# divisibility of polynomials over a finite field

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 $P_1\cdot\ldots\cdot P_{n-1}-1$.

• 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. Commented May 3, 2017 at 17:55
• Could you tell me the books that contain the statement, no matter if it is stated without a proof?
– Orat
Commented May 4, 2017 at 8:20

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 $f$.

• I don't get 'no such polynomial is self-inverse' part. Is it easy?
– Orat
Commented May 4, 2017 at 11:12
• @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. Commented May 4, 2017 at 11:13