Nakayama's lemma corollary I am just self-studying Justin Smith algebraic geometry. I have the following question. Consider the following lemma.

Let $m\subset R$ be a maximal ideal of a noetherian ring $R$ or an arbitrary ideal of a noetherian domain. Then
$$\bigcap_{j = 1}^{\infty} m^j = (0).$$

I just have a small question. In the proof they call the intersection $b$. Since $R$ is noetherian, $b$ is finitely generated as a module over $R$. Since
$$mb = b$$
Nakayama's Lemma implies...
Now I am confused how come we have $mb = b$? Can someone explain this?
 A: One direction is clear, that is, $mb \subseteq b$. So, we'll have to prove the other direction. The other direction is slightly tricky.
By the Artin-Rees Lemma, for every $n>0$, there exists some $k>0$ such that
$m^{k} \cap b \subseteq m^{n}b$
Choose $n=1$.
Now, note that $b \subseteq m^{k}$. This implies that $b \subseteq mb$ which means that $b=mb$
$m$ is maximal and $A$ is local, so using Nakayama's Lemma implies that $b=0$
A: @Shivering Soldier has already posted a link to a very nice direct proof of this fact using primary decomposition, and @RithvikReddy has already written a proof using the Artin-Rees lemma (+1), but here's another alternative proof.

Let $R$ be any (commutative) Noetherian ring and let $I$ be an ideal of $R$. We will show that $IJ=J$, where $J=\bigcap_{n=1}^\infty I^n$.

To see this, note that the set of ideals of $R$ whose intersection with $J$ is $IJ$ is non-empty, as it contains $IJ$. So, since $R$ is Noetherian, we may find an ideal $K\leqslant R$ maximal subject to $K\cap J=IJ$.
Claim: For any $x\in I$, there exists $k\in\mathbb{N}$ such that $x^k\in K$. To see this, consider the following ascending chain of ideals $$(K:x)\subseteq(K:x^2)\subseteq(K:x^3)\subseteq\dots.$$ Since $R$ is Noetherian, this chain must stabilize; let $k$ be such that $(K:x^l)=(K:x^k)$ for all $l\geqslant k$. Now we claim $x^k\in K$; otherwise, $K':=K+(x^k)$ is an ideal strictly containing $K$, and so by maximality of $K$ we have $K'\cap J\supsetneq IJ$, and so there exists $a\in K$ and $\lambda\in R$ such that $a+\lambda x^k\in J\setminus IJ$. In particular, note that $$xa+\lambda x^{k+1}=x(a+\lambda x^k)\in IJ\subseteq K,$$ so in fact, since $a\in K$, also $\lambda x^{k+1}\in K$, ie $\lambda\in (K:x^{k+1})$. But $(K:x^{k+1})=(K:x^k)$ by hypothesis, so in fact $\lambda x^k\in K$, whence $a+\lambda x^k\in K\cap J=IJ$, a contradiction. Thus indeed $x^k\in I$ and so we are done. $\blacksquare$
Now, since $R$ is Noetherian, $I$ is finitely generated, and so, by applying the claim to each of the generators of $I$ and taking the sum of the exponents, we can find $n\in\mathbb{N}$ such that $I^n\subseteq K$. Thus $I^n\cap J\subseteq K\cap J=IJ$. On the other hand, $J\subseteq I^n$ and so $I^n\cap J=J$, so we in fact have $J\subseteq IJ$ and thus we are done.
