- Prove that for any positive integer \(n\) there is an irreducible polynomial of degree \(n\) over \(\boldsymbol{F}_{p}\)
Solution
Fix a positive integer \(n\), and work in an algebraic closure \(\overline{\FF}_p\).
1. The set\[ K=\{a\in\overline{\FF}_p:a^{p^n}=a\} \]is a field with exactly \(p^n\) elements.
The polynomial \[ f(x)=x^{p^n}-x \] has derivative \(f'(x)=-1\), so it has no repeated roots. Since it splits in the algebraic closure and has degree \(p^n\), it has exactly \(p^n\) distinct roots; hence \(|K|=p^n\).
If \(a,b\in K\), then in characteristic \(p\), \[ (a+b)^{p^n}=a^{p^n}+b^{p^n}=a+b, \] and \[ (ab)^{p^n}=a^{p^n}b^{p^n}=ab. \] Also \(0,1\in K\), and if \(0\neq a\in K\), then \[ (a^{-1})^{p^n}=(a^{p^n})^{-1}=a^{-1}. \] Thus \(K\) is a subfield of \(\overline{\FF}_p\) with \(p^n\) elements.
2. The multiplicative group \(K^\times\) is cyclic of order \(p^n-1\).
The multiplicative group of every finite field is cyclic. Since \(|K|=p^n\), its nonzero elements form a cyclic group of order \(p^n-1\).
3. Let \(\alpha\) generate \(K^\times\). Then\[ [\FF_p(\alpha):\FF_p]=n. \]
Set \[ d=[\FF_p(\alpha):\FF_p]. \] Because \(\FF_p(\alpha)\subseteq K\), it is a finite field with \(p^d\) elements, so \[ d\le n. \] Its multiplicative group has order \(p^d-1\), and the order of \(\alpha\) must divide this number. But \(\alpha\) generates \(K^\times\), so \[ \operatorname{ord}(\alpha)=p^n-1. \] Hence \[ p^n-1\mid p^d-1. \] If \(d<n\), then \(p^d-1<p^n-1\), impossible. Therefore \(d=n\).
4. The minimal polynomial of \(\alpha\) over \(\FF_p\) is irreducible of degree \(n\).
By definition, the minimal polynomial \(m_\alpha(x)\in\FF_p[x]\) is irreducible, and \[ \deg m_\alpha=[\FF_p(\alpha):\FF_p]. \] By 3, this degree is \(n\).
5. Therefore for every positive integer \(n\) there exists an irreducible polynomial of degree \(n\) over \(\FF_p\).
The polynomial \(m_\alpha\) from 4 is such a polynomial.