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I'm reading through "$p$-adic Numbers, $p$-adic Analysis, and Zeta-Functions" by Koblitz to learn about p-adic numbers. In chapter 3, he describes the construction of $\Omega$ (a.k.a. $\Omega_p$), the completion of the algebraic closure $\overline{\mathbb Q}_p$ of $\mathbb Q_p$. I think this is also called $\mathbb C_p$.

Could one obtain (a field isomorphic to) $\Omega$ more directly by simply completing $\overline{\mathbb Q}$ with respect to the $p$-adic norm $|~~|_p$? The way described in Koblitz starts with $\mathbb Q$ and the norm $|~~|_p$, completes it, constructs the algebraic closure, and then completes it again.

More specifically, the norm $|~~|_p$ should extend from $\mathbb Q$ to $\overline{\mathbb Q}$ in the same way Koblitz describes it extending from $\mathbb Q_p$ to $\overline{\mathbb Q}_p$ (using $\mathbb N_{\mathbb Q(\alpha) / \mathbb Q}$). Then completing $\overline{\mathbb Q}$ with respect to $|~~|_p$ should preserve the property of being algebraically closed using essentially the same proof.

Will this work, or will something go wrong?

Zachary
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    The problem with the approach you suggest is that you forgot to define $|\cdot|_p$ on $\overline{\mathbf Q}$. It's true that $\overline{\mathbf Q}$ is dense in ${\mathbf C}_p$ (same thing as $\Omega_p$), so in a sense ${\mathbf C}_p$ is a $p$-adic completion of $\overline{\mathbf Q}$, but good luck giving a direct definition of $|\cdot|_p$ in one stroke on $\overline{\mathbf Q}$ without using ${\mathbf C}_p$. In case you don't see what I'm getting at, please tell us what $|1+2i|_5$ and $|1-2i|_5$ are. – KCd Jan 11 '14 at 03:02
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    Your idea that $|\cdot|_p$ extends from the non-complete field ${\mathbf Q}$ to its algebraic extension by using the norm is incorrect. Look at the proof that the definition of the extension of $|\cdot|_p$ from ${\mathbf Q}_p$ to its finite extensions using the norm is an absolute value and figure out where it breaks down if you replace ${\mathbf Q}_p$ with $\mathbf Q$. – KCd Jan 11 '14 at 03:07
  • Thanks for your speedy answer. The definition I had in mind was $|\alpha|p = |\mathbb N{\mathbb Q(\alpha) / \mathbb Q}(\alpha)|_p^{1/[\mathbb Q(\alpha) : \mathbb Q]}$ for $\alpha \in \overline{\mathbb Q}$. I see why the arguments in Koblitz' book do not apply here, but I'm curious if there is an example showing that what I'm proposing is not a norm on $\overline{\mathbb Q}$. – Zachary Jan 13 '14 at 18:22
  • So I guess this definition has $|1 + 2i|_5 = |1 - 2i|_5 = |5|_5^{1/2} = 5^{-1/2}$. – Zachary Jan 13 '14 at 18:29
  • Ah, I see what you mean now. We have $|2|_5 = 1$ yet $|(1 + 2i) + (1 - 2i)|_5 > |1 + 2i|_5 + |1-2i|_5$ with this definition of norm. Sorry it took so long to see it. – Zachary Jan 13 '14 at 18:42
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    Exactly. The point is that $|\cdot|_p$ has a unique extension from ${\mathbf Q}_p$ to any finite (or algebraic) extension, but this is false for finite extensions of ${\mathbf Q}$. In fact, the number of extensions of $|\cdot|_p$ from ${\mathbf Q}$ to a finite extension $K/{\mathbf Q}$ is equal to the number of prime ideals in ${\mathcal O}_K$ that lie over $p$. In particular, when $K = {\mathbf Q}(i)$ there are two extensions of $|\cdot|_p$ to $K$ when $p \equiv 1 \bmod 4$. – KCd Jan 13 '14 at 19:40
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    The two extensions of $|\cdot|_5$ to ${\mathbf Q}(i)$ are related to the two primes $1+2i$ and $1-2i$ that divide $5$. For one of the extensions, $|1+2i| = 1/5$ (not $1/\sqrt{5}$) and $|1-2i| = 1$, while for the other extension $|1-2i| = 1/5$ and $|1+2i| = 1$. These are the $(1+2i)$-adic and $(1-2i)$-adic absolute values. See http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/ostrowskiQ(i).pdf for a description of all the nontrivial absolute values on ${\mathbf Q}(i)$, up to equivalence. – KCd Jan 13 '14 at 19:43
  • Excellent, thank you! This has been very helpful. – Zachary Jan 13 '14 at 23:33
  • @KCd: You argument is exactly the answer to this question and, I am pretty sure that it has helped/will help people like me. So kindly consider post this as a solution. – Bumblebee Mar 23 '20 at 00:08
  • @Bumblebee you can collect my comments into an answer if you wish. I don't care about points/rewards/etc. – KCd Mar 23 '20 at 03:03
  • Related: https://math.stackexchange.com/q/4352037/96384 – Torsten Schoeneberg Jan 26 '22 at 06:34

1 Answers1

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This is a community wiki answer based on the comments, plus some remarks of my own.

In principle, yes, you could proceed by extending the $p$-adic valuation from $\mathbb Q$ to $\overline{\mathbb Q}$, and the extending from $\overline{\mathbb Q}$ to $\mathbb C$. But you have to be careful if you want to go this route. In particular, there is not a unique way to extend the $p$-adic valuation from $\mathbb Q$ to $\overline{\mathbb Q}$.

In the comments, it was suggested that perhaps one could define

$$|\alpha|_p := |N_{{\mathbb Q}(\alpha)/{\mathbb Q}}(\alpha)|_p^{1/[{\mathbb Q}(\alpha):{\mathbb Q}]}.$$ So for example, this would give $|1+2i|_5=|1−2i|_5=|5|^{1/2}_5=5^{−1/2}$. However, then we deduce that $$|(1+2i)+(1−2i)|_5 = |2|_5 = 1 > 5^{-1/2} = \max(|1+2i|_5,|1−2i|_5),$$ contradicting the ultrametric inequality.

In general, the number of extensions of $|\cdot|_p$ from $\mathbb Q$ to a finite extension $K$ is equal to the number of prime ideals in ${\cal O}_K$ that lie over $p$. The two extensions of $|\cdot|_5$ to ${\mathbb Q}(i)$ are related to the two primes $1+2i$ and $1−2i$ that divide $5$. For one of the extensions, $|1+2i|=1/5$ (not $1/\sqrt{5}$) and $|1−2i|=1$, while for the other extension $|1−2i|=1/5$ and $|1+2i|=1$.

However, the lack of uniqueness is not a dealbreaker; for every finite extension of $\mathbb Q$ you can pick an extension of $|\cdot|_p$, and use Zorn's lemma to ensure that you're making choices that are consistent with each other.

Once you have an extension to all algebraic numbers, you can extend to transcendental numbers $t$ by the so-called Gauss norm construction: $$ |a_n t^n+\cdots +a_1 t + a_0| := \max_i |a_i|$$ and extending to rational functions of $t$ by multiplicativity.

Although this works, it is generally considered less natural than the standard approach because the extension of the $p$-adic norm from $\mathbb{Q}_p$ to $\overline{\mathbb{Q}}_p$ is unique. This uniqueness is one way in which local fields are simpler than global fields.

Timothy Chow
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