What’s the field with 25 elements?
Solution
Up to isomorphism there is a unique field with \(25=5^2\) elements, namely \(\FF_{25}\). A concrete model is \[ \FF_{25}\cong \FF_5[t]/(t^2+2). \]
1. The polynomial \(t^2+2\) is irreducible over \(\FF_5\).
A quadratic over a field is reducible iff it has a root. The squares in \(\FF_5\) are \(0,1,4\), while a root of \(t^2+2\) would satisfy \(t^2=3\). Since \(3\) is not a square modulo \(5\), there is no root.
2. Therefore the quotient is a field with \(25\) elements.
Irreducibility makes \((t^2+2)\) maximal in \(\FF_5[t]\). Every residue class has a unique representative \(a+bt\) with \(a,b\in\FF_5\), giving \(5^2=25\) elements.
Thus one may write \[ \FF_{25}=\{a+b\alpha:a,b\in\FF_5,\ \alpha^2=3\}. \]