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Ideal of an ideal being an ideal itself
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Take a commutative ring with unity (a.k.a. ring) $R$ and let $I$ be an ideal in $R$. Are the following true??
a) If $J$ is an ideal in $I$ (observed as a ring) , $J$ is not ideal in $R$(if this is true, please give me example)
b) If $J$ is a maximal ideal in $I$, then $J$ is ideal in $R$
c) If $J$ is ideal in $I$, and $I$ is maximal ideal in $R$, then $J$ is ideal in $R$
commutative-algebra
asked Jul 16 at 8:35
nikola
557214
add a comment |Â
up vote
1
down vote
favorite
Take a commutative ring with unity (a.k.a. ring) $R$ and let $I$ be an ideal in $R$. Are the following true??
a) If $J$ is an ideal in $I$ (observed as a ring) , $J$ is not ideal in $R$(if this is true, please give me example)
b) If $J$ is a maximal ideal in $I$, then $J$ is ideal in $R$
c) If $J$ is ideal in $I$, and $I$ is maximal ideal in $R$, then $J$ is ideal in $R$
commutative-algebra
asked Jul 16 at 8:35
nikola
557214
2
What do you mean by ring? Some books insist rings contain "1", in which case no proper ideal will be a ring
– mathworker21
Jul 16 at 8:37
Some insist on that and others don't. Also, a subring can have an identity which is not an identity of the whole ring. Consider the 2x2 matrices. Those which are zero except for the top left element are a subring. This does not include the identity of the whole ring but it does have an identity. It is not an ideal though.
– badjohn
Jul 16 at 10:49
Another example worth considering is $mathbbZ times mathbbZ$.
– badjohn
Jul 16 at 11:20
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Take a commutative ring with unity (a.k.a. ring) $R$ and let $I$ be an ideal in $R$. Are the following true??
a) If $J$ is an ideal in $I$ (observed as a ring) , $J$ is not ideal in $R$(if this is true, please give me example)
b) If $J$ is a maximal ideal in $I$, then $J$ is ideal in $R$
c) If $J$ is ideal in $I$, and $I$ is maximal ideal in $R$, then $J$ is ideal in $R$
commutative-algebra
asked Jul 16 at 8:35
nikola
557214
Take a commutative ring with unity (a.k.a. ring) $R$ and let $I$ be an ideal in $R$. Are the following true??
a) If $J$ is an ideal in $I$ (observed as a ring) , $J$ is not ideal in $R$(if this is true, please give me example)
b) If $J$ is a maximal ideal in $I$, then $J$ is ideal in $R$
c) If $J$ is ideal in $I$, and $I$ is maximal ideal in $R$, then $J$ is ideal in $R$
commutative-algebra
asked Jul 16 at 8:35
nikola
557214
asked Jul 16 at 8:35
nikola
557214
asked Jul 16 at 8:35
nikola
557214
asked Jul 16 at 8:35
nikola
557214
557214
2
What do you mean by ring? Some books insist rings contain "1", in which case no proper ideal will be a ring
– mathworker21
Jul 16 at 8:37
Some insist on that and others don't. Also, a subring can have an identity which is not an identity of the whole ring. Consider the 2x2 matrices. Those which are zero except for the top left element are a subring. This does not include the identity of the whole ring but it does have an identity. It is not an ideal though.
– badjohn
Jul 16 at 10:49
Another example worth considering is $mathbbZ times mathbbZ$.
– badjohn
Jul 16 at 11:20
add a comment |Â
2
What do you mean by ring? Some books insist rings contain "1", in which case no proper ideal will be a ring
– mathworker21
Jul 16 at 8:37
Some insist on that and others don't. Also, a subring can have an identity which is not an identity of the whole ring. Consider the 2x2 matrices. Those which are zero except for the top left element are a subring. This does not include the identity of the whole ring but it does have an identity. It is not an ideal though.
– badjohn
Jul 16 at 10:49
Another example worth considering is $mathbbZ times mathbbZ$.
– badjohn
Jul 16 at 11:20
2
2
What do you mean by ring? Some books insist rings contain "1", in which case no proper ideal will be a ring
– mathworker21
Jul 16 at 8:37
What do you mean by ring? Some books insist rings contain "1", in which case no proper ideal will be a ring
– mathworker21
Jul 16 at 8:37
Some insist on that and others don't. Also, a subring can have an identity which is not an identity of the whole ring. Consider the 2x2 matrices. Those which are zero except for the top left element are a subring. This does not include the identity of the whole ring but it does have an identity. It is not an ideal though.
– badjohn
Jul 16 at 10:49
Some insist on that and others don't. Also, a subring can have an identity which is not an identity of the whole ring. Consider the 2x2 matrices. Those which are zero except for the top left element are a subring. This does not include the identity of the whole ring but it does have an identity. It is not an ideal though.
– badjohn
Jul 16 at 10:49
Another example worth considering is $mathbbZ times mathbbZ$.
– badjohn
Jul 16 at 11:20
Another example worth considering is $mathbbZ times mathbbZ$.
– badjohn
Jul 16 at 11:20
add a comment |Â
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2
What do you mean by ring? Some books insist rings contain "1", in which case no proper ideal will be a ring
– mathworker21
Jul 16 at 8:37
Some insist on that and others don't. Also, a subring can have an identity which is not an identity of the whole ring. Consider the 2x2 matrices. Those which are zero except for the top left element are a subring. This does not include the identity of the whole ring but it does have an identity. It is not an ideal though.
– badjohn
Jul 16 at 10:49
Another example worth considering is $mathbbZ times mathbbZ$.
– badjohn
Jul 16 at 11:20