# On a special type of ideal in local Artinian ring

Let $$(R,\mathfrak m)$$ be a local Artinian ring. If $$J$$ is a non-zero ideal of $$R$$ such that $$J^2=\mathfrak mJ$$, then is it true that $$J=\mathfrak m$$ ? or at least $$J^2=\mathfrak m^2$$ ?

NOTE: $$J^2=\mathfrak mJ\implies J^{n+1}=\mathfrak m^nJ,\forall n>1$$, and since $$\exists n_0>1$$ such that $$\mathfrak m^{n_0}=0$$, so we get $$J^n=0$$ for some $$n>1$$. But I am unable to proceed further.

No. For instance, let $$k$$ be a field and let $$R=k[x]/(x^3)$$, with maximal ideal $$\mathfrak{m}=(x)$$. Then $$J=(x^2)$$ satisfies $$J^2=\mathfrak{m}J=0$$ but $$J^2\neq \mathfrak{m}^2$$.