TITLE: Proving an ideal of a unital commutative ring
QUESTION [1 upvotes]: I am studying a writing on noetherian domain of dimension 1, although the question I am having now is not directly related to noetherian domain. The very first paragraph of this chapter begins like this:

Let $R$ be a commutative ring with 1, and let $T$ and $U$ be ideals of $R$. We define $$N_R(U/T) := \{r \in R \mid Ur \subseteq T\}.$$ Note that $N_R(U/T)$ is an ideal of $R$ and that $$T \subseteq N_R(U/T) \subseteq N_R(U^2/T)\subseteq N_R(U^3/T)\subseteq N_R(U^4/T) \subseteq \ ...$$

Notice the author very casually asserts that "$N_R(U/T)$ is an ideal of $R$." I would like to pick up the assertion's proof as part of my writing assignment. To prove it, I think I need to follow these steps:
(1) First I need to prove that $N_R(U/T)$ is an additive subgroup of $R$ by showing that (a) the identity element $0$ belongs to $N_R(U/T)$, and (b) each element has additive inverse and finally (c) the addition of two elements is closed under $N_R(U/T)$.
(2) That $N_R(U/T)$ absorbs multiplication by any element from $R.$
And here is my only question: Is this the only way to prove the above assertion? I suspect there is a shorter and more efficient way of proving it, judging from how casual the author makes the statement. Thank you for your time and effort.

REPLY [1 votes]: Another way to look at it is that $N_R(U/T)$ is the annihilator of the $R$-module $\frac{U+T}{T}$. 
This is useful if you believe the annihilator of an $R$-module is always an ideal in $R$ (which is an easy exercise you may or may not have already done.)
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