Finding essential ideal in a ring $R$












1












$begingroup$


Let $R$ be a ring and let $L$ be a left ideal of $R$. The left ideal $L$ is said to be essential if $L cap S neq lbrace 0 rbrace$, for any non-zero left ideal $S$ of $R$.



Now let $L$ be a random left ideal of $R$. I have to show that there exists a left ideal $L'$ of $R$ such that $L cap L' =lbrace 0 rbrace$ and $L oplus L'$ is an essential ideal of $R$.



Any tips?






ring-theory ideals







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I would go for a maximal $L'$ which is disjoint to $L$, using Zorn lemma..
    $endgroup$
    – Berci
    Dec 21 '18 at 12:13






  • 1




    $begingroup$
    Using Zorn's lemma, I have found a maximal element $M$ of the set $P=lbrace I mid I text{ideal and} I cap L = (0) rbrace$. If I take $L'=M$, then the first condition holds. For the second condition, if $S$ is a non-zero left ideal of $R$, then there are two options: $S cap L neq lbrace 0 rbrace$ (in this case $L oplus L' cap S neq lbrace 0 rbrace$) and $S cap L = lbrace 0 rbrace$. In this case we have that $S in P$. How do I continue from here?
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 13:51


















1












$begingroup$


Let $R$ be a ring and let $L$ be a left ideal of $R$. The left ideal $L$ is said to be essential if $L cap S neq lbrace 0 rbrace$, for any non-zero left ideal $S$ of $R$.



Now let $L$ be a random left ideal of $R$. I have to show that there exists a left ideal $L'$ of $R$ such that $L cap L' =lbrace 0 rbrace$ and $L oplus L'$ is an essential ideal of $R$.



Any tips?






ring-theory ideals







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I would go for a maximal $L'$ which is disjoint to $L$, using Zorn lemma..
    $endgroup$
    – Berci
    Dec 21 '18 at 12:13






  • 1




    $begingroup$
    Using Zorn's lemma, I have found a maximal element $M$ of the set $P=lbrace I mid I text{ideal and} I cap L = (0) rbrace$. If I take $L'=M$, then the first condition holds. For the second condition, if $S$ is a non-zero left ideal of $R$, then there are two options: $S cap L neq lbrace 0 rbrace$ (in this case $L oplus L' cap S neq lbrace 0 rbrace$) and $S cap L = lbrace 0 rbrace$. In this case we have that $S in P$. How do I continue from here?
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 13:51
















1












1








1





$begingroup$


Let $R$ be a ring and let $L$ be a left ideal of $R$. The left ideal $L$ is said to be essential if $L cap S neq lbrace 0 rbrace$, for any non-zero left ideal $S$ of $R$.



Now let $L$ be a random left ideal of $R$. I have to show that there exists a left ideal $L'$ of $R$ such that $L cap L' =lbrace 0 rbrace$ and $L oplus L'$ is an essential ideal of $R$.



Any tips?






ring-theory ideals







share|cite|improve this question









$endgroup$




Let $R$ be a ring and let $L$ be a left ideal of $R$. The left ideal $L$ is said to be essential if $L cap S neq lbrace 0 rbrace$, for any non-zero left ideal $S$ of $R$.



Now let $L$ be a random left ideal of $R$. I have to show that there exists a left ideal $L'$ of $R$ such that $L cap L' =lbrace 0 rbrace$ and $L oplus L'$ is an essential ideal of $R$.



Any tips?






ring-theory ideals




ring-theory ideals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 21 '18 at 11:49









J.BosserJ.Bosser

363210




363210








  • 1




    $begingroup$
    I would go for a maximal $L'$ which is disjoint to $L$, using Zorn lemma..
    $endgroup$
    – Berci
    Dec 21 '18 at 12:13






  • 1




    $begingroup$
    Using Zorn's lemma, I have found a maximal element $M$ of the set $P=lbrace I mid I text{ideal and} I cap L = (0) rbrace$. If I take $L'=M$, then the first condition holds. For the second condition, if $S$ is a non-zero left ideal of $R$, then there are two options: $S cap L neq lbrace 0 rbrace$ (in this case $L oplus L' cap S neq lbrace 0 rbrace$) and $S cap L = lbrace 0 rbrace$. In this case we have that $S in P$. How do I continue from here?
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 13:51
















  • 1




    $begingroup$
    I would go for a maximal $L'$ which is disjoint to $L$, using Zorn lemma..
    $endgroup$
    – Berci
    Dec 21 '18 at 12:13






  • 1




    $begingroup$
    Using Zorn's lemma, I have found a maximal element $M$ of the set $P=lbrace I mid I text{ideal and} I cap L = (0) rbrace$. If I take $L'=M$, then the first condition holds. For the second condition, if $S$ is a non-zero left ideal of $R$, then there are two options: $S cap L neq lbrace 0 rbrace$ (in this case $L oplus L' cap S neq lbrace 0 rbrace$) and $S cap L = lbrace 0 rbrace$. In this case we have that $S in P$. How do I continue from here?
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 13:51










1




1




$begingroup$
I would go for a maximal $L'$ which is disjoint to $L$, using Zorn lemma..
$endgroup$
– Berci
Dec 21 '18 at 12:13




$begingroup$
I would go for a maximal $L'$ which is disjoint to $L$, using Zorn lemma..
$endgroup$
– Berci
Dec 21 '18 at 12:13




1




1




$begingroup$
Using Zorn's lemma, I have found a maximal element $M$ of the set $P=lbrace I mid I text{ideal and} I cap L = (0) rbrace$. If I take $L'=M$, then the first condition holds. For the second condition, if $S$ is a non-zero left ideal of $R$, then there are two options: $S cap L neq lbrace 0 rbrace$ (in this case $L oplus L' cap S neq lbrace 0 rbrace$) and $S cap L = lbrace 0 rbrace$. In this case we have that $S in P$. How do I continue from here?
$endgroup$
– J.Bosser
Dec 21 '18 at 13:51






$begingroup$
Using Zorn's lemma, I have found a maximal element $M$ of the set $P=lbrace I mid I text{ideal and} I cap L = (0) rbrace$. If I take $L'=M$, then the first condition holds. For the second condition, if $S$ is a non-zero left ideal of $R$, then there are two options: $S cap L neq lbrace 0 rbrace$ (in this case $L oplus L' cap S neq lbrace 0 rbrace$) and $S cap L = lbrace 0 rbrace$. In this case we have that $S in P$. How do I continue from here?
$endgroup$
– J.Bosser
Dec 21 '18 at 13:51












1 Answer
1






active

oldest

votes


















1












$begingroup$


then there are two options: $S∩L≠{0}$ (in this case $L⊕L′∩S≠{0}$) and $S∩L={0}$. In this case we have that $S∈P$. How do I continue from here?




You're basically right on top of the answer. You've already chosen $L'$ to be maximal (in your comment). If $Loplus L'cap S={0}$, then $Loplus L'oplus S$ is direct as well.



So $L'oplus Sin P$, not merely $S$.



Now do you see the contradiction?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Then $L' oplus S= L'$, by the maximality of $L'$ and we're done.
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 14:29












  • $begingroup$
    @J.Bosser Well, if by "done" you mean "we have reached a contradiction", then yes.
    $endgroup$
    – rschwieb
    Dec 21 '18 at 14:32










  • $begingroup$
    I have a more general question. The definition of a direct sum of two ideals is clear to me. But if one writes $oplus_{i in I}J_i$, a 'general' direct sum of ideals, does this mean that $J_i cap J_j = lbrace 0 rbrace$, for $i,j in I$?
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 14:32












  • $begingroup$
    @J.Bosser NO, it is much stronger than that. It means that $(sum_{ineq j} I_j)cap I_j={0}$ for every index $j$. It is easy to formulate a counterexample in $mathbb R^2$ that has the property you mention, but which is obviously not a direct sum.
    $endgroup$
    – rschwieb
    Dec 21 '18 at 14:33












  • $begingroup$
    Okay, thanks for the clarification!
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 14:37












Your Answer








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',
autoActivateHeartbeat: false,
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%2f3048433%2ffinding-essential-ideal-in-a-ring-r%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$


then there are two options: $S∩L≠{0}$ (in this case $L⊕L′∩S≠{0}$) and $S∩L={0}$. In this case we have that $S∈P$. How do I continue from here?




You're basically right on top of the answer. You've already chosen $L'$ to be maximal (in your comment). If $Loplus L'cap S={0}$, then $Loplus L'oplus S$ is direct as well.



So $L'oplus Sin P$, not merely $S$.



Now do you see the contradiction?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Then $L' oplus S= L'$, by the maximality of $L'$ and we're done.
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 14:29












  • $begingroup$
    @J.Bosser Well, if by "done" you mean "we have reached a contradiction", then yes.
    $endgroup$
    – rschwieb
    Dec 21 '18 at 14:32










  • $begingroup$
    I have a more general question. The definition of a direct sum of two ideals is clear to me. But if one writes $oplus_{i in I}J_i$, a 'general' direct sum of ideals, does this mean that $J_i cap J_j = lbrace 0 rbrace$, for $i,j in I$?
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 14:32












  • $begingroup$
    @J.Bosser NO, it is much stronger than that. It means that $(sum_{ineq j} I_j)cap I_j={0}$ for every index $j$. It is easy to formulate a counterexample in $mathbb R^2$ that has the property you mention, but which is obviously not a direct sum.
    $endgroup$
    – rschwieb
    Dec 21 '18 at 14:33












  • $begingroup$
    Okay, thanks for the clarification!
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 14:37
















1












$begingroup$


then there are two options: $S∩L≠{0}$ (in this case $L⊕L′∩S≠{0}$) and $S∩L={0}$. In this case we have that $S∈P$. How do I continue from here?




You're basically right on top of the answer. You've already chosen $L'$ to be maximal (in your comment). If $Loplus L'cap S={0}$, then $Loplus L'oplus S$ is direct as well.



So $L'oplus Sin P$, not merely $S$.



Now do you see the contradiction?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Then $L' oplus S= L'$, by the maximality of $L'$ and we're done.
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 14:29












  • $begingroup$
    @J.Bosser Well, if by "done" you mean "we have reached a contradiction", then yes.
    $endgroup$
    – rschwieb
    Dec 21 '18 at 14:32










  • $begingroup$
    I have a more general question. The definition of a direct sum of two ideals is clear to me. But if one writes $oplus_{i in I}J_i$, a 'general' direct sum of ideals, does this mean that $J_i cap J_j = lbrace 0 rbrace$, for $i,j in I$?
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 14:32












  • $begingroup$
    @J.Bosser NO, it is much stronger than that. It means that $(sum_{ineq j} I_j)cap I_j={0}$ for every index $j$. It is easy to formulate a counterexample in $mathbb R^2$ that has the property you mention, but which is obviously not a direct sum.
    $endgroup$
    – rschwieb
    Dec 21 '18 at 14:33












  • $begingroup$
    Okay, thanks for the clarification!
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 14:37














1












1








1





$begingroup$


then there are two options: $S∩L≠{0}$ (in this case $L⊕L′∩S≠{0}$) and $S∩L={0}$. In this case we have that $S∈P$. How do I continue from here?




You're basically right on top of the answer. You've already chosen $L'$ to be maximal (in your comment). If $Loplus L'cap S={0}$, then $Loplus L'oplus S$ is direct as well.



So $L'oplus Sin P$, not merely $S$.



Now do you see the contradiction?






share|cite|improve this answer









$endgroup$




then there are two options: $S∩L≠{0}$ (in this case $L⊕L′∩S≠{0}$) and $S∩L={0}$. In this case we have that $S∈P$. How do I continue from here?




You're basically right on top of the answer. You've already chosen $L'$ to be maximal (in your comment). If $Loplus L'cap S={0}$, then $Loplus L'oplus S$ is direct as well.



So $L'oplus Sin P$, not merely $S$.



Now do you see the contradiction?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 21 '18 at 14:20









rschwiebrschwieb

108k12104253




108k12104253












  • $begingroup$
    Then $L' oplus S= L'$, by the maximality of $L'$ and we're done.
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 14:29












  • $begingroup$
    @J.Bosser Well, if by "done" you mean "we have reached a contradiction", then yes.
    $endgroup$
    – rschwieb
    Dec 21 '18 at 14:32










  • $begingroup$
    I have a more general question. The definition of a direct sum of two ideals is clear to me. But if one writes $oplus_{i in I}J_i$, a 'general' direct sum of ideals, does this mean that $J_i cap J_j = lbrace 0 rbrace$, for $i,j in I$?
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 14:32












  • $begingroup$
    @J.Bosser NO, it is much stronger than that. It means that $(sum_{ineq j} I_j)cap I_j={0}$ for every index $j$. It is easy to formulate a counterexample in $mathbb R^2$ that has the property you mention, but which is obviously not a direct sum.
    $endgroup$
    – rschwieb
    Dec 21 '18 at 14:33












  • $begingroup$
    Okay, thanks for the clarification!
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 14:37


















  • $begingroup$
    Then $L' oplus S= L'$, by the maximality of $L'$ and we're done.
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 14:29












  • $begingroup$
    @J.Bosser Well, if by "done" you mean "we have reached a contradiction", then yes.
    $endgroup$
    – rschwieb
    Dec 21 '18 at 14:32










  • $begingroup$
    I have a more general question. The definition of a direct sum of two ideals is clear to me. But if one writes $oplus_{i in I}J_i$, a 'general' direct sum of ideals, does this mean that $J_i cap J_j = lbrace 0 rbrace$, for $i,j in I$?
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 14:32












  • $begingroup$
    @J.Bosser NO, it is much stronger than that. It means that $(sum_{ineq j} I_j)cap I_j={0}$ for every index $j$. It is easy to formulate a counterexample in $mathbb R^2$ that has the property you mention, but which is obviously not a direct sum.
    $endgroup$
    – rschwieb
    Dec 21 '18 at 14:33












  • $begingroup$
    Okay, thanks for the clarification!
    $endgroup$
    – J.Bosser
    Dec 21 '18 at 14:37
















$begingroup$
Then $L' oplus S= L'$, by the maximality of $L'$ and we're done.
$endgroup$
– J.Bosser
Dec 21 '18 at 14:29






$begingroup$
Then $L' oplus S= L'$, by the maximality of $L'$ and we're done.
$endgroup$
– J.Bosser
Dec 21 '18 at 14:29














$begingroup$
@J.Bosser Well, if by "done" you mean "we have reached a contradiction", then yes.
$endgroup$
– rschwieb
Dec 21 '18 at 14:32




$begingroup$
@J.Bosser Well, if by "done" you mean "we have reached a contradiction", then yes.
$endgroup$
– rschwieb
Dec 21 '18 at 14:32












$begingroup$
I have a more general question. The definition of a direct sum of two ideals is clear to me. But if one writes $oplus_{i in I}J_i$, a 'general' direct sum of ideals, does this mean that $J_i cap J_j = lbrace 0 rbrace$, for $i,j in I$?
$endgroup$
– J.Bosser
Dec 21 '18 at 14:32






$begingroup$
I have a more general question. The definition of a direct sum of two ideals is clear to me. But if one writes $oplus_{i in I}J_i$, a 'general' direct sum of ideals, does this mean that $J_i cap J_j = lbrace 0 rbrace$, for $i,j in I$?
$endgroup$
– J.Bosser
Dec 21 '18 at 14:32














$begingroup$
@J.Bosser NO, it is much stronger than that. It means that $(sum_{ineq j} I_j)cap I_j={0}$ for every index $j$. It is easy to formulate a counterexample in $mathbb R^2$ that has the property you mention, but which is obviously not a direct sum.
$endgroup$
– rschwieb
Dec 21 '18 at 14:33






$begingroup$
@J.Bosser NO, it is much stronger than that. It means that $(sum_{ineq j} I_j)cap I_j={0}$ for every index $j$. It is easy to formulate a counterexample in $mathbb R^2$ that has the property you mention, but which is obviously not a direct sum.
$endgroup$
– rschwieb
Dec 21 '18 at 14:33














$begingroup$
Okay, thanks for the clarification!
$endgroup$
– J.Bosser
Dec 21 '18 at 14:37




$begingroup$
Okay, thanks for the clarification!
$endgroup$
– J.Bosser
Dec 21 '18 at 14:37


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3048433%2ffinding-essential-ideal-in-a-ring-r%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

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
Read more

Puebla de Zaragoza

Image
Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
Read more

Musa

Image
Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
Read more