= Solution
Choose $a_1,\ldots,a_d\in\mathfrak m$ whose classes form a <basis> of the <cotangent space of a local ring> $\mathfrak m/\mathfrak m^2$, and put $J=(a_1,\ldots,a_d)$. Spanning gives $\mathfrak m=J+\mathfrak m^2$. Thus the <quotient module> $N=\mathfrak m/J$ satisfies $N=\mathfrak mN$. It is a <finitely generated module>, since $A$ is a <Noetherian ring>.
Here is the needed form of the <Nakayama lemma>, with proof. Choose generators $n_1,\ldots,n_r$ of $N$ and write $n_i=\sum_jc_{ij}n_j$ with $c_{ij}\in\mathfrak m$. For the column vector $n$, $(I-C)n=0$. Multiply by the adjugate to obtain $\det(I-C)n=0$. Since $\det(I-C)\equiv1\pmod{\mathfrak m}$, this determinant is a <unit> in the <local ring> $A$, so every $n_i=0$. Therefore $N=0$, proving $\mathfrak m=J$. Conversely, any generators of $\mathfrak m$ must span its cotangent space, so at least $d$ are needed. Hence
$$
\boxed{\mu(\mathfrak m)=d,}
$$
Here $\mu(M)$ denotes the <minimal number of generators of a module> $M$.
For the additional minimum, the <maximal ideal> $\mathfrak m$ is itself $\mathfrak m$-primary, so it supplies an eligible <ideal> generated by $d$ elements. Consequently
$$
\boxed{\min\{\mu(Q):Q\text{ is }\mathfrak m\text{-primary}\}\leq d.}
$$
The minimum here is over <primary ideals>. \b[A bound $\mu(Q)\leq d$ for every $\mathfrak m$-primary <ideal> would be false.] Indeed, for $A=k[[u,v]]$ and $\mathfrak m=(u,v)$, $d=2$ but $Q=\mathfrak m^2=(u^2,uv,v^2)$ is primary and needs three generators. To verify primaryness, if $ab\in Q$ and $a\notin Q$, then $b$ cannot be a <unit>, so $b\in\mathfrak m$ and $b^2\in Q$. The three displayed monomials form a <basis> of $Q/\mathfrak mQ=\mathfrak m^2/\mathfrak m^3$, proving the lower bound of three, while they already generate $Q$. This illustrates that <primary ideals can need more generators than the maximal ideal>.
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