I am reading the theory of finitely generated modules over a PID. One of the applications of the the theory is that one can derive the theory of rational canonical form of a linear operator on a finite dimensional vector space.
I considered the following question:
Question. Given a linear operator $T$ on a finite dimensional vector space $V$, find a basis of $V$ with respect to which the matrix of $T$ is in the rational canonical form.
While attempting to solve this, I "proved" that given any basis $(v_1,\ldots,v_n)$ of $V$, there exist positive integers $k$ and $d_{k+1},\ldots,d_n$ such that $$ (v_{k+1},Tv_{k+1},\ldots,T^{d_{k+1}-1}v_{k+1},\ldots,v_n,Tv_n,\ldots,T^{d_n-1}v_n) $$ forms a basis of $V$.
This seems bizarre. But I can't find a mistake in my working.
I am sorry for the long post.
Notations and Some Standard Things:
Let $V$ be a finite dimensional vector space over a field $F$ and $T$ be a linear operator on $V$. Let $(v_1,\ldots,v_n)$ be a basis for $V$. We give $V$ a $F[x]$ module structure by defining $$p(x)\cdot(a_1v_1+\cdots+a_nv_n)=a_1p(T)v_1+\cdots+a_np(T)v_n$$ for all $a_i\in F$ and $p(x)\in F[x]$. Let $(e_1,\ldots,e_n)$ be the standard basis of $F[x]^n$ as an $F[x]$-module. Define a map $f:F[x]^n\to V$ as $$f\left(\sum_{i=1}^np_i(x)e_i\right)=p_1(T)v_1+\cdots+p_n(T)v_n$$ It is easy to see that $f$ is an $F[x]$-module homomorphism. Since $F[x]^n$ is a free module over a principal ideal domain, $\ker(f)$ too is a free $F[x]$-module of finite rank. In fact, the rank of $\ker(f)$ is also $n$. (If it weren't then the dimension of $F[x]^n/\ker(f)$ as an $F$-vector space would be infinite while the dimension of $V$, which is isomorphic to $F[x]^n/K$ as an $F$-vector space, is finite.) It is also known by Structure theorem for finitely generated modules over a PID that $$K=(F[x]e_1+\cdots+F[x]e_k)+(F[x]a_{k+1}(x)e_{k+1}+\cdots+F[x]a_n(x)e_n)$$ for some $1\leq k\leq j$ and each $a_i$ is a polynomial of degree at least $1$. Define a map $$ \varphi:F[x]^n\longrightarrow \bigoplus_{i=k+1}^n F[x]/\langle a_{i}(x)\rangle $$ as $$ \varphi\left(\sum_{j=1}^np_j(x)e_j\right)=(\bar p_{k+1}(x),\ldots,\bar p_n(x)) $$ It is easily seen that $\varphi$ is an $F[x]$-module homomorphism whose kernel is precisely $\ker(f)$. Let $\pi:F[x]^n\to F[x]^n/\ker(f)$ be the canonical projection homomorphism and $\bar f:F[x]^n/\ker(f)\to V$ and $\bar \varphi:F[x]^n/\ker \varphi\longrightarrow \bigoplus_{i=k+1}^n F[x]/\langle a_i(x)\rangle$ be the isomorphisms such that $f=\bar f\circ \pi$ and $\varphi=\bar\varphi\circ \pi$. Thus we have an $F[x]$-module isomorphism $$\bar f\circ\bar\varphi^{-1}:\bigoplus_{i=k+1}^nF[x]/\langle a_i(x)\rangle\to V$$ which explicitly reads $$\bar f\circ \bar\varphi^{-1}(\bar p_{k+1}(x),\ldots,\bar p_n(x))=p_{k+1}(T)v_{k+1}+\cdots+p_n(T)v_n$$ Since $\bar f\circ\bar \varphi^{-1}$ is an $F[x]$-module isomorphism, it is also an isomorphism $\bigoplus_{i=k+1}^nF[x]/\langle a_i(x)\rangle$ and $V$ as $F$-vector spaces.
The Confusion:
So a basis of $\bigoplus_{i=k+1}^nF[x]/\langle a_i(x)\rangle$ is mapped to a basis of $V$ via $\bar f\circ\bar \varphi^{-1}$. Say $\deg a_i(x)=d_i$. Then for each $k+1\leq i\leq n$, we can choose the basis $\mathcal B_i=(\bar 1,\bar x,\ldots,\bar x^{d_i-1})$ for $F[x]/\langle a_i(x)\rangle$. We can thus form a basis $\mathcal B=(\mathcal B_{k+1},\ldots,\mathcal B_n)$ for $\bigoplus_{i=k+1}^nF[x]/\langle a_i(x)\rangle$. Operating $\bar f\circ\bar \varphi^{-1}$ to the elements of $\mathcal B$, we get a basis $\mathcal B'$ of $V$ which reads $$\mathcal B'=(v_{k+1},Tv_{k+1},\ldots,T^{d_{k+1}-1}v_{k+1},\ldots,v_n,Tv_n,\ldots,T^{d_n-1}v_n)$$
It also seems that the above basis is a basis with respect to which the matrix of $T$ is in the rational canonical form.
Can anybody see where have I made a mistake?
Thanks.
I found my mistake.
I am wrong when I say that $$\ker f= (F[x]e_1+\cdots+F[x]e_k)+(F[x]a_1(x)e_{k+1}+\cdots+F[x]a_{n}(x)e_n)$$
There exists a basis $(e_1',\ldots,e_n')$ of $F[x]^n$ for which
$$ \ker f= (F[x]e_1'+\cdots+F[x]e_k')+(F[x]a_1(x)e_{k+1}'+\cdots+F[x]a_{n}(x)e_n') $$
But that basis need not be $(e_1,\ldots,e_n)$.