Let \(F\) be a field of 81 elements. For each of the following polynomials, determine the number of roots that lie in \(F\): \(x^{80}-1\), \(x^{81}-1\), \(x^{88}-1\).
Solution
The numbers of distinct roots in \(F\) are, respectively, \[ \boxed{80,\quad 1,\quad 8}. \]
1. The characteristic is \(3\), and every nonzero \(a\in F\)satisfies \(a^{80}=1\).
The additive order of \(1\) divides \(|F|=81\). It is a prime: if it were \(uv\) with \(1<u,v<uv\), then the two nonzero field elements \(u\cdot1\) and \(v\cdot1\) would have product zero. Thus the characteristic is the only prime dividing \(81\), namely \(3\). The multiplicative group \(F^\times\) has order \(80\), so Lagrange’s theorem gives \(a^{80}=1\) for every \(a\ne0\) [@DF04].
2. The polynomial \(x^{80}-1\) has \(80\) distinct roots in \(F\).
Every element of \(F^\times\) is a root by step 1, and zero is not a root. There are exactly \(80\) such elements. In particular, the monic degree-\(80\) polynomial factors as \[ x^{80}-1=\prod_{a\in F^\times}(x-a). \]
3. The polynomial \(x^{81}-1\) has exactly one root in \(F\).
In characteristic \(3\) the binomial theorem gives \((u-v)^3=u^3-v^3\). Iterating four times yields \[ x^{81}-1=(x-1)^{81}. \] Hence the only root is \(1\), with multiplicity \(81\).
4. The polynomial \(x^{88}-1\) has exactly \(8\) roots in \(F\).
A root \(a\) must be nonzero. Using \(a^{80}=1\), the equation \(a^{88}=1\) is equivalent to \(a^8=1\). Thus the roots in \(F\) are precisely those of \(x^8-1\).
2. Unique factorization in the polynomial ring overa field implies that this monic divisor is itself a product ofdistinct linear factors in \(F[x]\) [@DF04]. Its degree is \(8\),so it has exactly \(8\) distinct roots in \(F\).