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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!

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    $k[[t]]$ is the ring of formal power series2012-02-10
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    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.2012-02-10
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    Is $A$ nilpotent, or something? I don't see how you can evaluate.2012-02-10
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    @Lierre because its task from our exam2012-02-10
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    @DylanMoreland something2012-02-10
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    @Lierre finite modules over PID? But this $k[[t]]-$ module is infinite2012-02-10
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    @Aspirin But $V$ is finitely generated over $k$, so it's certainly finitely generated over $k[[t]]$.2012-02-10
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    @DylanMoreland you mean finitely generated?2012-02-10
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    @DylanMoreland now I'm preparing to exam, and looking through tasks from last exam, if you think that now I'm on exam. I'm sorry2012-02-10
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    @Aspirin No, I assumed better of you, because I've seen your questions before and it seems like you're pretty reasonable. I just wanted to avoid controversy. Does my comment about $V$ (a $6$-dimensional vector space over $k$ which is also a module over $k[[t]]$) being finitely generated over $k[[t]]$ make sense?2012-02-10
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    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).2012-02-10
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    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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