#multiplicative group

Not to start every one of these posts with ‘okay so’ but okay so with a field you naturally get two groups for free: its additive group and its multiplicative group (also known as its group of units). There are some fields for which these have just about the same group structure, and some for which they don’t.

Let’s first look at the field of real numbers ℝ. Let ℝ+ denote its additive group and ℝ* its multiplicative group. Using the familiar exponential function we have

so exp is a group homomorphism between these two groups. The exponential function only maps to the positive real numbers, so it isn’t quite an isomorphism. It’s as if the multiplicative group consists of two copies of ℝ+: one for the positives and one for the negatives (0 is not an element of ℝ*, as it has no multiplicative inverse).

For any field K define the group A(K) by

That is to say, the direct product of its additive group with the cyclic group of two elements, here conveniently represented by the multiplicative group of {±1}.

If we now map each element of A(ℝ) to ℝ* with the function (x,u) ↦ ue^x, then we get our desired isomorphism (which means these two groups have essentially the same structure).

Let’s now look at the rational numbers ℚ. The next question would be is A(ℚ) isomorphic to ℚ*? We cannot define the exponential function to take only rational values. Let’s restrict ourselves to proving that ℚ+ is not isomorphic to the positive multiplicative group ℚ*_{>0} (the proof works directly for A(ℚ) and ℚ* but it’s a bit more verbose).

Let’s assume for the sake of contradiction that there is such an isomorphism f: ℚ+ → ℚ*_{>0}. Then f must be surjective, so there is an x ∈ ℚ such that f(x) = 2. This means that

and this would mean that f(x/2) = sqrt(2), which is not rational, it means that no such isomorphism exists.

We conclude that A(ℝ) is isomorphic with ℝ*, but A(ℚ) is not with ℚ*. We can now define a minimal subfield of ℝ that satisfies this property. Define the field E_0 by