5

Let $A$ be a commutative regular local ring of dimension $d$ with maximal ideal $\mathcal m$ and $a \in A$ an element of the ring.

Suppose that $\mathcal m \cdot a \subset \mathcal m^2$, i.e. if I multiply the element $a$ by an arbitrary element of $\mathcal m$, then I am in the square ideal of $\mathcal m$.

Can I conclude from this that already $a \in \mathcal m$?

Cyril
  • 1,483
  • As Michael shows, you don't need regular, this is true for any local ring since any element in $A- \mathfrak m$ is a unit. – Parsa Apr 19 '12 at 22:17
  • @Parsa, Not for any local ring, at least, we need $\mathfrak{m}$ is finitely generated or something such that $\mathfrak{m}\subset \mathfrak{m}^2$ is impossible. – wxu Apr 20 '12 at 00:30

1 Answers1

3

If $a \notin \mathfrak{m}$, then $a$ must be a unit, so you would have $\mathfrak{m} \subset \mathfrak{m}^2$, which is not possible. So $a \in \mathfrak{m}$.

Edit: As pointed out by Matt E., this argument holds unless $\mathfrak{m} = 0$, in which case we would have $\mathfrak{m} = \mathfrak{m}^2$.

Michael Joyce
  • 14,126