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By abuse of notation, I'll call a complete, algebraically closed field as an "impressive field" in the article.

The smallest impressive field of characteristic 0 with Archimedean absolute value is well-known and it's $\mathbb{C}$. Also, we could compute the absolute value from the expression of elements in $\mathbb{C}$, as \begin{align} |x+iy| = x^2+y^2 \end{align} This is at least, possible to compute analytically because every real number has own decimal expression.

But if the absolute value is non-Archimedean, the existence of the smallest impressive field (of characteristic 0, containing the isomorphic copy of $\mathbb{Q}_p$) is not direct and constructive.

If the number is algebraic, we know the result is related to the norm of the number. However, for the general element in $\mathbb{C}_p$, is there any explicit formula for \begin{align} |x+iy|_p \end{align} or at least, method to compute(for example, using the algebraic closure of $\mathbb{Q}_p$ is dense in $\mathbb{C}_p$)?

NOTE: I use the fact that $\mathbb{C}_p \cong \mathbb{C}$. It may be hold when the Axiom of Choice holds. So, the calculation may not be related to the expression in $\mathbb{C}$ in some axiom systems. But, ZFC is widely used.

On the other hand, Let $K$ be the smallest impressive field containing isomorphic copy of $\mathbb{F}_p((t))$. This is the completion of algebraic closure of $\mathbb{F}_p((t))$

Is there any explicit expression or method of calculation for the absolute value of the field, $K$? The construction may be very similar to the case of $\mathbb{C}_p$, but I'm not sure.

The gists of my question are

If the element in $\mathbb{C}_p$ is expressed of some forms. Then, is there any method to compute the absolute value by using those forms?

And I ask same question for the impressive field containing the isomorphic copy of $\mathbb{F}_p((t))$(It is completion of algebraic closure of $\mathbb{F}_p((t))$).

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    $\begingroup$ One can express an element of $\Bbb C$ as $x+iy$ with unique $x,y \color{red}{\in \Bbb R}$. Now you seem to think that because (via AC) there exist isomorphisms $\Bbb C_p \simeq \Bbb C$, you can also express an arbitrary element of $\Bbb C_p$ in that form. Well you cannot, because for that you would have to know where that isomorphism sends your element of $\Bbb C_p$. Which you cannot say for most elements, because such isomorphisms cannot be constructed explicitly, they exist only for abstract reasons with AC. $\endgroup$ Sep 6 '18 at 18:56
  • $\begingroup$ @TorstenSchoeneberg Thanks for comment. Now I realize what I've been taking lightly. $\endgroup$
    – ChoMedit
    Sep 7 '18 at 0:22
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Let me make a stab at this, even though I’m hardly an expert.

We have $\Bbb Q_p\subset\overline{\Bbb Q_p}\subset\widehat{\overline{\Bbb Q_p}}=\Bbb C_p$. The second field is an algebraic closure of $\Bbb Q_p$, and the third is the completion of that, known to be algebraically closed (by a theorem).

The second field, $\overline{\Bbb Q_p}$ has a canonical absolute value $|\star|_p$ extending the standard absolute value on $\Bbb Q_p$. The only role of Axiom of Choice is in the construction of the algebraic closure, not in defining the absolute value on it.

From your question, I divine that you are not familiar with the definition of the absolute value on $\overline{\Bbb Q_p}$. To talk about it, I will (partly for typographical reasons) use the corresponding additive valuation $v_p(z)$, defined to be $-\log_p\bigl(|z|_p\bigr)$, so that $v_p(p)=1$, $v(zw)=v(z)+v(w)$, and $v(z+w)\ge\min\bigl(v(z),v(w)\bigr)$. The valuation of $0$ is set at $\infty$. Now, for every finite extension $K\supset\Bbb Q_p$, there is a unique extension of $v_p$ from $\Bbb Q_p$ to $K$, namely $$ v_K(z)=v_{\Bbb Q_p}\bigl(\mathbf N^K_{\Bbb Q_p}(z)\bigr)\,, $$ where the $\mathbf N^K_{\Bbb Q_p}$ is the field-theoretic norm from $K$ down to $\Bbb Q_p$. To show that this is a valuation, you need the properties of the norm, completeness, and Hensel’s Lemma.

Now $\overline{\Bbb Q_p}$ is the union (more precisely the direct limit with respect to inclusions) of the finite extensions $K$ of $\Bbb Q_p$. In particular, to find $v(z)$ for $z\in\overline{\Bbb Q_p}$, you find a finite extension $K$ of that contains $z$, and calculate $v$ of that, using your handy-dandy formula.

So that’s the absolute value on $\overline{\Bbb Q_p}$, which is not complete under this absolute value. For that, you have to pass to the completion, $\Bbb C_p$, whose elements are, in the usual way, represented by Cauchy sequences of elements of the algebraic closure.

Now comes the place where people can get confused; I know I was confused about this for years, back when I was in school. The function $v$ takes its values, most properly, in $\Bbb Q\cup\{\infty\}$, the infinite value corresponding to the zero-value of the absolute-value function. Here’s why: From the nature of the valuation as non-Archimedean, that is $v(z+w)\ge\min\bigl(v(z),v(w)\bigr)$, a Cauchy sequence $\{z_n\}$ whose limit is not zero has the sequence $\{v(z_n)\}$ becoming eventually constant. This constant, of course, will be in $\Bbb Q$.

In other words, to get the absolute value of an element $\xi\in\Bbb C_p$, you do not need to do any limiting process beyond knowing something about a sequence whose limit is $\xi$.

Let me give you an example, first of a series whose limit is not a $\Bbb C_p$-element, then one whose limit is. The series $\sum_1^\infty p^{1/n}$ is not $p$-adically Cauchy, because the individual terms don’t get small: $v(p^{1/n})=1/n$, not approaching $\infty$, (i.e. the absolute values don’t go to zero). But the series $\sum_1^\infty p^{n\,+\,1/n}$ is Cauchy, with terms getting small (valuations going to infinity), and so represents a good element of $\Bbb C_p$. And its absolute value is that of the very first term, $p^2$, whose valuation is $2$ and whose absolute value is $\frac1{p^2}$.

For the completion of an algebraic closure of $k((t))$, for a field $k$ of characteristic $p>0$, the story is precisely the same. It is surpassingly hard to understand just the algebraic closure, at least it is for me, but the same formulas work just fine, and you’re good to go.

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  • $\begingroup$ My first idea is possibility to find some explicit $p$-adic valuation in $\mathbb{C}$(For example, if it were possible to given, we could calculate $|\pi|_p$ in here.) But, now I realize that this is weird and impossible. Thanks for verification. $\endgroup$
    – ChoMedit
    Sep 7 '18 at 0:33
  • $\begingroup$ Right. There is only one $p-adic valuation in $\Bbb C_p$, and there is no $p$-adic $\pi$. $\endgroup$
    – Lubin
    Sep 7 '18 at 14:05

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