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Revision difference : non-isomorphic completions of $\mathbb{Q}$ |
| Version 4 |
Version 3 |
| No field $\mathbb{Q}_p$ of the {\em $p$-adic numbers} (\PMlinkname{$p$-adic rationals}{PAdicIntegers}) is isomorphic with the field $\mathbb{R}$ of the real numbers. |
No field $\mathbb{Q}_p$ of the {\em $p$-adic numbers} (\PMlinkname{$p$-adic rationals}{PAdicIntegers}) is isomorphic with the field $\mathbb{R}$ of the real numbers. |
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| {\em Proof.} \,Let's assume the existence of an isomorphy \,$f:\,\mathbb{R}\to \mathbb{Q}_p$\, for some positive prime number $p$. \,If we denote \,$f(\sqrt{p}) = a$, \,then we obtain |
{\em Proof.} \,Let's assume the existence of an isomorphy \,$f:\,\mathbb{R}\to \mathbb{Q}_p$\, for some positive prime number $p$. \,If we denote \,$f(\sqrt{p}) = a$, \,then we obtain |
| $$a^2 = (f(\sqrt{p}))^2 = f((\sqrt{p})^2) = f(p) = p,$$ |
$$a^2 = (f(\sqrt{p}))^2 = f((\sqrt{p})^2) = f(p) = p,$$ |
| because the isomorphy maps the elements of the prime subfield on themselves. \,Thus, if \,$|\cdot|_p$\, is the \PMlinkname{normed $p$-adic valuation}{PAdicValuation} of $\mathbb{Q}$ and of $\mathbb{Q}_p$, we get |
because the isomorphy maps the elements of the prime subfield on themselves. \,Thus, if \,$|\cdot|_p$\, is the \PMlinkname{normed $p$-adic valuation}{PAdicValuation} of $\mathbb{Q}$ and of $\mathbb{Q}_p$, we get |
| $$|a|_p = \sqrt{|a^2|_p} = \sqrt{|p|_p} = \sqrt{\frac{1}{p}} = |
$$|a|_p = \sqrt{|a^2|_p} = \sqrt{|p|_p} = \sqrt{\frac{1}{p}} = |
| \frac{1}{\sqrt{p}},$$ |
\frac{1}{\sqrt{p}},$$ |
| which value is an irrational number. \,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. |
which value is an irrational number. \,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. |
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