ochiai/Grobner の履歴(No.1)

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  • 1 (2016-10-18 (火) 06:56:36)

$F=(f_1,\ldots,f_m) \in M(1,m,R)$, $G=(g_1,\ldots,g_r) \in M(1,r,R)$.

There exists $C \in M(m,r,R)$ and $D \in M(r,m,R)$ such that $G=FC$ and $F=GD$. Put $A=I_m - CG \in M(m,m,R)$. Then $FA=O_{1,m}$ shows $\langle A \range \subset S_F$.

Suppose $B \in M(r,s,R)$ be a generating system of $S_G$. Put $B^* = CB \in M(m,s, R)$. Then $F B^*=O_{1,s}$ shows $\langle B^* \range \subset S_F$.

Now take $\mathbf{h} \in S_F \subset M(m,1,R)$. Put $\mathbf{h}_* = D \mathbf{h} \in M(r, 1, R)$ and $\mathbf{k} = C \mathbf{h}_* \in M(m,1,R)$. Then $\mathbf{h} - \mathbf{k} = A \mathbf{h} \in \langle A \rangle$. On the other hand, $G \mathbf{h}_* =GD \mathbf{h} = F \mathbf{h}=0$ shows that $\mathbf{h}_* \in S_G$. Then there exists $\mathbf{\alpha} \in M(s,1,R)$ such that $\mathbf{h}_* = B \mathbf{\alpha}$. This shows $\mathbf{k} = CB \mathbf{\alpha} = B^* \mathbf{\alpha} \in \langle B^* \rangle$.

Summing up these two consequences, we obtain $\mathbf{h} = \mathbf{h}-\mathbf{k} + \mathbf{k} \in \langle A \rangle + \langle B^* \rangle = \langle A \cup B^* \rangle$.