The existence of the exponential on $\mathbb{C}$ has a very basic, yet very strong consequence : $(\mathbb{C}^*,\cdot)$ is a quotient of $(\mathbb{C},+)$. This question is concerned with fields $K$ such that $K^*$ is a quotient of $K$ ; that is, with the existence of a surjective group morphism $K\to K^*$. I will refer to such morphisms as "exponentials" on $K$. (I know that the notion of exponential fields exists, but I only consider the surjective case.)

The existence of such a map is not benign : since the additive group $K$ is $q$-divisible for all $q\neq char(K)$, it implies that $K^*$ is also $q$-divisible. In characteristic $0$, this actually implies that $K$ is algebraically closed (I suspect that in characteristic $p$ this implies that $K$ is separably closed, but I'll focus on zero characteristic for now). EDIT : That was false, but it doesn't really matter since Eric's answer gives a construction when $K^*$ is divisible.

Question 1: It's a basic fact that algebraically closed fields of characteristic zero and with the same transcendance degree over $\mathbb{Q}$ are isomorphic. So $\mathbb{C}_p \simeq \mathbb{C}$ for all $p$, and thus $\mathbb{C}_p$ admits (at least one) exponential map. On the other hand, isomorphisms $\mathbb{C}\to \mathbb{C}_p$ are highly non-constructible objects, so this does not give us any clue about what an exponential on $\mathbb{C}_p$ may look like.

Can such an exponential map $\mathbb{C}_p\to \mathbb{C}_p^*$ be explicitly defined ? In particular, does it exist without the axiom of choice (or with only a weak version) ?

Question 2: If we put the axiom of choice back on :

For any countable cardinal $\kappa$, do algebraically closed fields of transcendance degree $\kappa$ over $\mathbb{Q}$ admit a surjective exponential ? (In particular, what about $\overline{\mathbb{Q}}$ ?)

We already know from $\mathbb{C}$ that this is true for $\kappa = 2^{\aleph_0}$ so since "algebraically closed fields of characteristic zero with a surjective exponential" may be countably axiomatized in first-order logic, Löwenheim-Skolem implies that there are models of all infinite cardinality, so in particular question 2 has a positive answer for all uncountable $\kappa$, and for some countable $\kappa$. But since all countable $\kappa$ give models of the same cardinality $\aleph_0$, we cannot use that remark to answer the question.

  • $\begingroup$ What's $\Bbb{C}_p$? $\endgroup$ – Rob Arthan Apr 25 '16 at 23:44
  • 2
    $\begingroup$ @RobArthan It's the completion of $\overline{\mathbb{Q}_p}$, and happens to be algebraically closed, so it's the smallest extension of $\mathbb{Q}$ that is both algebraically closed and complete (for a norm extending the $p$-adic norm). It's considered the natural equivalent of $\mathbb{C}$ when working with the $p$-adic norm instead of the archimedian one. $\endgroup$ – Captain Lama Apr 26 '16 at 8:54
  • 1
    $\begingroup$ It is consistent that any homomorphism between Polish groups is continuous. So if an exponential map can be explicitly defined, it means that it is in all likelihood continuous. Whether or not it helps, I don't know. Maybe there's a theorem that there cannot be a continuous map like this. I don't know. $\endgroup$ – Asaf Karagila Apr 29 '16 at 7:29
  • 1
    $\begingroup$ um.... why does $K^*$ being $q$-divisible for all $q>0$ imply that $K$ is algebraically closed ? $\endgroup$ – mercio Apr 29 '16 at 8:09
  • $\begingroup$ @mercio I had an argument that was a bit too long for putting it in the question, but after Eric's answer I thought about it again, and it was flawed. It doesn't really matter now, Eric's answer is completely satisfactory for question 2. $\endgroup$ – Captain Lama Apr 29 '16 at 8:43

I have no idea how to answer question 1, but in the presence of the axiom of choice, everything is very simple. If $K$ is an algebraically closed field of characteristic $0$, then the additive group of $K$ is isomorphic to a direct sum of $|K|$ copies of $\mathbb{Q}$ (this is trivial when $K$ is uncountable; when $K$ is countable note that it must have infinite degree over $\mathbb{Q}$ to be algebraically closed). The torsion subgroup of the multiplicative group (i.e., the roots of unity) is $\mathbb{Q}/\mathbb{Z}$, and since $\mathbb{Q}/\mathbb{Z}$ is injective, the multiplicative group splits as a direct sum $\mathbb{Q}/\mathbb{Z}\oplus A$ for some group $A$. The group $A$ is torsion-free and divisible and hence a $\mathbb{Q}$-vector space, and its dimension is easily seen to be $|K|$ (this is trivial when $K$ is uncountable; when $K$ is countable note that the prime integers are an infinite linearly independent set).

Thus $K^*$ is a direct sum of $\mathbb{Q}/\mathbb{Z}$ and $|K|$ copies of $\mathbb{Q}$. There is clearly a surjective homomorphism from $K$ to this direct sum, since there is a surjective homomorphism $\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}$ and $|K|$ is infinite. In fact, there is a surjective homomorphism whose kernel is cyclic, just like in the case of $\mathbb{C}$.

Note that all that this argument needs from the axiom of choice is for $K$ to be well-orderable. So if you know $K$ is well-orderable without the axiom of choice, you get a surjective exponential without choice. In particular, for instance, $\overline{\mathbb{Q}}$ has a surjective exponential which you can actually write down (in terms of explicit bases for $\overline{\mathbb{Q}}$ and $\overline{\mathbb{Q}}^*$), though this will of course still not be very "natural".

Note also that it is not actually true that $K^*$ being divisible implies $K$ is algebraically closed. Indeed, this just says $K$ is closed under taking radicals, but since not all polynomials can be solved by radicals, this will not generate all roots of polynomials. The above discussion, however, applies perfectly well to any field of characteristic $0$ such that $K^*$ is divisible (with the modification that the group of roots of unity might be a subgroup of $\mathbb{Q}/\mathbb{Z}$ rather than all of $\mathbb{Q}/\mathbb{Z}$; however, as a divisible subgroup, it will be a direct summand of $\mathbb{Q}/\mathbb{Z}$, and hence still a quotient of $\mathbb{Q}$).

Finally, I would note that almost no field $K$ of characteristic $p>0$ has a surjective exponential. Indeed, every element of $K$ is annihilated by $p$, so this would mean $x^p=1$ for all $x\in K^*$. This implies $x=1$ for all $x\in K^*$, so the only such field is $K=\mathbb{F}_2$.

  • $\begingroup$ Thanks, very nice answer ! I'll wait to see if someone has something interesting to say about question 1 before accepting your answer. $\endgroup$ – Captain Lama Apr 29 '16 at 8:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.