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







share|cite|improve this question


















  • 1




    What do you mean with ''each pair of zero-divisors has a nonzero annihilator''?
    – Wuestenfux
    13 hours ago










  • @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
    13 hours ago

















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







share|cite|improve this question


















  • 1




    What do you mean with ''each pair of zero-divisors has a nonzero annihilator''?
    – Wuestenfux
    13 hours ago










  • @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
    13 hours ago















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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 13 hours ago









13571

235




235








  • 1




    What do you mean with ''each pair of zero-divisors has a nonzero annihilator''?
    – Wuestenfux
    13 hours ago










  • @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
    13 hours ago
















  • 1




    What do you mean with ''each pair of zero-divisors has a nonzero annihilator''?
    – Wuestenfux
    13 hours ago










  • @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
    13 hours ago










1




1




What do you mean with ''each pair of zero-divisors has a nonzero annihilator''?
– Wuestenfux
13 hours ago




What do you mean with ''each pair of zero-divisors has a nonzero annihilator''?
– Wuestenfux
13 hours ago












@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
13 hours ago






@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
13 hours ago

















active

oldest

votes











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














 

draft saved


draft discarded


















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






























active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes
















 

draft saved


draft discarded



















































 


draft saved


draft discarded














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





















































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







-

Popular posts from this blog

QoS: MAC-Priority for clients behind a repeater

Image
1 1 I've setup a wireless network using a Linksys-Router (DD-WRT) as an AP and a TP-Link repeater for range extension. To manage bandwith for specific users, I've setup QoS-Rules on the AP (DD-WRT) using MAC-Priority: However the AP shows only the Repeater's MAC and the MACs of users inside the AP-Range. The MAC addresses of users behind the Repeater are not listed : The purple MAC is not listed in the AP-Clientlist. Its related to a device connected via the TP-Link repeater. Q: How can I setup MAC-Priority for repeater-clients? Do I simply add the purple MAC in AP-MAC priority list although it's not listed as a client? I want to setup specific priority instead of giving one priority for all clients behind the repeater by adding the repeater's MAC in the MAC priority list in the AP as
Read more

Ивакино (Тотемский район)

Image
У этого топонима есть и другие значения, см. Ивакино. Деревня Ивакино 59°42′01″ с. ш. 42°01′36″ в. д. H G Я O Страна Россия   Россия Субъект Федерации Вологодская область Муниципальный район Тотемский район Сельское поселение Погореловское История и география Тип климата умеренно-континентальный Часовой пояс UTC+3 Население Население 33 человека ( 2002 ) Национальности русские Цифровые идентификаторы Почтовый индекс 161327 Код ОКАТО 19 246 848 008 Код ОКТМО 19 646 448 136 Прочее Рег. номер 6261 Показать/скрыть карты Ивакино Ивакино Ивакино Ивакино — деревня в Тотемском районе Вологодской области. Входит в состав Погореловского сельского поселения [1] , с точки зрения административно-территориального деления — в Погореловский сельсовет. Расстояние по автодороге до районного центра Тотьмы — 61,5 км, до центра муниципального образования деревни Погорелово — 1,5 км.
Read more

Can't locate Autom4te/ChannelDefs.pm in @INC (when it definitely is there)

Image
up vote 1 down vote favorite I've been running into trouble running make for a build process that I know works on a 32-bit Ubuntu VM. I am running a 64-bit Ubuntu VM, and I have a feeling that the 64-bit may be the problem, but am not entirely sure. Basically, when I run the make command, I get the following error: Can't locate Autom4te/ChannelDefs.pm in @INC (@INC contains: [...]/staging_dir/host/share/autoconf /etc/perl /usr/local/lib/perl/5.14.2 /usr/local/share/perl/5.14.2 /usr/lib/perl5 /usr/share/perl5 /usr/lib/perl/5.14 /usr/share/perl/5.14 /usr/local/lib/site_perl .) at [...]/staging_dir/host/bin/autoreconf line 40. Now if I navigate to [...]/staging_dir/host/share/autoconf I can see that, contrary to what autoreconf thinks, Autom4te/ChannelDefs.pm definitely exists, so I don't really understand what
Read more