Every regular ideal contained in a maximal regular ideal?
$begingroup$
Let $A$ be a (not necessarily unital)commutative ring. We define a regular ideal $I$ if exists element $e in A$, such that $e$ is the unit in $A/I$.
Can we deduce, that every proper regular ideal is contained in a maximal regular ideal?
It seems that this is another Zorn's lemma argument: let $m_i$ be a chain of proper, regular ideals then $bigcup m_i$ is a proper ideal, but is it regular?
abstract-algebra ring-theory operator-theory
asked Dec 6 '18 at 9:17
CL.CL.
2,1992822
$endgroup$
add a comment |
$begingroup$
Let $A$ be a (not necessarily unital)commutative ring. We define a regular ideal $I$ if exists element $e in A$, such that $e$ is the unit in $A/I$.
Can we deduce, that every proper regular ideal is contained in a maximal regular ideal?
It seems that this is another Zorn's lemma argument: let $m_i$ be a chain of proper, regular ideals then $bigcup m_i$ is a proper ideal, but is it regular?
abstract-algebra ring-theory operator-theory
asked Dec 6 '18 at 9:17
CL.CL.
2,1992822
$endgroup$
add a comment |
$begingroup$
Let $A$ be a (not necessarily unital)commutative ring. We define a regular ideal $I$ if exists element $e in A$, such that $e$ is the unit in $A/I$.
Can we deduce, that every proper regular ideal is contained in a maximal regular ideal?
It seems that this is another Zorn's lemma argument: let $m_i$ be a chain of proper, regular ideals then $bigcup m_i$ is a proper ideal, but is it regular?
abstract-algebra ring-theory operator-theory
asked Dec 6 '18 at 9:17
CL.CL.
2,1992822
$endgroup$
Let $A$ be a (not necessarily unital)commutative ring. We define a regular ideal $I$ if exists element $e in A$, such that $e$ is the unit in $A/I$.
Can we deduce, that every proper regular ideal is contained in a maximal regular ideal?
It seems that this is another Zorn's lemma argument: let $m_i$ be a chain of proper, regular ideals then $bigcup m_i$ is a proper ideal, but is it regular?
abstract-algebra ring-theory operator-theory
abstract-algebra ring-theory operator-theory
asked Dec 6 '18 at 9:17
CL.CL.
2,1992822
asked Dec 6 '18 at 9:17
CL.CL.
2,1992822
asked Dec 6 '18 at 9:17
CL.CL.
2,1992822
asked Dec 6 '18 at 9:17
CL.CL.
2,1992822
asked Dec 6 '18 at 9:17
CL.CL.
2,1992822
2,1992822
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The set of ideals of $A/I$ is in bijection with the set of ideals of $A$ which contain $I$. Since $A/I$ is unital, there exists a maximal ideal in $A/I$ and this corresponds to a maximal ideal $J$ of $A$ containing $I$. Note that if $e in A$ is such that $e + I in A/I$ is the unit, then $e + J in A/J cong (A/I)/(J/I)$ is the unit, too. In fact, we have $ea - a, ae - a in I subseteq J$ for all $a in A$. Note that this shows that every ideal containing a regular ideal is already regular.
answered Dec 6 '18 at 9:29
Matthias KlupschMatthias Klupsch
6,1591227
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add a comment |
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1 Answer
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$begingroup$
The set of ideals of $A/I$ is in bijection with the set of ideals of $A$ which contain $I$. Since $A/I$ is unital, there exists a maximal ideal in $A/I$ and this corresponds to a maximal ideal $J$ of $A$ containing $I$. Note that if $e in A$ is such that $e + I in A/I$ is the unit, then $e + J in A/J cong (A/I)/(J/I)$ is the unit, too. In fact, we have $ea - a, ae - a in I subseteq J$ for all $a in A$. Note that this shows that every ideal containing a regular ideal is already regular.
answered Dec 6 '18 at 9:29
Matthias KlupschMatthias Klupsch
6,1591227
$endgroup$
add a comment |
$begingroup$
The set of ideals of $A/I$ is in bijection with the set of ideals of $A$ which contain $I$. Since $A/I$ is unital, there exists a maximal ideal in $A/I$ and this corresponds to a maximal ideal $J$ of $A$ containing $I$. Note that if $e in A$ is such that $e + I in A/I$ is the unit, then $e + J in A/J cong (A/I)/(J/I)$ is the unit, too. In fact, we have $ea - a, ae - a in I subseteq J$ for all $a in A$. Note that this shows that every ideal containing a regular ideal is already regular.
answered Dec 6 '18 at 9:29
Matthias KlupschMatthias Klupsch
6,1591227
$endgroup$
add a comment |
$begingroup$
The set of ideals of $A/I$ is in bijection with the set of ideals of $A$ which contain $I$. Since $A/I$ is unital, there exists a maximal ideal in $A/I$ and this corresponds to a maximal ideal $J$ of $A$ containing $I$. Note that if $e in A$ is such that $e + I in A/I$ is the unit, then $e + J in A/J cong (A/I)/(J/I)$ is the unit, too. In fact, we have $ea - a, ae - a in I subseteq J$ for all $a in A$. Note that this shows that every ideal containing a regular ideal is already regular.
answered Dec 6 '18 at 9:29
Matthias KlupschMatthias Klupsch
6,1591227
$endgroup$
The set of ideals of $A/I$ is in bijection with the set of ideals of $A$ which contain $I$. Since $A/I$ is unital, there exists a maximal ideal in $A/I$ and this corresponds to a maximal ideal $J$ of $A$ containing $I$. Note that if $e in A$ is such that $e + I in A/I$ is the unit, then $e + J in A/J cong (A/I)/(J/I)$ is the unit, too. In fact, we have $ea - a, ae - a in I subseteq J$ for all $a in A$. Note that this shows that every ideal containing a regular ideal is already regular.
answered Dec 6 '18 at 9:29
Matthias KlupschMatthias Klupsch
6,1591227
answered Dec 6 '18 at 9:29
Matthias KlupschMatthias Klupsch
6,1591227
answered Dec 6 '18 at 9:29
Matthias KlupschMatthias Klupsch
6,1591227
answered Dec 6 '18 at 9:29
Matthias KlupschMatthias Klupsch
6,1591227
6,1591227
add a comment |
add a comment |
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