Give a complete factorization of each of the following over the indicated rings. Explain why the factors you obtain are irreducible.
\(x^5-1\) over \(\mathbf{C}\).
\(x^6-1\) over \(\mathbf{Q}\).
\(x^7+3x^5-6x^3+9x-15\) over \(\mathbf{Q}\).
\(x^{11}-x\) over \(\mathbf{F}_{11}\).
\(x^{13}-x-1\) over \(\mathbf{F}_{13}\).
Solution
1. Over \(\mathbf C\), with \(\zeta=e^{2\pi i/5}\), the factorization is\[ x^5-1=\prod_{j=0}^{4}(x-\zeta^j). \]
The five numbers \(\zeta^j\) are distinct and each has fifth power \(1\). Thus the product of their linear factors divides \(x^5-1\); both polynomials are monic of degree \(5\), so they are equal. A polynomial of degree \(1\) over a field cannot be a product of two positive-degree polynomials, so every displayed factor is irreducible.
2. Over \(\mathbf Q\), the factorization is\[ x^6-1=(x-1)(x+1)(x^2+x+1)(x^2-x+1). \]
Factor \(x^6-1=(x^3-1)(x^3+1)\) and use the sum and difference of cubes. The linear factors are irreducible. The two quadratic factors have discriminant \(-3\), so neither has a rational root. A reducible quadratic over a field has a linear factor and hence a root, proving their irreducibility.
3. The polynomial \(x^7+3x^5-6x^3+9x-15\) is itself irreducible over\(\mathbf Q\), so there are no further factors.
Its leading coefficient is \(1\), all other coefficients are divisible by \(3\), and its constant coefficient is not divisible by \(9\). By Eisenstein’s criterion at \(3\) it is irreducible. In detail, by Gauss’s lemma, a proper factorization over \(\mathbf Q\) would give monic positive-degree factors \(g,h\in\mathbf Z[x]\) [@DF04]. Modulo \(3\), their product is \(x^7\), so each reduction is a positive power of \(x\). Consequently \(3\) divides both \(g(0)\) and \(h(0)\), forcing \(9\mid g(0)h(0)=-15\), a contradiction.
4. Over \(\mathbf F_{11}\), the factorization is\[ x^{11}-x=\prod_{a\in\mathbf F_{11}}(x-a). \]
Every nonzero \(a\) belongs to the multiplicative group of order \(10\), so \(a^{10}=1\) and \(a^{11}=a\). The identity also holds at \(a=0\). Thus the eleven distinct field elements give eleven distinct linear factors of the monic degree-\(11\) polynomial. Their product is the polynomial, and each factor is irreducible.
5. The polynomial \(x^{13}-x-1\) is irreducible over \(\mathbf F_{13}\).
Choose a root \(\alpha\) in an algebraic closure, and let \(d=[\mathbf F_{13}(\alpha):\mathbf F_{13}]\). Its minimal polynomial divides the given polynomial, so \(1\leq d\leq13\). The root equation gives \(\alpha^{13}=\alpha+1\). Induction, using the Frobenius homomorphism in characteristic \(13\), gives \[ \alpha^{13^j}=\alpha+j\cdot1\qquad(j\geq0). \] The field \(\mathbf F_{13}(\alpha)\) has \(13^d\) elements, so every element of it satisfies \(z^{13^d}=z\). Taking \(j=d\) in the displayed identity therefore gives \(d\cdot1=0\), and hence \(13\mid d\). Thus \(d=13\). The degree-\(13\) minimal polynomial divides the monic polynomial \(x^{13}-x-1\) of the same degree, so they are equal. This proves irreducibility and completes all five factorizations.