TITLE: Question about regular local rings
QUESTION [5 upvotes]: 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 $a \in \mathcal m$?

REPLY [3 votes]: 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 $\mathfrak{m} = \mathfrak{m}^2$.