Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\tau$?

share|cite|improve this question
(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
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
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:… – M Turgeon Jun 8 '12 at 23:30
THere is certainly no continuous isomorphism. Not even a Borel measurable isomorphism. – GEdgar Jun 9 '12 at 1:19
up vote 2 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.