# irreducible polynomials obtained from biquadratic fields

###### Corollary.

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

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

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

###### 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.

$\begin{array}{cc}\hfill x& =\sqrt{m}+\sqrt{n}\hfill \\ \hfill x-\sqrt{m}& =\sqrt{n}\hfill \\ \hfill {(x-\sqrt{m})}^{2}& =n\hfill \\ \hfill {x}^{2}-2\sqrt{m}x+m& =n\hfill \\ \hfill {x}^{2}+m-n& =2\sqrt{m}x\hfill \\ \hfill {({x}^{2}+m-n)}^{2}& =4m{x}^{2}\hfill \\ \hfill {x}^{4}+(2m-2n){x}^{2}+{(m-n)}^{2}& =4m{x}^{2}\hfill \\ \hfill {x}^{4}+(2m-2n-4m){x}^{2}+{(m-n)}^{2}& =0\hfill \\ \hfill {x}^{4}-2(m+n){x}^{2}+{(m-n)}^{2}& =0\mathit{\u220e}\hfill \end{array}$

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 |