Why is $(J : I ) = \{x \in R\mid xI \subset J \}$ an ideal?

Question

Let $R$ be a commutative ring with $1$. If $I , J \lhd R$ then we define $(J : I ) = \{x \in R\mid xI \subset J \}$.

Please help me to show that $$(J : I )\lhd R$$

Thanks in advance.

Answer 1

Some details for the last property:

If $y\in (J:I)$, and $x\in R$, then $xy\in (J:I)$.

By definition, for all $i\in I$, $y\mkern2mui\in J$, and we must show the same property is satisfied by $xy$. Now $$(xy)i=x(yi)$$ lies in $J$ because $y\mkern2mui$ lies in $J$ by the hypothesis on $y$, and $J$ is an ideal.

Edit:

To prove $\;\operatorname{Hom}(R/I, R/J)\simeq (J:I)/J$, you can prove this more general following assertion: for any $R$-module $M$, $$\operatorname{Hom}(R/I, M)\simeq(0:_M I)\qquad (\text{the annihilator of $I$ in $M$)}$$ Indeed, a homomorphism from $R/I$ to $M$ corresponds to a homomorphism $\varphi:R\to M$ which vanishes on $I$, and such a $\varphi$ is uniquely determined by the value of $\;\varphi(1)$.