Maximal ideal contained in an ideal There is probably a simple answer to this but I can't for the life of me figure it out. 
Every ring in this question has a unit but isn't necessarily commutative.
Let $R$ be a ring and let $I$ be a left ideal of $R$. In Basic Algebra II, Jacobson defines
$$(I:R)=\{ b \in R \mid bR \subseteq I\}.$$
This is an ideal because it equals, $\text{ann}_R R/I$, the annihilator of the left $R$-module $R/I$. Because $R$ has a $1$ it follows that $(I:R)$ is contained in $I$. The author goes on to claim that $(I:R)$ contains every left ideal of $R$ properly contained in $I$, but I'm not seeing it. 
If $J$ is a left ideal of $R$ contained in $I$ why should it be that $J \subseteq (I:R)$? The "obvious" thing would be to say that that since $J$ is an ideal $JR \subseteq J \subseteq I$, but the rings here aren't necessarily commutative and $J$ is only a left ideal so this doesn't work. Is there any sort of reason this actually works given that $J$ is properly contained in $I$?
 A: Basic Algebra II, edition of 1980, page 188:

Let $I$ be a left ideal of the ring $R$. We define
  $$
(I:R)=\{b\in R\mid bR\subset I\}.\tag{9}
$$
  It is clear that if we put $M=R/I$ and we regard this as a left $R$-module then
  $$
(I:R)=\operatorname{ann}_RR/I.\tag{10}
$$
  It follows from this or $(9)$ that $(I:R)$ is an ideal. Moreover, by $(9)$, $(I:R)\subset I$ and $(I:R)$ contains every ideal of $R$ contained in $I$. In other words, $(I:R)$ is the (unique) largest ideal contained in $I$.

I don't think this has changed in later editions, because it is generally false that $(I:R)$ contains every left ideal of $R$ (properly) contained in $I$. Consider the ring $R$ of $3\times 3$ matrices over the field $F$. Then $R=I_1\oplus I_2\oplus I_3$ as a direct sum of minimal left ideals. Take $I=I_1\oplus I_2$: then $(I:R)=0$, but $I$ properly contains the left ideal $I_1$.
Note that ideal (without the adjective left or right) means “two-sided ideal” and that $\subset$ denotes inclusion (not necessarily proper).
