Do there exist two non-isomorphic fields whose additive groups are isomorphic and whose multiplicative groups are isomorphic This Problem is due to Halmos's book.
He gave a solution $=>$ $\mathbb{Q}(x)$ and $\mathbb{Q}(x,y)$.
I understand the additive groups are isomorphic but he showed that the multiplicative group is the direct product of $\mathbb{Q}^*$ and the free abelian group on continuum many generators (namely irreducible polynomials). 
The two fields are not isomorphic, because they have transcendence bases consisting of different number of elements.
I don't understand why multiplicative groups are isomorphic here. Help me to understand this. 
 A: Consider the set $A$ of irreducible polynomials in $\Bbb Q[x]\subset \Bbb Q(x)$. On $A$, we have an equivalence relation $f\sim g\iff \exists q\in\Bbb Q^\times\colon f=qg$. Pick $A_0\subset A$ such that   it contains exactly one element of each equivalence class. Note that $A_0$ is countably infinite.
Every element $f\in \Bbb Q(x)^\times$ can be written in a unique manner as
$$ f(x)=q\cdot \prod_{p\in A_0}p(x)^{n_p}$$
with $q\in\Bbb Q^\times$, $n_p\in\Bbb Z$, and almost all $n_p$ are zero.
This establishes an isomorphism 
$$ \Bbb Q(x)^\times \to \Bbb Q^\times\oplus \bigoplus_{p\in A_0}\Bbb Z\approx \Bbb Q^\times\oplus \bigoplus_{j\in \Bbb N}\Bbb Z.$$
Similarly, consider the set $B$ of irreducible polynomials in $\Bbb Q[x,y]\subset \Bbb Q(x,y)$. As above, let $B_0$ contain representatives of the equivalence classes of $B$ under multiplication with a constant.
Then as above, $B_0$ is countably infinite and every $g\in\Bbb Q(x,y)$ can be written in a unique way as
$$g(x,y)=Q\cdot \prod_{P\in B_0} P(x,y)^{n_P} $$
with $Q\in \Bbb Q^\times$, $n_P\in \Bbb Z$, and almost all $n_P$ zero.
Again, this establishes an isomorphism 
$$ \Bbb Q(x,y)^\times \to \Bbb Q^\times\oplus \bigoplus_{P\in B_0}\Bbb Z\approx \Bbb Q^\times\oplus \bigoplus_{j\in \Bbb N}\Bbb Z.$$
Together, this shows that 
$$ \Bbb Q(x)^\times\approx \Bbb Q(x,y)^\times.$$
