# non-isomorphic completions of $\mathbb{Q}$

No field ${\mathbb{Q}}_{p}$ of the $p$-adic numbers ($p$-adic rationals (http://planetmath.org/PAdicIntegers)) is isomorphic with the field $\mathbb{R}$ of the real numbers.

Proof. Let’s assume the existence of a field isomorphism $f:\mathbb{R}\to {\mathbb{Q}}_{p}$ for some positive^{} prime number^{} $p$. If we denote $f(\sqrt{p})=a$, then we obtain

$${a}^{2}={(f(\sqrt{p}))}^{2}=f({(\sqrt{p})}^{2})=f(p)=p,$$ |

because the isomorphism maps the elements of the prime subfield^{} on themselves. Thus, if $|\cdot {|}_{p}$ is the normed $p$-adic valuation^{} (http://planetmath.org/PAdicValuation) of $\mathbb{Q}$ and of ${\mathbb{Q}}_{p}$, we get

$${|a|}_{p}=\sqrt{{|{a}^{2}|}_{p}}=\sqrt{{|p|}_{p}}=\sqrt{\frac{1}{p}},$$ |

which value is an irrational number as a square root of a non-square (http://planetmath.org/SquareRootOf2IsIrrationalProof) rational. But this is impossible, since the value group of the completion ${\mathbb{Q}}_{p}$ must be the same as the value group ${|\mathbb{Q}\setminus \{0\}|}_{p}$ which consists of all integer powers of $p$. So we conclude that there can not exist such an isomorphism.

Title | non-isomorphic completions of $\mathbb{Q}$ |
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Canonical name | NonisomorphicCompletionsOfmathbbQ |

Date of creation | 2013-03-22 14:58:17 |

Last modified on | 2013-03-22 14:58:17 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 9 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 13J10 |

Classification | msc 13A18 |

Classification | msc 12J20 |

Classification | msc 13F30 |

Related topic | PAdicCanonicalForm |

Defines | $p$-adic numbers |