0
$\begingroup$

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 $\operatorname{char}(K) \geq (n-1)!$?

  • 0
    $\operatorname{char}(K)$ 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?2012-05-09
  • 0
    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?2012-05-09
  • 0
    Yes. Hint: the sequence you have written is not the relevant one for checking if $V$ is cyclic. $1$ is not $\psi^{m-1}(t^{m-1})$.2012-05-09
  • 0
    The cyclic basis should now be fixed. But I don't see why it fails if $\mathrm{char}(K) \neq 0$..2012-05-09
  • 3
    $t^{n-1}/(n-1)!$ is nonsense if $\operatorname{char}(K)|(n-1)!$.2012-05-09
  • 0
    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

1 Answers 1