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Try looking at the proof of Theorem 17.2. -Kristie | Try looking at the proof of Theorem 17.2. -Kristie | ||
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| + | 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. | ||
Revision as of 09:55, 13 November 2008
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.
