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 $m\neq m^2$?