# 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 $\mathbb{Q}$ NonisomorphicCompletionsOfmathbbQ 2013-03-22 14:58:17 2013-03-22 14:58:17 pahio (2872) pahio (2872) 9 pahio (2872) Theorem msc 13J10 msc 13A18 msc 12J20 msc 13F30 PAdicCanonicalForm $p$-adic numbers