Proving an ideal of a unital commutative ring

Question

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.

Answer 1

I can't think of any easier way to prove the statement, but the proof should not require very much effort (which is why the author was so casual about it). First, I claim that it is very obvious that $N_R(U/T)$ is an additive subgroup of $R$:

1. $U0=0\subset T$

2. If $Ur\subset T$, then $U(-r)\subset T$

3. If $Ur, Us \subset T$, then $U(r+s)=Ur+Us\subset T+T=T$

Now, to the second point, if $a\in R$ and $r\in N_r(U/T)$, then $U(ra)=(Ur)a\subset Ta\subset T$.

Answer 2

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.)