## You are here

Homenon-isomorphic completions of $\mathbb{Q}$

## Primary tabs

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

No field $\mathbb{Q}_{p}$ of the $p$-adic numbers ($p$-adic rationals) 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 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 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.

## Mathematics Subject Classification

13J10*no label found*13A18

*no label found*12J20

*no label found*13F30

*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff