- Consider the polynomial \(f=x^{3}-x+1\) in \(\FF_3[x]\). Prove that \(f\) is irreducible. Calculate the degree of the splitting field of \(f\) over \(\FF_3\) and the cardinality of the splitting field of \(f\).
Solution
Let \[ f(x)=x^3-x+1\in\FF_3[x]. \]
1. The polynomial \(f\) has no root in \(\FF_3\).
Evaluate at the three elements of \(\FF_3\): \[ f(0)=1, \qquad f(1)=1-1+1=1, \qquad f(2)=8-2+1=7\equiv1\pmod3. \] Thus \(f(a)\neq0\) for every \(a\in\FF_3\).
2. Hence \(f\) is irreducible over \(\FF_3\).
A reducible cubic over a field has a linear factor, hence a root in that field. By 1, \(f\) has no root in \(\FF_3\), so it is irreducible.
3. If \(\alpha\) is a root of \(f\) in an algebraic closure, then\[ [\FF_3(\alpha):\FF_3]=3 \]and therefore\[ |\FF_3(\alpha)|=3^3=27. \]
By 2, the minimal polynomial of \(\alpha\) over \(\FF_3\) is \(f\), which has degree \(3\). Hence the simple extension has degree \(3\). A degree-\(3\) extension of the field with three elements is a three-dimensional vector space over \(\FF_3\), so it has \(3^3=27\) elements.
4. The field \(\FF_3(\alpha)\) is the splitting field of \(f\).
The field \(\FF_3(\alpha)\) has \(27\) elements. Every element \(a\) of this field satisfies \[ a^{27}=a, \] so every element is a root of \(x^{27}-x\).
Since \(f\) is irreducible of degree \(3\), it divides \(x^{27}-x\) over \(\FF_3\): equivalently, the Frobenius orbit of any root of \(f\) has length \(3\), and its roots are \[ \alpha,\alpha^3,\alpha^9. \] All three lie in \(\FF_3(\alpha)\). Thus \(f\) splits there. Since any splitting field contains a root \(\alpha\), it must contain \(\FF_3(\alpha)\), so this field is the splitting field.
5. Therefore the splitting field has degree \(3\) over \(\FF_3\) and cardinality \(27\).
This is exactly <1>3–<1>4.