Finding essential ideal in a ring $R$
$begingroup$
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
asked Dec 21 '18 at 11:49
J.BosserJ.Bosser
363210
$endgroup$
add a comment |
$begingroup$
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
asked Dec 21 '18 at 11:49
J.BosserJ.Bosser
363210
$endgroup$
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
add a comment |
$begingroup$
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
asked Dec 21 '18 at 11:49
J.BosserJ.Bosser
363210
$endgroup$
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
asked Dec 21 '18 at 11:49
J.BosserJ.Bosser
363210
asked Dec 21 '18 at 11:49
J.BosserJ.Bosser
363210
asked Dec 21 '18 at 11:49
J.BosserJ.Bosser
363210
asked Dec 21 '18 at 11:49
J.BosserJ.Bosser
363210
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
add a comment |
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
add a comment |
1 Answer
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$begingroup$
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?
answered Dec 21 '18 at 14:20
rschwiebrschwieb
108k12104253
$endgroup$
$begingroup$
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
|
show 4 more comments
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$begingroup$
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?
answered Dec 21 '18 at 14:20
rschwiebrschwieb
108k12104253
$endgroup$
$begingroup$
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
|
show 4 more comments
$begingroup$
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?
answered Dec 21 '18 at 14:20
rschwiebrschwieb
108k12104253
$endgroup$
$begingroup$
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
|
show 4 more comments
$begingroup$
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?
answered Dec 21 '18 at 14:20
rschwiebrschwieb
108k12104253
$endgroup$
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?
answered Dec 21 '18 at 14:20
rschwiebrschwieb
108k12104253
answered Dec 21 '18 at 14:20
rschwiebrschwieb
108k12104253
answered Dec 21 '18 at 14:20
rschwiebrschwieb
108k12104253
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
|
show 4 more comments
$begingroup$
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
|
show 4 more comments
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$begingroup$
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