Revision as of 10:55, 13 November 2008 by Pmongkol (Talk)

Does anyone have any ideas about this one?

Try looking at the proof of Theorem 17.2. -Kristie

Let r = p/q and f(x) is primitive

then h(x) = f(x) / (x + p/q)

Note that h(x) is also in Z[x]

f(x) = h(x) * (x + p/q) q* f(x) = h(x) * (q*x + p)

The left side is not primitive therefore it is not in Z[x] while the right still is because p and q are relatively prime.

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