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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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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 0:30

1 Answer 1

up vote 3 down vote accepted

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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Dear Michael, regarding "which is not possible" --- unless $\mathfrak m = 0$ and $A$ is a field. Regards, –  Matt E Apr 19 '12 at 23:53
    
@Matt E: Thanks! I've added this observation to the answer. –  Michael Joyce Apr 20 '12 at 2:03
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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 $n$. –  Cyril Apr 20 '12 at 7:10

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