4.2 Proof of Cyclic Decomposition Theorem

Theorem (Cyclic Decomposition Theorem)

Let \(V\) such that \(\dim V=n\), \(f:V\to V\) is \(k\)-linear map. Let \(W_0\subsetneq V\) be an \(f\)-admissible subspace

Then \(\exists v_1,...,v_r\in V\) such that

\(V=W_0\oplus \lang v_1\rang _f\oplus...\oplus \lang v_r\rang_f\)
\(m_{f,v_i}(x)\mid m_{f,v_{i-1}}(x)\), \(\forall i,2\leq i\leq r\)

Furthermore, \(r\) and \(m_{f,v_1}(x),...,m_{f,v_r}(x)\) are unique satisfying conditions (1) and (2)

I give up!!!

Proof

We will prove that \(\exists z_1,...,z_r\in V\)such that
1. \(V=W_{0}+\lang z_{1}\rang_{f}+...+\lang z_{r}\rang_{f}\)
2. If \(1\leq j\leq r\), denote \(W_{j}=W_{0}+\langle z_{1}\rangle_{f}+...+\langle z_{j}\rangle_{f}\). Then the \(f\)-conductor \(p_j(x)\) of \(z_j\) into \(W_{j-1}\) has maximal degree among \(f\)-conductor into \(W_{j-1}\)
  
  In other words, \(\deg p_{j}(x)=\max\limits_{v\in V}\{\deg \phi:\phi (x)=C_{f,v,W_{j-1}}(x)\}\) which is bounded by \(m_f(x)\)
  
  For proving this, we will note that \(W_0\) is \(f\)-invariant
Proof

Fix \(j\), \(1\leq j\leq r\), note \(0<\max\limits_{v\in V}\{\deg \phi:\phi (x)=C_{f,v,W_{j-1}}(x)\}\leq\deg m_f(x)\leq n\)

For \(j=1\), we can choose \(z_1\) as a vector in \(V\) such that \(\deg p_{1}(x)=\max\{\deg\phi(x):\phi(x)=C_{f,v,W_0}(x)\}\)

Now consider the subspace \(W_1=W_{0}+\lang z_{1}\rang_{f}\supsetneq W_{0}\), \(W_1\) is \(f\)-invariant (easy proof)

Then if \(\dim V=\dim W_1\), we are done. If not, \(\dim W_{1}<\dim V\)

Then since \(\dim W_{1}>\dim W_{0}\), doing the same process we can choose \(z_2\in V:\deg p_2(x)\) is maximum

Let \(W_2=W_0+\lang z_1\rang_f+\lang z_2\rang_f\) which is \(f\)-invariant, then if \(\dim W_2=\dim V\), we are done.

If not, we continue this process and choose \(z_3\)...

Then \(\dim W_0<\dim W_1<\dim W_2<...<\dim V=n\)

In a finite number \(r\) of steps, we will have chosed vectors \(z_1,...,z_r:V=W_0+\lang z_1\rang_f+...+\lang z_r\rang_f\) (1)

And (2) will also be satisfied because we have chosen the \(z_1,...,z_r\) in the same way
Let \(z_1,...,z_r\) be as in (1). Fix \(j,1\leq j\leq r\). Let \(v\in V\), consider \(h(x)=C_{f,v,W_{j-1}}\left(x\right)\)

NTP: If \(h(f)(v)=w_{0}+\sum_{1\leq i<j}g_{i}\left(f\right)\left(w_{i}\right)\) with \(w_{0}\in W_{0},w_{i}\in W_{i},\forall i,1\leq i<r\), \(g_i(x)\in k[x],\forall i\)

Then \(h(x)\mid g_i(x),1\leq i<j\) and \(w_{0}=h(f)(\gamma_0)\) with \(\gamma_0\in W_0\)

If \(j=1\), then the statement is as follows: \(\forall v\in V\), if \(h(f)(v)=w_0\) for \(w_0\in W_0\), then \(w_0=h(f)(\gamma_0)\) for some \(\gamma_0\in W_0\)

That is we are saying that \(W_0\) is \(f\)-admissible. This is true by hypothesis

In order to prove the statement for \(j>1\), we will first use the division algorithm

While \(\forall i,1\leq i\leq j\), then \(g_{i}(x)=h(x)q_{i}(x)+r_{i}(x)\) (*) with \(r_i(x)=0\) or \(\deg r_i(x)<\deg h(x)\)

We want to prove that \(r_{i}(x)=0,\forall i,1\leq i\leq r\)

Let \(\gamma=v-\underbrace{\sum_{i=1}^{j-1}\underbrace{q_{i}\left(f\right)\left(w_{i}\right)}_{\in W_i}}_{W_{j-1}}\) (construct the right part of (*)). Look at \(\gamma-v\in W_{j-1}\)

This implies that the f-conductor of \(v\) into \(W_{j-1}\) is equal to the f-conductor of \(\gamma\) into \(W_{j-1}\)

Moreover, \(h(f)(\gamma)=h(f)(v)-\sum_{i=1}^{j-1}h\left(f\right)q_{i}\left(f\right)\left(w_{i} \right)=h(f)\left(v\right)-\sum_{i=1}^{j-1}\left(g_{i}\left(x\right)-r_{i}\left(x \right)\right)\left(w_{i}\right)\)

\(=h(f)(v)-\sum_{i=1}^{j-1}g_{i}\left(x\right)\left(w_{i}\right)+\sum_{i=1}^{j-1}r _{i}\left(x\right)\left(w_{i}\right)=w_{0}+\sum_{i=1}^{j-1}g_{i}\left(f\right)\left (w_{i}\right)-\sum_{i=1}^{j-1}g_{i}\left(x\right)\left(w_{i}\right)+\sum_{i=1}^{j-1} r_{i}\left(x\right)\left(w_{i}\right)\)

\(=w_{0}+\sum_{i=0}^{j-1}r_{i}\left(f\right)\left(w_{i}\right)\), suppose for some \(i\), \(r_i(x)\neq 0\)

Then \(\underbrace{h(f)(\gamma)}_{\in W_{j-1}}=w_{0}+\sum_{i=1}^{j-1}r_{i}\left(f\right )\left(w_{i}\right)\) we will deduce a contradiction

Let \(s=\max_{1\leq i<j}\left\lbrace i:r_{i}\left(x\right)\neq0\right\rbrace\neq\emptyset\). Then consider \(h(f)(\gamma)=w_{0}+\sum_{i=1}^{s}r_i(f)(w_i)\) (*)

Where \(r_s(x)\neq 0,\deg r_s(x)<\deg h(x)\)

Let \(p(x)\) be the \(f\)-conductor of \(\gamma\) into \(W_{s-1}\). Since \(W_{j-1}\supseteq W_{s-1}\), so \(h(x)\mid p(x)\)

Let \(g(x)\in k[x]\) such that \(p(x)=h(x)g(x)\). Apply \(g(f)\) to both sides of (*)

\(g(f)\circ h(f)(\gamma)=\underbrace{p(f)\left(\gamma\right)}_{\in W_{s-1}}=\underbrace{g\left(f\right)\left(w_{0}\right) +\sum_{i=1}^{s-1}g\left(f\right)\circ r_{i}\left(f\right)\left(w_{i}\right)}_{\in W_{s-1}}+g\left (f\right)\circ r_{s}\left(f\right)\left(w_{s}\right)\)

Then \(g\left(f\right)\circ r_{s}\left(f\right)\left(w_{s}\right)\in W_{s-1}\)

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\(\deg r_s(x)<\deg h(x)\) and \(\deg g(x)r_s(x)<\deg g(x)h(x)=\deg p(x)\)

Then \(g(x)r_s(x)\) is \(f\)-conductors \(\gamma\) into \(W_{s-1}\), but \(\deg g(x)\cdot r_s(x)<\deg p(x)\) which is a contradiction

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Now let's use condition (2) of step1, \(\deg g(x)r_{s}(x)=\deg g(x)+\deg r_s(x)\geq\deg\{f\text{-conductor of }w_{s}\text{ into }W_{s-1}\}=\deg p_{s}\left(x\right)\geq\deg\{f\text{-conductor of }v\text{ into }W_{s-1}\}=\deg p (x)=\deg g(x)+\deg h(x)\)

Then \(\deg r_s(x)\geq \deg h(x)\) contradiction, so \(r_i(x)=0,\forall i\), then \(h(x)\mid g_i(x)\)

\(h(f)(v) = w_0 + \sum_i g_i(f)(w_i)\)

\(h(f) (\sum_i q_i(f)(w_i))\)

\(\Rightarrow h(f)(v - \sum_i q_i(f)(w_i)) = w_3 \in W_0\)

Since \(W_0\) is f-admissible.

then \(\exists r_0 \in W_0 / w_3 = h(f)(r_0)\)
\(\exists v_1,...,v_r\in V\) satisfying conditions (1) and (2) of the theorem

We have \(z_1,...,z_r\) like in step 1. Fix j, \(1\leq j\leq r\). Apply step 2 to \(z_j\) and the \(f\) conductor of \(z_j\), that is \(p_j(x)\)

Then \(p_{j}(f)(z_{j})\in W_{j-1},p_{j}(f)(z_{j})=p_{j}\left(f\right)\left(\gamma_{0}\right )+\sum_{1\leq i<j}p_{j}\left(f\right)q_{i}\left(f\right)\left(z_{i}\right)\)

Let \(v_{j}=z_{j}-\gamma_{0}-\sum_{1\leq i<j}q_{i}\left(f\right)\left(z_{i}\right)\), since \(v_j-z_j\in W_{j-1}\), so the \(f\)-conductor of \(v_j\) into \(W_{j-1}\) coincides with the \(f\)-conductor \(z_j\) into \(W_{j-1}\), that is \(p_j(x)\) A

And \(p_{j}(f)(v_{j})=0_{V}\in W_{j-1}\), we get \(W_{j-1}\cap \lang v_j\rang _f=\{0_V\}\) B

We have that \(v_j\) satisfies A and B, so it follows that \(W_{j}=W_{0}\oplus\langle v_{1}\rangle_{f}\oplus\langle v_{2}\rangle_{f}\oplus\ldots \oplus\langle v_{j}\rangle_{f}\)

And \(p_j(x)=m_{f,v_j}(x)\)

We have \(v_1,...,v_r\) and \(p_{1}(x)=m_{f,v_1}(x),...,p_{r}(x)=m_{f,v_r}(x)\) satisfying condition 1 and step 2 item 2
Unicity

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