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Let $V$ be the space of all $f(t) \in K[t]$ with $\mathrm{deg} f \leq n-1$ and let $\psi: V \to V$ with $\psi(f) = f'$. Further $\mathrm{char}(K) = 0$.

Then $\psi$ is nilpotent. Since one can take the derivative over and over again until $f^{(m)} = 0$. Further more $V$ is cyclic since $\frac{t^{n-1}}{(n-1)!}, \left(\frac{t^{n-1}}{(n-1)!}\right)', ..., t, 1$ form a basis of $V$.

But why is it important that $\mathrm{char}(K) = 0$?

If $\operatorname{char}(K) \neq 0$ the term $\frac{t^{n-1}}{(n-1)!}$ is nonsense but how to prove that there can't be another cyclic base if $\operatorname{char}(K) \neq 0$?

Would it work if $\operatorname{char}(K) \geq (n-1)!$?

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$\operatorname{char}(K)$ is not relevant in showing that $\psi$ is nilpotent. Perhaps it's relevant in some part of whatever you're reading that you haven't mentioned here? –  Chris Eagle May 9 '12 at 11:48
    
No this example comes on it's own after we introduced the term nilpotent. I wondered if it's relevant but couldn't see why..or wait: is it relevant in showing that $V$ is cyclic? –  jemnix May 9 '12 at 11:49
    
Yes. Hint: the sequence you have written is not the relevant one for checking if $V$ is cyclic. $1$ is not $\psi^{m-1}(t^{m-1})$. –  Chris Eagle May 9 '12 at 11:53
    
The cyclic basis should now be fixed. But I don't see why it fails if $\mathrm{char}(K) \neq 0$.. –  jemnix May 9 '12 at 12:03
3  
$t^{n-1}/(n-1)!$ is nonsense if $\operatorname{char}(K)|(n-1)!$. –  Chris Eagle May 9 '12 at 12:09

1 Answer 1

If $p=\operatorname{char}(K) \geq n$ then $t^{n-1}$ and all its derivatives form a cyclic basis of $V$. However if $p\neq0$ one has $\psi^p=0$ and no cyclic submodule of $V$ can have dimension${}>p$. In particular if $0<p<n$ then $V$ is not cyclic.

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