The following post is a digested version of a question I asked on math.SE a few months ago.

To every field $F = (F, {+}, {*})$, we can associate two natural groups. These are the additive group $F^+ = (F, {+})$ and the multiplicative group of units $F^\times = (F \smallsetminus \{0\}, {*})$. A fun question to ask, especially of someone just getting started on basic group and ring theory, is whether or not these two groups are ever isomorphic for any field.

If you haven’t seen this question before, feel free to try it yourself! I can wait.

The snappiest proof I’ve seen goes like this. Suppose for a contradiction that $\phi : F^+ \mathrel{\tilde\longrightarrow} F^\times$ is an isomorphism of groups. Then the equation $x + x = 0$ has exactly as many solutions as $y^2 = 1$. Now proceed in two cases.

Clearly $x = 0$ is a solution to $2x = 0$. If there are any other solutions, then $F$ must have characteristic two, i.e. $2\cdot1 = 2 = 0$. However, in characteristic two, $y^2 - 1 = (y-1)^2$ has only a single repeated root, $y = 1$. If $x = 0$ is the only solution, then the characteristic of $F$ is not two, and then $y^2 - 1 = (y-1)(y+1)$ has two roots, $1$ and $-1$.

In either case, the number of solutions disagrees, so $\phi$ cannot exist. Thus, for all fields $F$, $F^+ \not\cong F^\times$.

Alright, cool. Here’s a different question. Does there exist a pair of fields $E$ and $F$ such that $E^+ \cong F^\times$? Well, that’s easy. We can find an example in the finite fields: $\def\F{\mathbb F}\F_2^+ \cong \F_3^\times$!

That’s a bit unsatisfying, so let’s take a stab at characterizing all of them. We can use our casework from the first question to start off.

If the characteristic of $E$ is 2, then there are $\def\card#1{\lvert#1\rvert}\card E$ solutions to $2x = 0$, so the only thing that works is giving $F$ characteristic 2 and $\card E = 2$. Then $\card F = 3$ and we see precisely the example I gave above.

Now we may assume that $F$ has characteristic two and $E$ doesn’t. If $E$ has characteristic $p \ge 3$, then every element satisfies $px = 0$, so the nonzero elements of $F$ must be solutions to $y^p = 1$. From here it is easy to finish: $\card F = p+1 = 2^n$ for some $n$, so whenever $p$ is a Mersenne prime, we have $\F_p^+ \cong \F_{2^n}^\times$.

This leaves the case when $E$ has characteristic zero. This is where things get a bit jargony, so I recommend some knowledge of group theory and field theory, and maybe a bit of comfort around model theory. If you wish to leave now then the punchline is this: we can construct such a pair of fields, thanks to Adler (1978), but it takes a lot of high-powered tech.

$E$ has characteristic zero, so it is torsion-free. All fields of characteristic zero contain (a copy of) $\def\Q{\mathbb Q}\Q$, so that $E/\Q$ is a field extension, i.e. $E^+ \cong \Q^{(I)} := \bigoplus_{i \in I} \Q$ is a rational vector space, with basis indexed by some set $I$. The Wikipedia page for divisible group makes for some fun reading at this point.

At this point there’s not a lot we can say without invoking some fancy technology. So we’ll do that, by taking a look at a lemma from the following paper.

Adler, Allan. On the multiplicative semigroups of rings. Comm. Algebra 6 (1978), no. 17, 1751-1753. doi:10.1080/00927877808822318

This paper is not about this problem, but instead concerns itself with the model theory of multiplicative groups of fields. Here is more or less the result we care about.

Lemma 2. There exists a field $F$ such that $F^\times$ is torsion-free and divisible.

To prove this, we’re going to need ultraproducts. Let $P = \{2,3,5,...\}$ be the set of primes. A nonprincipal ultrafilter on $P$ is a collection $U \subseteq 2^P$ of subsets satisfying:

1. for all $A \subseteq P$, exactly one of $A$ and $P \smallsetminus A$ belongs to $U$,
2. if $A \in U$ and $A \subseteq B$, then $B \in U$,
3. if $A,B \in U$ then $A \cap B \in U$,
4. no finite set belongs to $U$.

Intuitively $U$ is a schema for classifying every partition of $P$ as having a small side and a big side (that’s condition 1) in a coherent way (that’s 2 and 3). Condition 4 is the additional stipulation that you’re not cheating, where by cheating I mean saying something like “a subset is big iff it contains my favourite element $p_0 \in P$”. Proving the existence of nonprincipal ultrafilters on arbitrary infinite sets requires some form of the Axiom of Choice. But we don’t care about that so we’ll just assume it.

Now for each $p \in P$, let $K_p = \F_{2^p}$. We’re going to use the ultrafilter to take a quotient on the ring-theoretic direct product $\prod_p K_p$, whose elements are $P$-indexed sequences $a = (a_p)_{p \in P}$. Namely, we’ll say two elements $a,b \in \prod_p K_p$ are $U$-mostly equal if they agree on a $U$-big set, that is, $\{ p \in P : a_p = b_p \} \in U$. The quotient $F_0 = \prod_p K_p \big/ U$ of the direct product by this equivalence relation is called an ultraproduct, and can be endowed with ring structure of its own.

What makes ultraproducts useful is the Fundamental Theorem of Ultraproducts, due to Łoś. It states that a first-order formula is true in $F_0$ iff it is true for a $U$-big collection of the $K_p$. The proof is an unenlightening structural induction, stemming from the definition of the quotient. In any case, we see that $F_0$ is a field, because for every element $a \in F_0$, either it has $U$-many zeroes and hence is equal to zero, or it has $U$-many nonzeroes and we only have to invert those. In fact, we can identify $F_0^\times$ with $\prod_p K_p^\times/U$ by the same reasoning.

If it wasn’t yet clear from context, this ultraproduct is our candidate for field whose multiplicative group is torsion-free and divisible. It remains to show that for all $n \ge 2$, $\forall x\mathpunct:(x^n = 1 \to x = 1)$ and $\forall x\mathpunct:\exists y\mathpunct: x = y^n$. And this is where we use the cleverness of our definition of $K_p$.

Proposition. If $p$ and $q$ are distinct primes, then $\card{K_p} = 2^p - 1$ and $\card{K_q} = 2^q - 1$ are relatively prime.

Proof. Run the Euclidean algorithm on $p > q$. Because they are prime, they are relatively prime, and the sequence of remainders will end in 1.

Then we can run a parallel computation that implies the GCD of $2^p-1$ and $2^q-1$ is equal to $2^{\gcd(p,q)} - 1 = 2^1 - 1 = 1$, as shown above. ∎

From the proposition, it follows that for all $n \ge 2$, there are only finitely many $p \in P$ such that the $K_p$ has nontrivial $n$-torsion or $n$-th roots don’t exist. (This is some easy group theory, and if you really want to be pedantic then it’s an exercise.) All finite subsets of $P$ are $U$-small, so by the Fundamental Theorem of Ultraproducts, $F_0^\times$ satisfies both $\forall x\mathpunct:(x^n = 1 \to x = 1)$ and $\forall x\mathpunct:\exists y\mathpunct: x = y^n$ for all $n \ge 2$. It follows that $F_0^\times$ is a rational vector space. This concludes the proof of Lemma 2. ∎

There easily exists a field $E$ whose additive group is isomorphic to $F_0^\times$—just take a field extension of $\Q$ of appropriate degree. So we have found one pair of infinite fields with the desired property.

Exercise. (For those of you who like ultraproducts.) Show that there is an ultrafilter $U'$ on the set of prime powers $Q$ such that the multiplicative group of the ultraproduct $\prod_q \F_q \big/ U'$ is divisible and torsion-free.

Are there any more? Well, this is where it gets a bit silly.

The theory of fields whose multiplicative groups are torsion-free and divisible is a countable first-order theory. And we have a model $F_0$ of cardinality $\card{F_0} = 2^{\aleph_0}$. So by Löwenheim–Skolem, we have models of every infinite cardinality.

If the cardinality is uncountable, then the cardinality equals the dimension of the multiplicative group as a rational vector space. For the model of countable cardinality, we need to take a bit more care to figure out its dimension.

Exercise. Let $a$ be algebraic over a finite field $\F_p$. Then $a$ has finite multiplicative order.

Proposition. For finite $n \ge 1$, there does not exist an $F$ such that $F^\times \cong \Q^n$.

Proof. (Due to math.SE user Starfall.) We have seen that $F$ must have characteristic two. If $a \in F^\times$ is algebraic over the prime field $\F_2$, then by the exercise it has finite order, and hence must be 1. So there exists a transcendental over $\F_2$. Thus there is an embedding of fields $\F_2(t) \hookrightarrow F$.

$\F_2[t]$ is a PID, so $\F_2(t)^\times \cong \def\Z{\mathbb Z}\Z^{(\aleph_0)}$, the direct sum of countably many copies of $\Z$. So the field embedding induces an embedding of groups $\Z^{(\aleph_0)} \cong \F_2(t)^\times \hookrightarrow F^\times \cong \Q^n$. The domain has $\Z$-linearly independent sets of arbitrary size, but the codomain is too small for that, so this is a contradiction. ∎

Using this proposition we can conclude that the countable model $F$ must have $F^\times \cong \Q^{(\aleph_0)}$.

As mentioned previously, finding fields $E$ with $E^+$ isomorphic to any particular rational vector space is very easy—just toss in an appropriate number of indeterminates. So we can say that we have characterized all pairs $E^+ \cong F^\times$ up to group isomorphism:

• If $E$ has characteristic $p > 0$ and $p+1$ is a prime power, then $E = \F_p$ and $F = \F_{p+1}$.
• if $E$ has characteristic $0$ and $\dim E/\Q \ge \aleph_0$, then there exists an $F$ by the magic of Löwenheim–Skolem.

Let me close with a silly unrelated coincidence. While researching ultraproducts, in order to digest Allan Adler’s paper, I came across another short paper by a contemporary, Andrew Adler, on cardinalities of ultraproducts.

Adler, Andrew. The cardinality of ultrapowers—an example. Proc. Amer. Math. Soc. 28 (1971), 311-312. doi:10.1090/S0002-9939-1971-0280361-9.