Let \(F\) be a finite field. Show for any positive integer \(n\) that there are irreducible polynomials of degree \(n\) in \(F[x]\).
<1>2. Assume \(n\ge2\), and let \(\overline F\) be an algebraic closure of \(F\). Define \[ E=\{\alpha\in\overline F:\alpha^{q^n}=\alpha\}. \] Then \(E\) is a field containing \(F\).
<1>3. The field \(E\) has exactly \(q^n\) elements.
<1>4. For each divisor \(d\) of \(n\), there is at most one subfield of \(E\) having \(q^d\) elements.
<1>5. Every proper intermediate field \[ F\subseteq K\subsetneq E \] has \(q^d\) elements for some proper divisor \(d<n\) of \(n\).
<1>6. The union of all proper intermediate fields between \(F\) and \(E\) has fewer than \(q^n\) elements.
<1>7. Choose \[ \alpha\in E \] outside every proper intermediate field. Then \[ F(\alpha)=E. \]
<1>8. The minimal polynomial of \(\alpha\) over \(F\) is irreducible of degree \(n\).
Solution
1. Write\[ |F|=q. \]For \(n=1\), an irreducible polynomial of degree \(1\) is immediate, for example \(x-a\) with \(a\in F\).
The field \(F\) has characteristic \(p\) for some prime \(p\), and \(q\) is a power of \(p\). If \(a,b\in E\), then Frobenius gives \[ (a+b)^{q^n}=a^{q^n}+b^{q^n}=a+b, \] and \[ (ab)^{q^n}=a^{q^n}b^{q^n}=ab. \] Also \(0,1\in E\). If \(a\in E\) is nonzero, then \[ (a^{-1})^{q^n}=(a^{q^n})^{-1}=a^{-1}. \] Hence \(E\) is a subfield of \(\overline F\).
For every \(a\in F\), we have \(a^q=a\), and therefore \(a^{q^n}=a\). Thus \(F\subseteq E\).
The polynomial \[ P(x)=x^{q^n}-x \] has derivative \[ P'(x)=q^n x^{q^n-1}-1=-1 \] in characteristic \(p\), because \(p\mid q^n\). Thus \(P\) has no repeated roots. Since \(\overline F\) is algebraically closed, \(P\) splits completely there into exactly \(q^n\) distinct roots. By definition these roots are precisely the elements of \(E\). Hence \[ |E|=q^n. \]
Let \(K\subseteq E\) be a subfield with \(|K|=q^d\). Every element \(a\in K\) satisfies \[ a^{q^d}=a. \] Therefore \(K\) is contained in the root set of \[ x^{q^d}-x. \] That polynomial has at most \(q^d\) roots, while \(K\) already has exactly \(q^d\) elements. Hence \(K\) is exactly its root set. Thus any subfield of \(E\) with \(q^d\) elements is uniquely determined.
Since \(E/F\) is a finite extension with \[ [E:F]=n, \] the tower law gives \[ [E:F]=[E:K][K:F]. \] Thus \[ d=[K:F] \] divides \(n\). Because \(K\ne E\), we have \(d<n\). Since \(|F|=q\), a \(d\)-dimensional vector space over \(F\) has \(q^d\) elements, so \[ |K|=q^d. \]
By 4 and 5, there is at most one proper intermediate field for each proper divisor \(d\) of \(n\), and such a field has \(q^d\) elements. Every proper divisor of \(n\) is at most \(n/2\). Hence the union has cardinality at most \[ \sum_{\substack{d\mid n\\ d<n}}q^d \le \sum_{j=1}^{\lfloor n/2\rfloor}q^j. \] For \(q\ge2\) and \(n\ge2\), \[ \sum_{j=1}^{\lfloor n/2\rfloor}q^j<q^n. \] Indeed, the left side is less than the full geometric sum \[ 1+q+\cdots+q^{n-1}=\frac{q^n-1}{q-1}, \] and this is at most \(q^n-1<q^n\). Thus the proper intermediate fields cannot cover \(E\).
The field \(F(\alpha)\) is an intermediate field between \(F\) and \(E\). By the choice of \(\alpha\), it cannot be proper. Hence it equals \(E\).
By 7, \[ F(\alpha)=E. \] Therefore \[ \deg m_{\alpha,F} =[F(\alpha):F] =[E:F] =n. \] By definition, the minimal polynomial \(m_{\alpha,F}\) is irreducible in \(F[x]\). Thus \(F[x]\) contains an irreducible polynomial of degree \(n\).