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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!

asldjk
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    Typically, the completion of $\Bbb Q(t)$ at the absolute value given by $v = (t)$ is $\Bbb Q((t))$, so it is not a local field. – Watson Mar 26 '20 at 14:42
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    A key property that we want for global field is that the corresponding ring of integers has finite (non-trivial) quotients. But neither $\Bbb Q[t]$ nor $\Bbb Z[t]$ have this property. – Watson Mar 26 '20 at 14:44
  • Interesting. Could you say a bit more about what it means to complete a function field (for instance, why is $v=(t)$ an absolute value?) and the bit about the ring of integers of $\mathbb{Q}(t)$ having finite quotients? Or could you direct me to a reference? – asldjk Mar 26 '20 at 14:57
  • You give different definitions of a global field, but what definition of local field do you want us to assume? – Bib-lost Apr 28 '20 at 12:26
  • I'm using the definition that it is a field $K$ with a nontrivial absolute value that is locally compact under the induced topology. I am aware that we can also define a local field as as discrete valued field with finite residue field and is complete. – asldjk Apr 29 '20 at 13:57

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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.

reuns
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  • Ah thanks, the distinction you bring up btwn absolute values trivial on $\mathbb{Q}$ vs. absolute values not trivial on $\mathbb{Q}$ is an impt one. However, I don't fully see how we have an uncountable set of places. You've given many examples of absolute values induced by various embeddings of $\mathbb{Q}(t)$, and certainly there are uncountably many ways to define a transcendental extension, but how do we know that the completions wrt these absolute values aren't topologically equivalent? I think I can see how the Euclidean and $p$-adic ones are different, I'm less sure abt the rest. – asldjk Mar 27 '20 at 21:46
  • Also, forgive me if this is a basic question, but for the completions wrt the absolute values that are non-trivial on $\mathbb{Q}$, do they all fail to be locally compact? – asldjk Mar 27 '20 at 21:53
  • Each complex conjugate pair $z,\overline{z} \in \Bbb{C}-\overline{\Bbb{Q}}$ gives a different place – reuns Mar 28 '20 at 13:00
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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.

Geoffrey Trang
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    I know it's a transcendental extension of $\mathbb{Q}$, and it doesn't fit definition 1 - my question really has to do with why $\mathbb{Q}(t)$ doesn't fit definition 2. Really, I'm asking a more fundamental question as to why global fields are defined the way that they are defined. – asldjk Mar 26 '20 at 14:12
  • "Why are global fields defined the way that they are defined?" - This has to do with the words "global" and "local" - see this post. – Dietrich Burde Mar 26 '20 at 14:17
  • Yes, I saw that post, but that post is more about why we use the terms "global" and "local" to describe fields, and the accepted answer just says that the product formula characterises global fields. I understand this. So my question is: why exactly does $\mathbb{Q}(t)$ fail, e.g. the product formula? – asldjk Mar 26 '20 at 14:23
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    A rational function field can also satisfy the product formula, see here. – Dietrich Burde Mar 26 '20 at 14:55
  • Ahhh interesting, thanks. So that means that the reason why we don't consider $\mathbb{Q}(t)$ to be a global field must be because its completions aren't local fields. – asldjk Mar 26 '20 at 14:58
  • @DietrichBurde The product formula for discrete valuations trivial on $\Bbb{Q}$ (which is just saying that a polynomial of degree $d$ has $d$ zeros), but it is a very small subset of all the absolute values. – reuns Mar 27 '20 at 10:16