Isomorphism class of field is not determined by isomorphism class of additive and multiplicative group

Statement
It is possible to have two fields $$K$$ and $$L$$ such that:


 * $$K$$ is not isomorphic to $$L$$ as a field
 * The additive group of $$K$$ is isomorphic to the additive group of $$L$$.
 * The multiplicative group of $$K$$ is isomorphic to the multiplicative group of $$L$$.

Related facts

 * Any finite field is determined up to isomorphism by its size, so there can be no examples of the above involving finite fields.

Example in characteristic zero
Suppose $$K$$ is the field obtained by taking the closure of $$\mathbb{Q}$$ under all radical operations for positive numbers. In other words, $$K$$ is the smallest subfield of $$\R$$ with the property that for every positive element of $$K$$, every rational power of that element in $$\R$$ is also in $$K$$. Note that $$K$$ is contained in $$\overline{\mathbb{Q}} \cap \R$$. However, due to the unsolvability of certain polynomial equations by radicals, $$K$$ is not equal to all of $$\overline{\mathbb{Q}} \cap \R$$.

Suppose $$t$$ is an element of $$\R$$ not in $$K$$. Note that, by the above, we could choose $$t$$ to be inside $$\overline{\mathbb{Q}} \cap \R$$ if we wished; however, this is not necessary. Suppose $$L$$ is the field obtained by taking the closure of $$K(t)$$ under all radical operations.


 * $$K$$ is not isomorphic to $$L$$: This is because $$K$$ is the radical closure of $$\mathbb{Q}$$, and $$L$$ has elements not in the radical closure.
 * The additive group of $$K$$ is isomorphic to the additive group of $$L$$: Both of these are countable-dimensional vector spaces over $$\mathbb{Q}$$.
 * The multiplicative group of $$K$$ is isomorphic to the multiplicative group of $$L$$: Both of these are isomorphic to the direct product of cyclic group:Z2 (in the form of the subgroup $$\{ -1, 1 \}$$) and a countable-dimensional vector space over $$\mathbb{Q}$$ (namely, the subgroup of positive elements). To see that the subgroup of positive elements has this structure, apply the logarithm map, then note that it is a vector space over $$\mathbb{Q}$$, then use the original construction of $$K$$ to argue that it is countable-dimensional. (To note that it is not finite-dimensional, note that the logarithms of the prime numbers are linearly independent).