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In proving Monsky's Theorem, it is required that we define the $2$-adic absolute value on an arbitrary finitely generated extension of $\mathbb{Q}$ say $\mathbb{K}=\mathbb{Q}(\alpha_1,\ldots,\alpha_n)$ where $\alpha_1,\ldots,\alpha_n\in\mathbb{R}$ in such a way that it extends the established definition for $\mathbb{Q}$: for any $a/b\in\mathbb{Q}$ we write $a/b=2^na'/b'$ where $a',b'\in\mathbb{Z}$ are odd and $n\in\mathbb{Z}$ then $|a/b|_2=2^{-n}$.

I understand that any finitely generated extension of $\mathbb{Q}$ arising in this way can be written in the form $\mathbb{K}=\mathbb{Q}(\beta_1,\ldots,\beta_k,\gamma)$ where $\beta_1,\ldots,\beta_k\in\{\alpha_1,\ldots,\alpha_n\}$ form a transcendence basis for $\mathbb{K}$ over $\mathbb{Q}$ and $\gamma\in\mathbb{K}$ is algebraic over $\mathbb{Q}(\beta_1,\ldots,\beta_k)$. As such, it suffices to describe how to perform an extension of the $2$-adic absolute value when the extension is simple.

In the case that we want a simple transcendental extension of the $2$-adic absolute value (with primitive element $t$) I have seen in multiple sources that we can use $$ |a_0+a_1t+\cdots+a_mt^m|_2=\max_i|a_i|_2 $$ and extend to rational functions by taking quotients of this.

But I am stuck for the case of simple algebraic extensions because all known proofs I have found rely on the base field being complete to deduce that the function $$ |x|_2=|N(x)|_2^{1/n} $$ defines a valid extension where $n$ is the degree of the extension. But a purely transcendental extension of the above form is not complete. Does anyone know how we can construct an extension to the $2$-adic absolute value in this case?

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  • $\begingroup$ I'm confused by your question. You ask about simple algebraic extensions, and about transcendental extensions. Which one do you want? $\endgroup$
    – LSpice
    Mar 6, 2023 at 23:44
  • $\begingroup$ In, e.g., Fesenko and Vostokov, Chapter 2, Theorem 2.6, they say that the extensions of a discrete valuation $v$ from $F$ to a simple algebraic extension by the root of an irreducible polynomial $f$ are parameterised by the factors of $f$ over the completion $\bar F$, with the extensions of $v$ being restrictions of the extensions of $\bar v$ to the relevant extension of $\bar F$. Theorem 2.8 is a weakening of completeness: if $F$ is Henselian, then there is a unique such extension. But that does not seem to apply here. $\endgroup$
    – LSpice
    Mar 7, 2023 at 13:28
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    $\begingroup$ Isn't it just the same as for an extension $K/k$ of number fields? Complete $k$ with respect to your valuation $v$. Then $K\otimes k_v$ is a product of fields, each having an extension by the local argument. Pick one and restrict it to $K$. $\endgroup$ Mar 7, 2023 at 14:45

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You are asking how to extend $|\cdot|_2$ from $\mathbf Q$ to a finite extension $L = \mathbf Q(\gamma)$. In the ring of integers $\mathcal O_L$, let $(2) = \mathfrak p_1^{e_1}\cdots \mathfrak p_g^{e_g}$. Then on $L$ we can define a $\mathfrak p$-adic absolute value where $|\alpha|_\mathfrak p = (1/2)^{{\rm ord}_\mathfrak p(\alpha)}$ for each prime $\mathfrak p = \mathfrak p_i$. Since ${\rm ord}_\mathfrak p(2) = e$, where $e = e(\mathfrak p|2)$, we have $|2|_\mathfrak p = (1/2)^{e}$. We want this to be $1/2$ in order to be compatible with $|\cdot|_2$ on $\mathbf Q$, so take $e$-th roots: redefine $|\cdot|_\mathfrak p$ to be $|\alpha|_\mathfrak p = (1/2)^{{\rm ord}_\mathfrak p(\alpha)/e}$ for $\alpha \in L$. This is a non-archimedean absolute value on $L$ that restricts to $|\cdot|_2$ on $\mathbf Q$, so to each $\mathfrak p$ we get an extension of $|\cdot|_2$. (In fact all extensions of $|\cdot|_2$ from $\mathbf Q$ to $L$ arise in this way, but that isn't important for you.)

Here is a second method that takes advantage of knowing already how to extend $|\cdot|_2$ from $\mathbf Q_2$ to finite extensions of $\mathbf Q_2$. Let $f(x)$ be the minimal polynomial of $\gamma$ over $\mathbf Q$ and $g(x)$ be an irreducible factor of $f(x)$ in $\mathbf Q_2[x]$. The field $\mathbf Q_2(\gamma')$, where $\gamma'$ is a root of $g(x)$, admits an extension of $|\cdot|_2$ to it from $\mathbf Q_2$ and you can embed $\mathbf Q(\gamma)$ into $\mathbf Q_2(\gamma')$ by mapping $\gamma$ to $\gamma'$. This is an isomorphism between $\mathbf Q(\gamma)$ and $\mathbf Q(\gamma')$, so the $2$-adic absolute value on $\mathbf Q(\gamma')$ as a subfield of $\mathbf Q_2(\gamma)$ can be turned into an absolute value on $\mathbf Q(\gamma)$ via that isomorphism. Similar reasoning shows you can extend an absolute value $|\cdot|$ on an arbitrary field $K$ to any simple algebraic extension of $K$ by first extending $|\cdot|$ from $K$ to its completion under $|\cdot|$ in the usual way, then to a suitable simple algebraic extension of that completion by the norm formula that you know about, and then finally to the original simple algebraic extension of $K$ by using a field embedding of it into the finite extension of the completion of $K$. This method is a more detailed version of the method suggested in a comment by Wuthrich above.

You can build up your field $\mathbf K$ in two ways: using the tower $$ \mathbf Q \subset \mathbf Q(\gamma) \subset \mathbf Q(\gamma,\beta_1) \subset \cdots \subset \mathbf Q(\gamma,\beta_1,\ldots,\beta_k) = \mathbf K $$ where you adjoin $\gamma$ as the first step or by the tower $$ \mathbf Q \subset \mathbf Q(\beta_1) \subset \cdots \subset \mathbf Q(\beta_1,\ldots,\beta_k) \subset \mathbf Q(\beta_1,\ldots,\beta_k,\gamma) = \mathbf K $$ where you adjoin $\gamma$ as the last step. Each step in either tower adjoins to a field a single element that is either algebraic over it or transcendental over it, so you just need to be able to extend a non-archimedean absolute value from a field to a simple algebraic extension or a simple transcendental extension. The algebraic extension case is described above. For the transcendental extension case, you can use the max-formula you already mentioned in your question: see Theorem 8 here for the extension of $|\cdot|_p$ from $\mathbf Q$ to $\mathbf Q(T)$ (it focuses on extending $|\cdot|_p$ from $\mathbf Q$ to $\mathbf Q[T]$, but then it's easy to go to the fraction field $\mathbf Q(T)$), and the argument there works with $(\mathbf Q,|\cdot|_p)$ replaced by any field equipped with a non-archimedean absolute value.

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    $\begingroup$ I think the question is how to extend from a purely transcendental extension of $\mathbb Q$. $\endgroup$
    – LSpice
    Mar 8, 2023 at 9:03
  • $\begingroup$ @LSpice look at the question again. It specifically says in the last paragraph "But I am stuck for the case of simple algebraic extensions". $\endgroup$
    – KConrad
    Mar 8, 2023 at 13:21
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    $\begingroup$ @‍KConrad, re, yes, but, given that the aim is to construct an extension on an arbitrary finitely generated extension field, and that the approach as described is to think of such an extension field as a simple algebraic extension of a transcendental extension of $\mathbb Q$, I still think that the goal is to extend from a purely transcendental extension of $\mathbb Q$, to a simple algebraic extension of that transcendental extension. $\endgroup$
    – LSpice
    Mar 8, 2023 at 13:55
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    $\begingroup$ @LSpice I updated my reply to account for that possibility. $\endgroup$
    – KConrad
    Mar 8, 2023 at 14:13
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    $\begingroup$ The first argument applies to the case of a general algebraic extension $L/F$ of a field $F$ with a discrete absolute value if one defines $\mathcal O_F$ to consist of elements with absolute value at most $1$, a DVR, and $\mathcal O_L$ to consist of elements integral over $\mathcal O_F$. $\endgroup$
    – Will Sawin
    Mar 8, 2023 at 14:48

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