x_{4}^{u_{4}+a_{24}}$ is a term in the sum $\sum_{i=1}^{5}H_{i}f_{i}$ and it should be canceled with another term of the above sum. Thus $x_{2}^{a_{12}}x_{4}^{a_{34}}$ divides $-T$, so $u_{2} -a_{2} \geq a_{12}$ a contradiction.\\






\begin{example} {\rm Consider $n_{1}=627$, $n_{2}=1546$, $n_{3}=1662$ and $n_{4}=3377$. The toric ideal $I(C)$ is minimally generated by the set $$G=\{x_1^{18}- x_{2}^{3}x_{3}^{4}, x_{2}^{25}- x_{3}^{7}x_{4}^{8}, x_3^{11}-x_{1}^{13}x_{4}^{3}, x_{4}^{11}-x_{1}^{5}x_{2}^{22}, x_{1}^{5}x_{3}^{7}-x_{2}^{3}x_{4}^{3}\}.$$ Here $a_{24}=8$, $a_{34}=3$, $a_{12}=3$, $a_{1}=18$, $a_{2}=25$ and $a_{13}=4<7=a_{23}$. Note that $a_{3}=11<14=a_{31}+a_{34}$ and $a_{12}+a_{34}=6<12=a_{41}+a_{23}$. We have that $a_{2}+a_{12}=28<29=a_{1}+a_{23}-a_{13}+a_{24}$. Thus $C$ has Cohen-Macaulay tangent cone at the origin. Remark that $x_{2}^{28}-x_{1}^{5}x_{3}^{14}x_{4}^{5} \in I(C)$, but ${\rm deg}(x_2^{28})=28>24={\rm deg}(x_{1}^{5}x_{3}^{14}x_{4}^{5})$.}

\end{example}



\begin{remark} {\rm Suppose that $I(C)$ is given as in case 2(b).\\ (1) It holds that $a_{1}>a_{12}+a_{13}$, $a_{4}<a_{41}+a_{43}$ and $a_{24}+a_{13}<a_{41} +a_{32}$.\\ (2) If $a_{24} \leq a_{34}$, then $x_{3}^{a_{3}+a_{13}}-x_{1}^{a_{41}}x_{2}^{2a_{32}}x_{4}^{a_{34}-a_{24}} \in I(C)$.\\(3) If $a_{12} \leq a_{32}$, then the binomial $x_{3}^{a_{3}+a_{13}}-x_{1}^{a_{1}}x_{2}^{a_{32}-a_{12}}x_{4}^{a_{34}}$ belongs to $I(C)$.}
\end{remark}


\begin{proposition} \label{Case 2(b) basic} Suppose that $I(C)$ is given as in case 2(b). Then $C$ has Cohen-Macaulay tangent cone at the origin if and only if \begin{enumerate} \item $a_{2} \leq a_{21}+a_{24}$ and \item for every monomial $x_{2}^{u_{2}}x_{3}^{u_{^3}}x_{4}^{u_{4}}$, where $u_{2}<a_{12}$, $u_{3} \geq a_{3}$ and $u_{4}<a_{24}$, with $u_{2}n_{2}+u_{3}n_{3}+u_{4}n_{4} \in n_{1}+\mathcal{S}$ there exists a monomial $N$ with $1 \in {\rm supp}(N)$ such that $M-N \in I(C)$ and also ${\rm deg}(M) \leq {\rm deg}(N)$.
\end{enumerate}

\end{proposition}

\noindent \textbf{Proof.} In this case $I(C)$ is minimally generated by the set $$G=\{f_{1} = x_1^{a_1}- x_2^{a_{12}} x_3^{a_{13}}, f_2 = x_{2}^{a_2}- x_{1}^{a_{21}}x_{4}^{a_{24}}, f_3 = x_3^{a_{3}}-x_{2}^{a_{32}}x_{4}^{a_{34}},$$ $$f_4 = x_{4}^{a_4}-x_{1}^{a_{41}}x_{3}^{a_{43}}, f_5 = x_{1}^{a_{41}}x_{2}^{a_{32}}-x_{3}^{a_{13}}x_4^{a_{24}}\}.$$
Suppose that $C$ has Cohen-Macaulay tangent cone at the origin. Since $I(C)$ is generated by the indispensable binomials, every binomial $f_i$, $1 \leq i \leq 5$, is indispensable of $I(C)$. In particular the binomial $f_2$ is indispensable of $I(C)$. Therefore the inequality $a_{2} \leq a_{21}+a_{24}$ holds. By Theorem \ref{BasicHer} condition (2) is also true.

Conversely, from Proposition \ref{1(b)} (iii), it is enough to consider a monomial $M=x_{2}^{u_2}x_{3}^{u_3}x_{4}^{u_4}$, where $u_2<a_{2}$, $u_3 \geq 0$ and $u_4<a_{4}$, with the property: there exists at least one monomial $P$ such that $1 \in {\rm supp}(P)$ and also ${\rm deg}_{\mathcal{S}}(M)={\rm deg}_{\mathcal{S}}(P)$. Suppose that $u_{3} \geq a_{3}$. If $u_{2} \geq a_{12}$, then we let $P=x_{1}^{a_{1}}x_{2}^{u_{2}-a_{12}}x_{3}^{u_{3}-a_{13}}x_{4}^{u_{4}}$. So we have that $M-P \in I(C)$ and also ${\rm deg}(M)<{\rm deg}(P)$. Similarly if $u_{4} \geq a_{24}$, then we let $P=x_{1}^{a_{41}}x_{2}^{u_{2}+a_{32}}x_{3}^{u_{3}-a_{13}}x_{4}^{u_{4}-a_{24}}$. So we have that $M-P \in I(C)$ and also ${\rm deg}(M)<{\rm deg}(P)$. If both conditions $u_{2}<a_{12}$ and $u_{4}<a_{24}$ hold, then condition (2) implies that there exists a monomial $N$ with $1 \in {\rm supp}(N)$ such that $M-N \in I(C)$ and also ${\rm deg}(M) \leq {\rm deg}(N)$. Suppose now that $u_3<a_3$. Then $M$ is divided by at least one of the monomials $x_{2}^{a_{12}}x_3^{a_{13}}$, $x_{3}^{a_{13}}x_4^{a_{24}}$ and $x_{2}^{a_{32}}x_4^{a_{34}}$. If $M$ is divided by $x_{2}^{a_{12}}x_3^{a_{13}}$, then $M=x_2^{a_{12}+p}x_3^{a_{13}+q}x_{4}^{u_4}$, for some non-negative integers $p$ and $q$, so $M-x_1^{a_{1}}x_2^{p}x_3^{q}x_{4}^{u_{4}} \in I(C)$ and also ${\rm deg}(M)<{\rm deg}(x_1^{a_{1}}x_2^{p}x_3^{q}x_{4}^{u_{4}})$. If $M$ is divided by $x_{3}^{a_{13}}x_4^{a_{24}}$, then $M=x_{2}^{u_2}x_3^{a_{13}+p}x_4^{a_{24}+q}$, for some non-negative integers $p$ and $q$, and therefore the binomial $M-x_1^{a_{41}}x_2^{u_{2}+a_{32}}x_3^{p}x_4^{q} \in I(C)$ and also ${\rm deg}(M) \leq {\rm deg}(x_1^{a_{41}}x_2^{u_{2}+a_{32}}x_3^{p}x_4^{q})$. Assume that neither $x_{2}^{a_{12}}x_3^{a_{13}}$ nor $x_{3}^{a_{13}}x_4^{a_{24}}$ divides $M$. Then necessarily $x_{2}^{a_{32}}x_4^{a_{34}}$