irreducible polynomials obtained from biquadratic fields

Corollary.

Let $m$ and $n$ be distinct squarefree integers, neither of which is equal to $1$ . Then the polynomial

x^{4}-2(m+n)x^{2}+(m-n)^{2}

is irreducible (http://planetmath.org/IrreduciblePolynomial2) (over $\mathbb{Q}$ ).

Proof.

By the theorem stated in the parent entry (http://planetmath.org/PrimitiveElementOfBiquadraticField), $\sqrt{m}+\sqrt{n}$ is an algebraic number of degree (http://planetmath.org/DegreeOfAnAlgebraicNumber) $4$ . Thus, a polynomial of degree $4$ that has $\sqrt{m}+\sqrt{n}$ as a root must be over $\mathbb{Q}$ . We set out to construct such a polynomial.

$\begin{array}[]{rl}x&=\sqrt{m}+\sqrt{n}\\ x-\sqrt{m}&=\sqrt{n}\\ (x-\sqrt{m})^{2}&=n\\ x^{2}-2\sqrt{m}\,x+m&=n\\ x^{2}+m-n&=2\sqrt{m}\,x\\ (x^{2}+m-n)^{2}&=4mx^{2}\\ x^{4}+(2m-2n)x^{2}+(m-n)^{2}&=4mx^{2}\\ x^{4}+(2m-2n-4m)x^{2}+(m-n)^{2}&=0\\ x^{4}-2(m+n)x^{2}+(m-n)^{2}&=0\qed\end{array}$

Title	irreducible polynomials obtained from biquadratic fields
Canonical name	IrreduciblePolynomialsObtainedFromBiquadraticFields
Date of creation	2013-03-22 17:54:22
Last modified on	2013-03-22 17:54:22
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	5
Author	Wkbj79 (1863)
Entry type	Corollary
Classification	msc 12F05
Classification	msc 12E05
Classification	msc 11R16
Related topic	ExamplesOfMinimalPolynomials
Related topic	BiquadraticEquation2