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I sometimes see arguments that begin by choosing an isomorphism of fields $\tau:\overline{\mathbf{Q}}_p\simeq\mathbf{C}$, and then defining some property in terms of this isomorphism. I'm not so familiar with the technical properties; must there exist continuous isomorphisms or isometries? Where can I read up on the basic properties of such isomorphisms or how this technique is deployed?

Sample question: suppose I say that $P\in\mathbf{Q}_p[X]$ is 'pure of weight $i\in\mathbf{Z}$' if every root $\lambda\in\overline{\mathbf{Q}}_p$ of $P$ satisfies $|\tau(\lambda)|_{\mathbf{C}}=i$. Does this notion depend on $\tau$?

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(There is no continuous isomorphism, and much less an isometric one: consider the restriction of $\tau$ to $\mathbb Q$) –  Mariano Suárez-Alvarez Jun 8 '12 at 22:42
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It is only a matter of notation, but let me note that $\overline{\mathbb{Q}}_p$ is not complete, and the completion is usually denoted $\mathbb{C}_p$. –  M Turgeon Jun 8 '12 at 23:28
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In any case, if you want to learn about this field, check Koblitz's P-adic numbers, p-adic analysis, and zeta functions, especially Chapter III: books.google.ca/… –  M Turgeon Jun 8 '12 at 23:30
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THere is certainly no continuous isomorphism. Not even a Borel measurable isomorphism. –  GEdgar Jun 9 '12 at 1:19
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Your notion of purity does not depend on $\tau$ for polynomials with $\mathbb{Q}$ coefficients but does depend on $\tau$ for arbitrary polynomials with $\mathbb{Q}_p$ coefficients. Any two isomorphisms from $\overline{\mathbb{Q}_p}$ to $\mathbb{C}$ will differ by an automorphism of $\mathbb{C}$. An automorphism of $\mathbb{C}$ must interchange roots of any polynomial with $\mathbb{Q}$ coefficients, so if $P$ is pure with respect to one choice of $\tau$ it is pure with respect to an arbitrary $\tau$.

To see that it doesn't work in general, let's work over $\mathbb{Q}_7$. This field has two square roots of 2(Hensel's lemma); we call them $\pm \alpha$. We'll reserve the symbol $\sqrt{2}$ to mean the usual positive element of $\mathbb{R}$. There's a $\tau$ that takes $j$ to $\sqrt{2}$ and there's a $\tau$ that takes $j$ to $-\sqrt{2}$ (simply compose $\tau$ with any extension of the unique automorphism of $\mathbb{Q}(\sqrt{2})$ to $\mathbb{C}$. Now consider the polynomial $$ (x-1)^2 - \alpha. $$ If $\tau(\alpha) = \sqrt{2}$ then the roots of this polynomial are $1 + 2^{1/4}$ and $1 - 2^{1/4}$, which have different complex absolute values. If $\tau(\alpha) = -\sqrt{2}$ the roots are $1 + 2^{1/4}i$ and $1 - 2^{1/4}i$, which have the same complex absolute value.

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Ah, sorry, I just noticed that the absolute value has to belong to $\mathbb{Z}$. That's a pretty serious restriction; I'm agnostic as to whether it's true with that restriction. –  user29743 Jun 8 '12 at 23:50
    
This would be fine if you just replaced “5” with “7”: in fact ${\mathbb{Z}}/5{\mathbb{Z}}$ doesn’t even have a square root of 2. –  Lubin Jun 9 '12 at 3:19
    
Ok still it's a nice example. –  vgty6h7uij Jun 9 '12 at 10:07
    
Ha! fixed, sorry. –  user29743 Jun 9 '12 at 14:50
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