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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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    Oh, gosh ... alright, so the characteristic is not relevant in showing that $\psi$ is nilpotent, but in showing that $V$ is cyclic. Does it work if $\mathrm{char}(K) \geq (n-1)!$?2012-05-09

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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 then $V$ is not cyclic.