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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.

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  • $\begingroup$ You mean $\Bbb Q(x)$, not $\Bbb Q[x]$ etc. as the latter is not a field. $\endgroup$ Commented Apr 18, 2018 at 17:37
  • $\begingroup$ yes! sorry...my fault $\endgroup$ Commented Apr 18, 2018 at 17:39
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    $\begingroup$ Also, I see only countably infinitely many irreducibles (in fact, both fields are countable ) $\endgroup$ Commented Apr 18, 2018 at 17:40
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    $\begingroup$ Write each rational function as a product of all irreducible polynomials to some integer powers, and a non-zero rational. Here, almost all the exponents of the irreducible polynomials are zero. Multiplying two rational functions consists in multiplying the non-zero rational and adding the exponents. $\endgroup$
    – user551819
    Commented Apr 18, 2018 at 17:43
  • $\begingroup$ Both $\Bbb{Q}[x]$ and $\Bbb{Q}[x,y]$ are UFDs. Both have countably infinitely many irreducible polynomials. That should explain the part about free abelian groups on countably (not continuum!) many generators. $\endgroup$ Commented Apr 18, 2018 at 17:44

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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 $$ \Bbb Q(x)^\times\approx \Bbb Q(x,y)^\times.$$

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