More VW hadout

by Wolstenholme, Nov 26, 2014, 6:21 PM

$ 1.1.h) $ OK when I saw this two methods immediately came to mind. One was finding some bijective function $ f: \mathbb{Z}_p \rightarrow \mathbb{Z}_p $ such that $ \alpha $ is a root of the polynomial (from now on call it $ P(x) $) only if $ f(\alpha) $ is not a root. In this case, (by Occam's razor) the function would most likely be something like $ f(x) = p - x. $ However, this didn't work which brings me to method two - finding lots of double roots. Now the best way to do this is to take derivatives (which is further motivated by the structure of $ P $ itself). So let's do that!


From now on we work in $ \mathbb{Z}_p. $ Some important things to keep in my mind will be the simple generalization of Wilson's Theorem given by $ (p - k)!(k - 1)! \equiv (-1)^k \pmod{p} $ and Fermat's Little Theorem. Now note that for any $ \alpha \in \mathbb{Z}_p $ we have $ \sum_{k = 0}^{p - 1}k!\alpha^k = \sum_{k = 0}^{p - 1}(p - k - 1)!\alpha^{p - k - 1} = -\sum_{k = 0}^{p - 1}(-1)^k\frac{\alpha^{-k}}{k!}. $

Noting that the maps $ \alpha \rightarrow \frac{1}{\alpha} $ and $ \alpha \rightarrow -\alpha $ are bijective in $ \mathbb{Z}_p $ we have that $ \sum_{k = 0}^{p - 1}k!\alpha^k = 0 \Longleftrightarrow \sum_{k = 0}^{p - 1}\frac{\alpha^k}{k!} = 0 \Longleftrightarrow \sum_{k = 0}^{p - 1}\frac{\alpha^k}{k!} + \alpha^p - \alpha = 0. $

Now consider the polynomial $ Q(x) = \sum_{k = 0}^{p - 1}\frac{x^k}{k!} + x^p - x $ in $ \mathbb{Z}_p[x]. $ We have for any $ \alpha \in \mathbb{Z}_p $ that $ Q(\alpha) = Q'(\alpha) $ which implies that every root of $ Q $ is a double root which implies the same for $ P. $ Hence we have the desired result.

$ 1.1.i) $ Hmm this is pretty hard - I can rewrite it in terms of generating functions but I don't know if it helps

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oops darn what's the motivation for adding the frobenius endomoprhism (x^p-x) factor on the end
this is pretty tricky

by pi37, Dec 7, 2014, 1:06 AM

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^ To get $ Q(\alpha) = Q'(\alpha) $ cheaply

by Wolstenholme, Dec 8, 2014, 3:49 AM

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