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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 $R$

nikola
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    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 '18 at 08:37
  • 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 '18 at 10:49
  • Another example worth considering is $\mathbb{Z} \times \mathbb{Z}$. – badjohn Jul 16 '18 at 11:20
  • Consider $R=\mathbb{Z}$ and $I=(2)$, $J=(4)$, then $J$ is an ideal in $I$ and $J$ is an ideal in $\mathbb{Z}$, therefore a) is false – Gabriel Canedo Apr 28 '21 at 05:06
  • Observe if b) is true then a) is false. – Gabriel Canedo Apr 28 '21 at 05:08

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