I am reading Huneke, Sega, and Vraciu's paper Vanishing of Ext and Tor over Cohen-Macaulay local rings. There is a statement in their paper which I can't understand.
Let $(R,\mathfrak{m},k)$ be an artinian local ring with $\mathfrak{m}^3=0$. All modules are assumed to be finitely generated. For a module $X$, denote $b_i(X)$ to be the $i$-th Betti number of $X$. That is, $b_i(X)=\text{dim}_k\text{Ext}_R^i(X,k)$. Let $\text{Soc}(X)$ denote the socle of $X$. It is equal to $\{x\in X\mid \mathfrak{m}x=0\}$. Let $X_i$ denote the $i$-th syzygy of $X$.
The statement in their paper which I can't understand is
I don't know why $\text{Soc}(M_{i-1})=\text{Soc}(R^{b_{i-2}~(M)})$.
Let's assume $\text{Tor}_3^R(M,N)=0$.The aim is to prove $\text{Soc}(M_{2})=\text{Soc}(R^{b_{1}(M)})$. It is trivial that $\text{Soc}(M_{2})\subseteq\text{Soc}(R^{b_{1}(M)})$. It remains to show the other direction. Consider the short exact sequence $$ 0\rightarrow M_2\rightarrow R^{b_1(M)}\xrightarrow{\pi}M_1\rightarrow 0. $$ The other direction is equivalent to show that for $x\in \text{Soc}(R^{b_1(M)})$, then $\pi(x)=0$. I don't know how to prove this. By the assumption $\mathfrak{m}^3=0$ and $M_1$ is the first syzygy, we have $\mathfrak{m}^2M_1=0$. If $\text{Soc}(R)\subseteq \mathfrak{m}^2$, then we are done. But this is what the authors want to show in their paper.
Do I miss something? Thank you in advance.