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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!

  • 1
    IIRC Jordan normal form really just needs the characteristic polynomial to split, and here it definitely does! See also [this Terry Tao post](http://terrytao.wordpress.com/2007/10/12/the-jordan-normal-form-and-the-euclidean-algorithm/).2012-02-10

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