Let $(A, \mathfrak{m}, k)$ be a Noetherian local ring of dimension $d$. I would like to prove, or rather understand, why the following holds:
If the associated graded ring $\text{gr}_{\mathfrak{m}}(A)$ is isomorphic to $k[t_1, \ldots, t_d]$, then $\text{dim}_{k}(\mathfrak{m}/\mathfrak{m}^2)=d$ (see, for example, Atiyah-MacDonald 11.22).
My thoughts: The given isomorphism of graded rings, call it $\varphi$, induces a group isomorphism $\mathfrak{m}/\mathfrak{m}^2 \cong (t_1,\ldots,t_d)/(t_1,\ldots,t_d)^2$. If we could conclude that $\mathfrak{m}/\mathfrak{m}^2$ and $(t_1,\ldots,t_d)/(t_1,\ldots,t_d)^2$ are isomorphic as $k$-vector spaces, we would be done. However, I don't see that it is necessary for $\varphi$ to be $k$-linear, i.e. $\varphi(a+\mathfrak{m})=a+\mathfrak{m} \in A/\mathfrak{m}=k[t_1,\ldots,t_d]/(t_1,\ldots,t_d)$.
The degree zero part of the isomorphism $\varphi$ gives you an isomorphism $\varphi_0:k\cong A/\mathfrak{m}$. Without loss of generality, you may assume this isomorphism is the identity (otherwise, just compose $\varphi$ with the automorphism $k[t_1,\dots,t_d]\to k[t_1,\dots,t_d]$ that applies $\varphi_0^{-1}$ to every coefficient, and now you have an isomorphism $\varphi'$ whose degree $0$ part is the identity). The $k$-vector space structure on the degree $1$ part of $k[t_1,\dots,t_d]$ is just induced by the multiplication by elements in the degree $0$ part of the ring $k[t_1,\dots,t_d]$. Similarly the $k$-vector space structure on $\mathfrak{m}/\mathfrak{m}^2$ coincides with the multiplication by degree $0$ elements in the graded ring $\text{gr}_{\mathfrak{m}}(A)$. Since $\varphi$ is a ring homomorphism and is the identity on the degree $0$ part, it follows that $\varphi$ preserves this $k$-vector space structure.