Let \(p\) be prime, \(n\) be a positive integer, \(f(X) = X^{p^n} - X + 1\), and \(K\) be an algebraic closure of \(\mathbb{F}_p\).
If \(\alpha \in K\) is a root of \(f(X)\) and \(a \in \mathbb{F}_{p^n} \subseteq K\), show that \(\alpha + a\) is a root of \(f(X)\) and \(\alpha^p - \alpha + 1 \in \mathbb{F}_{p^n}\).
Show that \(X^p - X + 1\) is irreducible in \(\mathbb{F}_p[X]\).
Show that \(X^{2^2} - X + 1\) is irreducible in \(\mathbb{F}_2[X]\).
Show that if \(f(X)\) is irreducible in \(\mathbb{F}_p[X]\), then \(n = 1\) or \(n = 2 = p\).
Solution
We will prove that the two cases in part (d) are exactly the irreducible cases.
1. For each positive integer \(m\), the roots in \(K\) of\(X^{p^m}-X\) form a field with \(p^m\) elements.
The polynomial splits in the algebraically closed field \(K\), and its derivative is \(-1\), so its \(p^m\) roots are distinct. In characteristic \(p\), the binomial theorem and iteration give \((u+v)^{p^m}=u^{p^m}+v^{p^m}\). The fixed elements of the map \(u\mapsto u^{p^m}\) are therefore closed under addition and subtraction, and also under multiplication and nonzero inversion. They contain \(0,1\), so they form a field with exactly \(p^m\) elements. We denote this subfield of \(K\) by \(\mathbb F_{p^m}\). Its degree over \(\mathbb F_p\) is \(m\), by counting elements in a finite-dimensional vector space over \(\mathbb F_p\).
2. The two assertions in part (a) hold.
Write \(q=p^n\). Since \(f(\alpha)=0\), we have \(\alpha^q=\alpha-1\). For \(a\in\mathbb F_q\), step 1 gives \(a^q=a\). Therefore \[ f(\alpha+a)=\alpha^q+a^q-\alpha-a+1=0. \] Put \(\beta=\alpha^p-\alpha+1\). Taking \(q\)th powers gives \[ \begin{aligned} \beta^q &=(\alpha^q)^p-\alpha^q+1\\ &=(\alpha-1)^p-(\alpha-1)+1\\ &=\alpha^p-\alpha+1=\beta. \end{aligned} \] Hence \(\beta\in\mathbb F_q\), again by step 1.
3. Every root of \(f\) lies in \(\mathbb F_{p^{np}}\).
Starting with \(\alpha^{p^n}=\alpha-1\), induction on \(j\geq1\) gives \[ \alpha^{p^{jn}}=\alpha-j, \] where the integer \(j\) is viewed in the prime field. Indeed, raising the equality to the \(p^n\)th power subtracts one more, because every element of the prime field is fixed by Frobenius. At \(j=p\) we obtain \(\alpha^{p^{np}}=\alpha\), proving the claim.
4. The polynomial \(X^p-X+1\) is irreducible over \(\mathbb F_p\),proving part (b).
Let \(\alpha\) be any root. With \(n=1\), step 3 puts \(\mathbb F_p(\alpha)\) inside \(\mathbb F_{p^p}\). Its degree \(d\) over \(\mathbb F_p\) divides \(p\) by the tower law, so \(d=1\) or \(p\). But \(a^p-a+1=1\) for every \(a\in\mathbb F_p\), so \(\alpha\notin\mathbb F_p\) and \(d\ne1\). Thus \(d=p\). The monic minimal polynomial of \(\alpha\) divides \(X^p-X+1\) and has the same degree, so it equals that polynomial.
5. The polynomial \(X^4+X+1\) is irreducible over \(\mathbb F_2\),proving part (c).
It takes value \(1\) at both \(0\) and \(1\), so it has no linear factor. The only monic irreducible quadratic over \(\mathbb F_2\) is \(q(X)=X^2+X+1\): a monic quadratic \(X^2+aX+b\) without a root at zero must have \(b=1\), and absence of a root at one then forces \(a=1\).
Modulo \(q\), the residue class \(x\) satisfies \(x^2=x+1\) and \(x^4=(x+1)^2=x^2+1=x\). Consequently \(x^4+x+1=1\), so \(q\) does not divide the polynomial. Any reducible quartic without a linear factor is a product of two irreducible quadratics. The only possible quadratic factor has just been excluded, so the quartic is irreducible.
6. No other cases are irreducible, proving part (d).
If \(f\) is irreducible and \(\alpha\) is a root, then \([\mathbb F_p(\alpha):\mathbb F_p]=\deg f=p^n\). Step 3 and the tower law imply \[ p^n\mid np,\qquad\text{hence }p^{n-1}\mid n. \] For \(n\geq3\), the inequality \(2^{n-1}>n\) holds: it is true at \(n=3\), and doubling the left side preserves strict inequality at the next integer since \(2n\geq n+1\). Then \(p^{n-1}\geq2^{n-1}>n\), contradicting the divisibility. Thus \(n=1\) or \(2\). In the latter case the divisibility gives \(p\mid2\), hence \(p=2\). Steps 4 and 5 establish irreducibility in both surviving cases, completing every part.