# Ideal is different from its square

Assume $m<R$ is a non-zero ideal of the commutative ring $R$. When is $m\neq m^2$? Would it suffice if $m$ was a prime ideal?

Context: I am trying to prove that if the dimension of a variety $X$ is $d\geq 1$, then the Zariski tangent space to any point is non-zero. But the tangent space is isomorphic to the factor $m/m^2$ of the maximal ideal $m$ of the local coordinate ring $O$ by its square $m^2$. Since $d\geq1$, $m$ is non-zero, but still why is $m\neq m^2$?

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One version of the Nakayama lemma says that for a local ring $A$ with maximal ideal $\mathfrak m$ and finitely generated module $M$, if $\mathfrak mM = M$ then $M = 0$. You'd like to apply this to the module $M = \mathfrak m$ so what you want is that $\mathfrak m$ is a finitely generated ideal. If $A$ is, for example, Noetherian (as any localization of a quotient of a polynomial ring will be) then this is true.