distinct zero-divisors like $a$ and $b$ we have a nonzero element $cin R$ s.t. $ca=cb=0$.
ring-theory commutative-algebra ideals
asked Nov 15 at 8:18
13571
235
|
show 2 more comments
up vote
1
down vote
favorite
I want to know if
"zero-divisors of a ring constitute an ideal iff each pair of zero-divisors of the ring has a nonzero annihilator?"
the crucial point for zero-divisors of a ring to constitute an ideal is to check if the sum of each pair of zero-divisors is again a zero-divisor. so one direction is trivial:
if each pair of zero-divisors of the ring has a nonzero annihilator then
zero-divisors constitute an ideal.
what about the converse?
thanks.
Update
by ''each pair of zero-divisors has a nonzero annihilator'' i mean "for any pair of distinct zero-divisors like $a$ and $b$ we have a nonzero element $cin R$ s.t. $ca=cb=0$.
ring-theory commutative-algebra ideals
asked Nov 15 at 8:18
13571
235
2
What do you mean with ''each pair of zero-divisors has a nonzero annihilator''?
– Wuestenfux
Nov 15 at 8:38
@13571 Is it a commutative ring? Otherwise, right or left annihilator? Also as commented above, your writing is not clear, 'pair' in what sense?
– AnyAD
Nov 15 at 8:48
2
@AnyAD when the commutative algebra tag is used, we usually assume the ring in question is commutative.
– rschwieb
Nov 16 at 12:12
by ''each pair of zero-divisors has a nonzero annihilator'' i mean "any distinct zero-divisors $a$ and $b$ has a nonzero common annihilator"
– 13571
Nov 17 at 8:21
@rschwieb Thanks for that.
– AnyAD
Nov 18 at 2:44
|
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to know if
"zero-divisors of a ring constitute an ideal iff each pair of zero-divisors of the ring has a nonzero annihilator?"
the crucial point for zero-divisors of a ring to constitute an ideal is to check if the sum of each pair of zero-divisors is again a zero-divisor. so one direction is trivial:
if each pair of zero-divisors of the ring has a nonzero annihilator then
zero-divisors constitute an ideal.
what about the converse?
thanks.
Update
by ''each pair of zero-divisors has a nonzero annihilator'' i mean "for any pair of distinct zero-divisors like $a$ and $b$ we have a nonzero element $cin R$ s.t. $ca=cb=0$.
ring-theory commutative-algebra ideals
asked Nov 15 at 8:18
13571
235
I want to know if
"zero-divisors of a ring constitute an ideal iff each pair of zero-divisors of the ring has a nonzero annihilator?"
the crucial point for zero-divisors of a ring to constitute an ideal is to check if the sum of each pair of zero-divisors is again a zero-divisor. so one direction is trivial:
if each pair of zero-divisors of the ring has a nonzero annihilator then
zero-divisors constitute an ideal.
what about the converse?
thanks.
Update
by ''each pair of zero-divisors has a nonzero annihilator'' i mean "for any pair of distinct zero-divisors like $a$ and $b$ we have a nonzero element $cin R$ s.t. $ca=cb=0$.
ring-theory commutative-algebra ideals
ring-theory commutative-algebra ideals
asked Nov 15 at 8:18
13571
235
asked Nov 15 at 8:18
13571
235
edited Dec 2 at 3:56
asked Nov 15 at 8:18
13571
235
asked Nov 15 at 8:18
13571
235
asked Nov 15 at 8:18
13571
235
235
2
What do you mean with ''each pair of zero-divisors has a nonzero annihilator''?
– Wuestenfux
Nov 15 at 8:38
@13571 Is it a commutative ring? Otherwise, right or left annihilator? Also as commented above, your writing is not clear, 'pair' in what sense?
– AnyAD
Nov 15 at 8:48
2
@AnyAD when the commutative algebra tag is used, we usually assume the ring in question is commutative.
– rschwieb
Nov 16 at 12:12
by ''each pair of zero-divisors has a nonzero annihilator'' i mean "any distinct zero-divisors $a$ and $b$ has a nonzero common annihilator"
– 13571
Nov 17 at 8:21
@rschwieb Thanks for that.
– AnyAD
Nov 18 at 2:44
|
show 2 more comments
2
What do you mean with ''each pair of zero-divisors has a nonzero annihilator''?
– Wuestenfux
Nov 15 at 8:38
@13571 Is it a commutative ring? Otherwise, right or left annihilator? Also as commented above, your writing is not clear, 'pair' in what sense?
– AnyAD
Nov 15 at 8:48
2
@AnyAD when the commutative algebra tag is used, we usually assume the ring in question is commutative.
– rschwieb
Nov 16 at 12:12
by ''each pair of zero-divisors has a nonzero annihilator'' i mean "any distinct zero-divisors $a$ and $b$ has a nonzero common annihilator"
– 13571
Nov 17 at 8:21
@rschwieb Thanks for that.
– AnyAD
Nov 18 at 2:44
2
2
What do you mean with ''each pair of zero-divisors has a nonzero annihilator''?
– Wuestenfux
Nov 15 at 8:38
What do you mean with ''each pair of zero-divisors has a nonzero annihilator''?
– Wuestenfux
Nov 15 at 8:38
@13571 Is it a commutative ring? Otherwise, right or left annihilator? Also as commented above, your writing is not clear, 'pair' in what sense?
– AnyAD
Nov 15 at 8:48
@13571 Is it a commutative ring? Otherwise, right or left annihilator? Also as commented above, your writing is not clear, 'pair' in what sense?
– AnyAD
Nov 15 at 8:48
2
2
@AnyAD when the commutative algebra tag is used, we usually assume the ring in question is commutative.
– rschwieb
Nov 16 at 12:12
@AnyAD when the commutative algebra tag is used, we usually assume the ring in question is commutative.
– rschwieb
Nov 16 at 12:12
by ''each pair of zero-divisors has a nonzero annihilator'' i mean "any distinct zero-divisors $a$ and $b$ has a nonzero common annihilator"
– 13571
Nov 17 at 8:21
by ''each pair of zero-divisors has a nonzero annihilator'' i mean "any distinct zero-divisors $a$ and $b$ has a nonzero common annihilator"
– 13571
Nov 17 at 8:21
@rschwieb Thanks for that.
– AnyAD
Nov 18 at 2:44
@rschwieb Thanks for that.
– AnyAD
Nov 18 at 2:44
|
show 2 more comments
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999404%2fzero-divisors-of-a-ring-constitute-an-ideal%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
What do you mean with ''each pair of zero-divisors has a nonzero annihilator''?
– Wuestenfux
Nov 15 at 8:38
@13571 Is it a commutative ring? Otherwise, right or left annihilator? Also as commented above, your writing is not clear, 'pair' in what sense?
– AnyAD
Nov 15 at 8:48
2
@AnyAD when the commutative algebra tag is used, we usually assume the ring in question is commutative.
– rschwieb
Nov 16 at 12:12
by ''each pair of zero-divisors has a nonzero annihilator'' i mean "any distinct zero-divisors $a$ and $b$ has a nonzero common annihilator"
– 13571
Nov 17 at 8:21
@rschwieb Thanks for that.
– AnyAD
Nov 18 at 2:44