irreducible polynomials obtained from biquadratic fields
Corollary.
Let and be distinct squarefree integers, neither of which is equal to . Then the polynomial
is irreducible (http://planetmath.org/IrreduciblePolynomial2) (over ).
Proof.
By the theorem stated in the parent entry (http://planetmath.org/PrimitiveElementOfBiquadraticField), is an algebraic number of degree (http://planetmath.org/DegreeOfAnAlgebraicNumber) . Thus, a polynomial of degree that has as a root must be over . We set out to construct such a polynomial.