s n_i}(B\otimes_AR)$. This equation should be interpreted as follows. We formally compute the right hand side then it is of the form $\pi\cdot X$, where $X$ involves $\tilde{m}_{i,i}^{\ast}$ and $\tilde{m}_{i,i}^{\ast\ast}$. The left hand side $\mathcal{X}_{i,i}^{\ast}(\tilde{m})$ is then defined to be $X$. Then by using an argument similar to the paragraph just before Equation (\[ea20\]) of Step (1), the image of $\mathcal{X}_{i,i}^{\ast}(\tilde{m})$ in $M_{1\times n_i}(B\otimes_AR)/(\pi\otimes 1)M_{1\times n_i}(B\otimes_AR)$ is independent of the choice of the lift $\tilde{m}$ of $m$. Therefore, we may denote this image by $\mathcal{X}_{i,i}^{\ast}(m)$. As for Equation (\[ea20\]) of Step (1), we need to express $\mathcal{X}_{i,i}^{\ast}(m)$ as matrices. By using conditions (5) and (6) of the description of an element of $(\mathrm{Ker~}\tilde{\varphi}/\tilde{M}^1)(R)$ given in Remark \[ra5\], we have that $$\mathcal{X}_{i,i}^{\ast}(m)=m_{i,i}^{\ast}+  m_{i,i}^{\ast\ast}\cdot \bar{h}_i+\mathcal{P}^{\ast}_i.$$ Here, $\mathcal{P}^{\ast}_i$ is a polynomial with variables in the entries of $m$ not including $m_{i,i}^{\ast}$ and $m_{i,i}^{\ast\ast}$. Since $m$ actually belongs to $\mathrm{Ker~}\varphi(R)/\tilde{G}^1(R)$, we have the following equation by the argument made at the beginning of this paragraph: $$\label{ea22}
     \mathcal{X}_{i,i}^{\ast}(m)=m_{i,i}^{\ast}+  m_{i,i}^{\ast\ast}\cdot \bar{h}_i+\mathcal{P}^{\ast}_i=\bar{f}_{i,i}^{\ast}=0.$$ Thus we get polynomials $\mathcal{X}_{i,i}^{\ast}(m)$ on $\mathrm{Ker~}\tilde{\varphi}/\tilde{M}^1$, vanishing on the subscheme $\mathrm{Ker~}\varphi/\tilde{G}^1$.\

3.  Assume that $i$ is odd and that $L_i$ is *free of type I*. By Equations (\[ea7\]), (\[ea8\]), (\[ea9\]), and (\[ea10\]) which involve an element of $\tilde{M}^1(R)$, each entry of $b_i', e_i', d_i'$ has $\pi$ as a factor and $f_i-f_i'$ has $\pi^2$ as a factor so that $b_i'\equiv b_i=0, e_i'\equiv e_i=0, d_i'\equiv d_i=0$ mod $(\pi\otimes 1)(B\otimes_AR)$, and $f_i'= f_i=\bar{\gamma}_i$. Let $\tilde{m}\in 