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?
Best Answer
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 $b=0$