ochiai/Grobner - PukiWiki

Becker, Weispfenning, GTM, 141, p247--p249. Proof of Theorem 6.4. The book gives the proof with many summation symbol $\sum$. In the first alternative proof below, we give a proof with matrix/vector notation. In the second alternative proof below, we give a proof with mapping notation with kernel and image. Accessibility on these three proofs depends.


Notation. Let $M(p,q,R)$ be the set of matrices of size $r \times s$ with entries in $R$, where $R=K[\underline{X}]$.

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

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

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

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 $\mathbf{h} - \mathbf{k} = (I_m -CD) \mathbf{h}= A \mathbf{h} \in \langle A \rangle$. 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 $\mathbf{h} = \mathbf{h}-\mathbf{k} + \mathbf{k} \in \langle A \rangle + \langle B^* \rangle = \langle A \cup B^* \rangle$.


Consider the following five maps $F: R^m \to R, G: R^r \to R, C: R^r \to R^m, D: R^m \to R^r, B: R^s \to R^r$ satisfying the following three conditions $G=FC, F=GD$ and ker($G$)= Im($B$). We claim that ker($F$) = Im($I-CD$) + Im($CB$). The inclusion $\supset$ follows from $F(I-CD)=F-GD=F-F=O$ and $FCB=GB=O$. The other inclusion $\subset$ is discussed as follows: $GD($ker$F$) $=F$(ker$F)=0$ shows $D($ker$F) \subset $ker $G=$Im $B$. Then $CD($ker$F) \subset$ Im$(CB)$. On the other hand, $(I-CD)($ker$F) \subset $Im$(I-CD)$. This completes the proof of claim.

Note that we can come back to the book if we put $A=I-CD$ and $B^*=CB$.


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Last-modified: 2016-10-18 (火) 16:45:36 (400d)