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Let's fix linear operator $A$ in $n-$ dimension space $V$. Define on $V$ structure of $k[[t]]$-module by $$f(t)\cdot v := f(A)v,$$ where $$f\in k[[t]],~~v\in V.$$ Task: How many different $k[[t]]-$ modules, which dimension as vector space over $k$ is equal to six ($n = 6$).

I don't know the solution, but I've formulated and proved next lemma (maybe it will be useful):

Lemma: if operator $A$ fixed on $M$, $B$ fixed on $N$. And $\Phi : M\to N$ is isomorphism of $k[[t]]-$ modules then $$B = \Phi A \Phi^{-1}$$ $\blacktriangleright$ Let's $f\in k[[t]]$, $v\in M$ then $$\Phi(fv)=f(\Phi(v)).$$ If $$f(t) = 1+0\cdot t+0\cdot t^2+0\cdot t^3+0\cdot t^4+\ldots = 1$$ then $$\Phi(fv) = (\Phi\circ A)v$$ and $$f(\Phi(v)) = (B\circ\Phi)v.$$ So $B = \Phi A \Phi^{-1}$ $\blacktriangleleft$

My questions:

  1. From my lemma follows that over $k = \mathbb{C}$ there are infinite quantity of such modules (Jordan decomposition). Is it true?
  2. Is this module well defined? (There are such $f\in k[[t]]$ that operator $f(A)$ isn't defined)
  3. Can you help me to solve task for any field $k$.

Thanks a lot!

share|cite|improve this question
$k[[t]]$ is the ring of formal power series – Aspirin Feb 10 '12 at 14:53
Why do you consider $k[[t]]$ instead of $k[t]$ ? As is, your module is not well-defined. But anyway $k[[t]]$ and $k[t]$ are both principal rings. Have a look on the theory of finite modules over PID : this will answer all your questions. – Lierre Feb 10 '12 at 15:04
Is $A$ nilpotent, or something? I don't see how you can evaluate. – Dylan Moreland Feb 10 '12 at 15:06
I agree that A should be chosen to be nilpotent (so that f(A) makes sense), and then the solution is pretty easy using Jordan form (which doesn't need algebraic closure when all the eigenvalues are 0). – Jack Schmidt Feb 10 '12 at 17:49
IIRC Jordan normal form really just needs the characteristic polynomial to split, and here it definitely does! See also this Terry Tao post. – Dylan Moreland Feb 10 '12 at 18:16

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