Let $R$ be a regular local ring with maximal ideal $\mathfrak{m}$. It is known that any elements $r_1, \ldots, r_d$ of $\mathfrak{m}$ which map to a basis for $\mathfrak{m}/\mathfrak{m}^2$ as an $R/\mathfrak{m}$-vector space form an $R$-regular sequence.
I am interested in a somehow reverse question:
If $r_1,\ldots,r_d \in \mathfrak{m}$ form an $R$-regular sequence, when do they map to a basis of $\mathfrak{m}/\mathfrak{m}^2$?
Obviously one needs $r_1,\ldots,r_d \not\in \mathfrak{m}^2$, but I am not sure if this is sufficient this is not sufficient as Mohan pointed out.
I am looking for a condition along the lines of "the ideal generated by $(r_1,\ldots,r_d)$ has to be $\mathfrak{m}$-primary and ...".
Edit: Mohan's comment suggests that one has to look at the ideal generated by $(r_1,\ldots,r_d)$ and his example suggests that this ideal has to be radical. Together with being $\mathfrak{m}$-primary, this already implies $(r_1,\ldots,r_d) = \mathfrak{m}$.