Let $K$ be a finite field.
Let $f$ be an irreducible polynomial over $K$ of degree $n\ge 2$.
Let $P_i$ be the product of all polynomials of degree $i$ over $K$.
Prove that $f$ divides $P_1\cdot\ldots\cdot P_{n-1}-1$.
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$\begingroup$ probably this is well known and posted before, since the proof of this fact is left for reader in many books, but I didn't manage to find it here. Sorry for doubling, if it happened. $\endgroup$– larry01Commented May 3, 2017 at 17:55
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$\begingroup$ Could you tell me the books that contain the statement, no matter if it is stated without a proof? $\endgroup$– OratCommented May 4, 2017 at 8:20
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1 Answer
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This is essentially Wilson's theorem for $ K[X]/(f) $. To prove it, simply note that any polynomial of degree $ 1 \leq \deg < n $ will admit an inverse of degree $ 1 \leq \deg < n $ modulo $ f $, and no such polynomial is self-inverse, so you may pair together inverses in the product to obtain that every factor in the product cancels, leaving you with $ 1 $ modulo $ f $.
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1$\begingroup$ I don't get 'no such polynomial is self-inverse' part. Is it easy? $\endgroup$– OratCommented May 4, 2017 at 11:12
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2$\begingroup$ @Orat $ K[X]/(f) $ is a field, so if you have $ a^2 = 1 $ for some $ a $, it follows that $ (a-1)(a+1) = 0 $ and thus $ a = 1 $ or $ a = -1 $. Both of these are constant polynomials, so they are not in the product. $\endgroup$ Commented May 4, 2017 at 11:13