Question 6

We just need to find \(i \in \left\{1,\dots,12\right\}\), such that \(i^{12} \equiv 1 \pmod{13}\) and \(12\) is the minimal such exponent. With computer assistance, one such solution is \(2\) with \(o(2) = 12\).

Question 7

(a) \(\mathbb{Z}\)

\[ x^8-1 = (x^4-1)(x^4+1) = (x-1)(x+1)(x^2+1)(x^4+1) \] and \[ x^6-1 = (x^3-1)(x^3+1) = (x-1)(x^2+x+1)(x+1)(x^2-x+1). \] the degree \(2\) terms all have no root in \(\mathbb{Q}\) and are as a result irreducible in \(\mathbb{Q}[x]\) and by Gauss criterion also irreducible in \(\mathbb{Z}[x]\).

(b) \(\mathbb{Z}/(2)\)

In \(\mathbb{Z}/(2)[x]\) as the base field has characteristic \(2\), \(x-1 = x+1\) and by Freshmen’s dream \[\begin{align*} x^2 + 1 &= (x + 1)^2 \\ x^4 + 1 &= (x^2 + 1)^2 \\ &= (x + 1)^4 \end{align*}\] so \[ x^8 - 1 = (x+1)(x+1)(x+1)^2(x+1)^4 = (x+1)^8. \] Also since \(-1 = 1\), \[ x^2 + x + 1 = x^2 - x - 1 \] so \[ x^6 - 1 = (x+1)^2 (x^2+x+1)^2 \] and we can check (by substituting \(x=0,1\)) that \(x^2+x+1\) is satisfies irreducibility criterion in \(\mathbb{Z}/2\).

(c) \(\mathbb{Z}/(3)\)

We can also sub \(x=0,1,2\) and check that \(x^2+1\) is irreducible, by inspection \[ x^4 + 1 = (x^2 + x + 2)(x^2 + 2x + 2) \] so we factorise \[ x^8 - 1 = (x+1)(x+2)(x^2+1)(x^2+x+2)(x^2+2x+2) \] which we can similarly check to be irreducible in \(\mathbb{Z}/(3)\).

Also \[\begin{align*} x^2 + x + 1 &= (x-1)^2 \\ x^2 - x + 1 &= (x+1)^2 \end{align*}\] so we have \[ x^6 - 1 = (x+1)^3(x-1)^3. \]