Why should we not consider $\mathbb{Q}(t)$ to be a global field?

Question

I'm trying to properly understand the definition of the global field. I understand that there are two typical definitions of the global field:

1) $K$ is a global field if it is either a finite extension of $\mathbb{Q}$ or a finite extension of $\mathbb{F}_p(t)$, where $\mathbb{F}_p(t)$ is the field of rational functions in one variable over $\mathbb{F}_p$.

2) $K$ is a global field if all its completions $K_v$ at each place $v$ of $K$ is a local field, and these completions satisfy the product formula (Artin-Whaples)

---

Artin-Whaples proved the equivalence between (1) and (2). An outline of this proof is also (partially) presented in question #2 from this problem set from a graduate MIT course in number theory, which is helpful. https://math.mit.edu/classes/18.785/2017fa/ProblemSet7.pdf

I'm trying to get a better understanding of this definition of global fields and of the Artin-Whaples equivalence. In particular, why is $\mathbb{Q}(t)$ (i.e. the field of rational functions over $\mathbb{Q}$) not a global field? While it obviously fails Definition 1, why does it fail Definition 2?

Are its completions no longer local fields? Or do they fail the product formula? Or is it something more basic, e.g. I've read on Wikipedia that global fields can be realised as the fraction field of Dedekind domains in which every ideal is of finite index. Perhaps $\mathbb{Q}(t)$ fails that?

Also, if someone can give me a better intuition of how I should understand completions of rational function fields that would be very much appreciated. In particular, while I'm aware of the function field analogy, I'm not sure how to think of the completions of $\mathbb{F}_p(t)$ or $\mathbb{Q}(t)$. I also would like to know why taking a transcendental extension over a finite field gives you a global field but a transcendental extension over a countable field apparently does not. (Or is it because the two fields have different characteristics, rather than different sizes...?)

Thanks!

Answer 1

The field $\mathbb{Q}(t)$ has characteristic zero, but it is not a finite extension of $\mathbb{Q}$. Hence, it is not a global field.

Answer 2

There are many absolute values on $\Bbb{Q}(t)$.

• The trivial absolute value

• The absolute values trivial on $\Bbb{Q}$, coming from the discrete valuations telling the order of the zero/pole at each algebraic number $a\in \overline{\Bbb{Q}}$ (the product formula holds when restricting to those)

• The euclidean absolute values coming from an embedding $t\to z, \Bbb{Q}(t)\to\Bbb{C}$ where $z\in \Bbb{C}-\overline{\Bbb{Q}}$

• The $p$-adic absolute value coming from an embedding $t\to z, \Bbb{Q}(t)\to\Bbb{C}_p$ where $z\in \Bbb{C}_p-\overline{\Bbb{Q}}$

• The $p$-adic absolute values coming from an embedding $t\to p^c+a,\Bbb{Q}(t)\to \Bbb{Q}_p[a](p^c)$ where $a\in \overline{\Bbb{Q}}$ and $c\in \Bbb{R}-\Bbb{Q}$ and $|p^c|_v=p^{-c}$

• The $p$-adic absolute value coming from an embedding $\Bbb{Z}[t]\to A\subset \Bbb{Z}_p[[t]]$ where $A= \{ \sum_n a_n t^n, |a_n|_p \to 0\}$ with the absolute value $|\sum_n a_n t^n|_v=\sup_n |a_n|_p$

• $?$

As you see the set of places is uncountable.