Let $A = \{a_0, a_1, \ldots, a_{n-1}\} \subset \mathbb{N}$ be a set of non-negative integers where we assume that $a_0 < \cdots < a_{n-1}$ and set $d := a_{n-1}$. For every $s \in \mathbb{N}$, the $s$-fold sumset of $A$, $sA$, is defined by $0A := \{0\}$ and for $s \geq 1$, $$sA := \{a_{i_1} + \cdots + a_{i_s} : 0 \leq i_1 \leq \cdots \leq i_s \leq n - 1\}.$$ Without loss of generality, we can assume that $a_0 = 0$ and $\gcd(a_1, \ldots, a_{n-1}) = 1$. When this occurs, $A$ is said to be in normal form. Consider now the points $\mathbf{a_0} = (0, d), \mathbf{a_1} = (a_1, d - a_1), \ldots, \mathbf{a_{n-1}} = (d, 0) \in\mathbb{N}^2$, the set $\mathbf{A} = \{\mathbf{a_0}, \mathbf{a_1}, \ldots, \mathbf{a_{n-1}}\}$, and the subsemigroup $\mathcal{S}$ of $\mathbb{N}^2$ generated by $\mathbf{A}$. Given an arbitrary infinite field $k$, one can associate to $A$ the projective monomial curve $\mathcal{C}_{A}$ parametrized by $\mathbf{A}$: $$\mathcal{C}_A = \{(v^d : u^{a_1}v^{d-a_1} : \cdots : u^{a_{n-2}}v^{d-a_{n-2}} : u^d) \ | \ (u:v) \in \mathbb{P}^1_k\} \subset \mathbb{P}^{n-1}_k.$$ If $A$ is in normal form, it is an algebraic curve of degree $d$ and its defining ideal $I(\mathcal{C}_A)$ is the kernel of the homomorphism of $k$-algebras $\varphi : k[x_0, \ldots, x_{n-1}] \to k[u, v]$ induced by $\varphi(x_i) = u^{a_i}v^{d-a_i}$. The ideal $I(\mathcal{C}_A)$ is homogeneous, binomial, and prime, i.e., it is a homogeneous toric ideal. Denoting by $k[A] := k[x_0, \ldots, x_{n-1}]/I(\mathcal{C}_A)$ the homogeneous coordinate ring of $\mathcal{C}_A$. If we denote $\mathrm{HF}_{A}$ (respectively, $\mathrm{HP}_{A}$) as the Hilbert function (respectively, the Hilbert polynomial) of $k[A]$, then there exists $t_0\in\mathbb{N}$ such that $\mathrm{HF}_{A}(t) = \mathrm{HP}_{A}(t), \, \forall t \geq t_0$.
Question: If we consider the set $A$ in normal form such that $k[A]$ is not Cohen-Macaulay ring, and $t_0$ as above, then is $k[t_0A]$ Cohen-Macaulay ring? We can illustrate this through the following example with $$A = \{ 0, 2, 3, 5, 6, 8 \}.$$ In this case, by using Macaulay2, we compute that $\mathrm{HF}_{A}(t) = \mathrm{HP}_{A}(t)$ for all $t \geq 2 =: t_0$, and $k[A]$ is not Cohen-Macaulay ring. Moreover, $k[2A]$ is Cohen-Macaulay ring.
