Let $A$, $B$, $m$ be ideals in a commutative ring $R$. Is $m \cdot (A \cap B) = m \cdot A \cap m \cdot B$?
Let $(A_{\alpha})_{\alpha \in I}$ be a set of ideals. Is $m \cdot \bigcap_I A_{\alpha} = \bigcap_I m \cdot A_{\alpha}$?
I'm trying to prove that in a local Noetherian commutative ring $R$, the intersection $M = \bigcap_n m^n$ of powers of the maximal ideal is $\{0 \}$. My technique of proof was using Nakayama's lemma and the fact that $m \cdot M = M$; proving the non-trivial intersection of the last equality using
\begin{align*} m \cdot M &= m \cdot \bigcap_n m^n \\
&= \bigcap_1^{\infty} m \cdot m^n \\
&= \bigcap_2^{\infty} m^n \\
&\supseteq M.
\end{align*}
However, I am unsure if I am allowed to use this operation on intersections and products of ideals.
Best Answer
No, ideal product doesn't usually distribute over intersection. You'll need something like the Artin-Rees lemma (see [Stacks 00IN]) to prove that a local Noetherian ring $(R, \mathfrak{m})$ is $\mathfrak{m}$-adically separated.
In your notation, when $m$ is a flat ideal and $A_\alpha$ is a finite set of ideals, then multiplication by $m$ does distribute over the intersection $\bigcap A_\alpha$ (more generally its true if the family $A_\alpha$ is in an appropriate sense compact). For a reference, see e.g. [Stacks 0BBY].
When we don't require that $m$ is flat, this can really fail, even when $m$ is the maximal ideal of a local Noetherian ring.
As an example, consider the ring $k[x,y]$ with $k$ a field and the maximal ideal $\mathfrak{m} = (x,y)$. Let $R = k[x,y]_\mathfrak{m}$. This is a local Noetherian ring with maximal ideal $(x,y)$. The ideals $m = (x,y)$, $A = (x), B = (y)$ satisfy the following identies:
$$mA \cap mB = (xy)$$ $$m(A \cap B) = xy(x,y)$$
As a closing remark, let's note the maximal ideal being flat is a big deal. Observe that in a Noetherian local ring a flat ideal is automatically free (Noetherian implies f.g. flat, and local implies f.g. flat is free), and in general an ideal is free iff it is principally generated by a non-zero divisor.
So in a Noetherian local ring, the maximal ideal is flat iff it is principally generated by a non-zero divisor. This is a lot to ask for. It implies that every ideal is a power of $m$. Hence the ring is either a discrete valuation ring or a field.