Proof. We will use that $\bigcap \mathfrak m^ n = 0$ by Lemma 10.51.4. Let $f, g \in R$ such that $fg = 0$. Suppose that $f \in \mathfrak m^ a$ and $g \in \mathfrak m^ b$, with $a, b$ maximal. Since $fg = 0 \in \mathfrak m^{a + b + 1}$ we see from the result of Lemma 10.106.1 that either $f \in \mathfrak m^{a + 1}$ or $g \in \mathfrak m^{b + 1}$. Contradiction. $\square$

There are also:

• 4 comment(s) on Section 10.106: Regular local rings

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).