Let R be a commutative ring with identity and I, J be ideals in R.
$I:J$ = {$r∈R\mid rj∈I, \forall j∈J$}
Prove that $I:J$ is an ideal in $R$, and contains $I$.
To begin this, I wrote $I:J$ in the form
$I:J$ = {$r∈R\mid rJ \subset I$}
Let $r_1,r_2∈ I:J$. Clearly $r_1-r_2∈R$, and $r_1J, r_2J \subset I$.
Because $I$ is a ring,
if $r_1j, r_2j∈I$, then $(r_1 – r_2)j∈I$, so that $(r_1-r_2)J \subset I$. This implies that $I:J$ is an ideal. Is this the correct line of thinking? Furthermore, I'm unsure how to prove that $I \subset I:J$ in a general manner
Best Answer
You're working too hard, and your work is not quite right. You need to show two things to show that $I:J$ is an ideal; (1) $I:J$ is closed under addition and (2) $I:J$ is closed under multiplication by elements of the ring $R$.
For (1), suppose $r_{1}, r_{2} \in I:J$. Then for any $j \in J$, $(r_{1}+r_{2})j = r_{1}j + r_{2}j \in I$. This follows since $r_{1}j, r_{2}j \in I$ by hypothesis, and $I$ is an ideal, hence closed under addition.
I will leave (2) for you to try. A hint: you will again need to use the fact that $I$ is an ideal, but this time, you need to use the second property of ideals referenced above.
To see that $I \subset I:J$, remember that ideals are closed under multiplication by elements of the ring. Thus, for any $j \in J$ and any $i \in I$, we know that $ij = ji \in RI = I$.