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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 $a \in \mathcal m$?

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    @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.2012-04-20

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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 $\mathfrak{m} = \mathfrak{m}^2$.

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    An interesting question would be if $\mathcal m \cdot a \subset \mathcal m^{n+1}$ implies $a \in \mathcal m^n$ for all $n$.2012-04-20