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Becker, Weispfenning, GTM, 141, p247--p249. Proof of Theorem 6.4. The book gives the proof with many summation symbol . 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(kerF) =F(kerF)=0 shows D(kerF) \subset ker G=Im B. Then CD(kerF) \subset Im(CB). On the other hand, (I-CD)(kerF) \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