Let $(R, \mathfrak m)$ be a local Cohen-Macaulay ring. How to show that $R$ has minimal multiplicity (i.e. $e(R)=\mu (\mathfrak m)- \dim R +1$ ) if and only if $\mathfrak m^2=(\overline x)\mathfrak m$ for some regular sequence $\overline x$ in $R$ ?
If this is not true in general is it at least true if we assume $R/\mathfrak m$ is infinite ?
I do not know if it is true when the residue field is finite, but let me show you why it holds if we assume that $R/\mathfrak{m}$ is infinite.
First note that the condition $\mathfrak{m}^2=(\overline{x})\mathfrak{m}$ for some regular sequence $\overline{x}$ is equivalent to say that $\mathfrak{m}^2=I\mathfrak{m}$ for some minimal reduction $I$ of $\mathfrak{m}$, because a minimal reduction is generated by $\dim R$ elements that are a regular sequence. Consider a minimal reduction $I$ of $\mathfrak{m}$. The length of the $R$-module $R/\mathfrak{m}I$ can be computed in these two ways: \begin{align} \ell_R(R/\mathfrak{m}I)&=\ell_R(R/\mathfrak{m})+\ell_R(\mathfrak{m}/\mathfrak{m}^2)+ \ell_R(\mathfrak{m^2}/\mathfrak{m}I)=1+\mu(\mathfrak{m})+\ell_R(\mathfrak{m^2}/\mathfrak{m}I), \\ \ell_R(R/\mathfrak{m}I)&= \ell_R(R/I)+ \ell_R(I/\mathfrak{m}I)=e(R)+ \dim(R). \end{align} Hence, it follows that $\mu(\mathfrak{m})-\dim(R)+1+\ell_R(\mathfrak{m}^2/\mathfrak{m}I)=e(R)$ and $R$ has minimal multiplicity if and only if $\mathfrak{m}^2=\mathfrak{m} I$.
If the residue field is finite and $\mathfrak{m}^2=(\overline{x})\mathfrak{m}$ for some regular sequence $\overline{x}$, the proof above shows that $R$ has minimal multiplicity, but I do not believe that the converse is true.
