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.
commutative-algebra ideals
add a comment |
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.
commutative-algebra ideals
2
What do you mean with ''each pair of zero-divisors has a nonzero annihilator''?
– Wuestenfux
yesterday
@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
yesterday
@AnyAD when the commutative algebra tag is used, we usually assume the ring in question is commutative.
– rschwieb
10 hours ago
add a comment |
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.
commutative-algebra ideals
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.
commutative-algebra ideals
commutative-algebra ideals
asked yesterday
13571
235
235
2
What do you mean with ''each pair of zero-divisors has a nonzero annihilator''?
– Wuestenfux
yesterday
@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
yesterday
@AnyAD when the commutative algebra tag is used, we usually assume the ring in question is commutative.
– rschwieb
10 hours ago
add a comment |
2
What do you mean with ''each pair of zero-divisors has a nonzero annihilator''?
– Wuestenfux
yesterday
@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
yesterday
@AnyAD when the commutative algebra tag is used, we usually assume the ring in question is commutative.
– rschwieb
10 hours ago
2
2
What do you mean with ''each pair of zero-divisors has a nonzero annihilator''?
– Wuestenfux
yesterday
What do you mean with ''each pair of zero-divisors has a nonzero annihilator''?
– Wuestenfux
yesterday
@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
yesterday
@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
yesterday
@AnyAD when the commutative algebra tag is used, we usually assume the ring in question is commutative.
– rschwieb
10 hours ago
@AnyAD when the commutative algebra tag is used, we usually assume the ring in question is commutative.
– rschwieb
10 hours ago
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
yesterday
@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
yesterday
@AnyAD when the commutative algebra tag is used, we usually assume the ring in question is commutative.
– rschwieb
10 hours ago