How do I extend the $2$-adic absolute value to prove Monsky's Theorem?

Question

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?

Answer 1

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.