[Math] Suppose that $I$ is an ideal of $J$ and that $J$ is an ideal of $R$. prove that if $I$ has a unity then $I$ is an ideal of $R$

abstract-algebraidealsring-theory

Suppose that $I$ is an ideal of $J$ and that $J$ is an ideal of $R$. prove that if $I$ has a unity then $I$ is an ideal of $R$

Attempt:

Given that $I$ is an ideal of $J$ which means :

$(i)~~i_1-i_2 \in I ~~\forall~~i_1,i_2 \in I$

$(ii)~~ji \in I,~~ij \in I~~\forall~~i \in I,\forall~~j \in I$

Given that $J$ is an ideal of $R$ which means :

$(iii)~~j_1-j_2 \in J ~~\forall~~j_1,j_2 \in I$

$(iv)~~rj \in J,~~jr \in J~~\forall~~j \in J,\forall~~r \in R$

We need to prove that $I$ is an ideal of $R$ which means :

$(v)~~i_1-i_2 \in I ~~\forall~~i_1,i_2 \in I$

$(vi)~~ri \in I,~~ir \in I~~\forall~~i \in I,\forall~~r \in R$

The first condition is easily realized from $(i)$. Now to realize the second condition :

from $(ii),(iv),~~ji~rj \in I \implies i~'rj \in I $

From here, how do I prove that $~~ri \in I,~~ir \in I~~\forall~~i \in I,\forall~~r \in R$

Thank you for your help …

Best Answer

Let $x\in I$ and $r \in R$ and $e$ the unity of $I$ then $re \in J$ and so $rex \in I$ but $rex=rx$