In my lecture notes it was left as an exercise to show that for any $R$ $R[X,Y]\simeq R[X]$ as R-modules. To show this you need a bijective function $\phi$ such that $\phi (x+y)=\phi(x)+\phi(y)$ and $\phi(rx)=r\phi(x)$. I don't know how to do this or if its even right since it seems hard to construct phi in such a way not to loose information when going from some $f_0(x,y)$ to $f_1(x)$.
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Forgot to mention but R should be any ring. – Erik Mar 04 '21 at 09:27
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1Maybe this helps: try to write down an isomorphism of $R$-modules between $R[X]$ and a countable direct sum of copies of $R$. Then try to do the same thing with $R[X, Y]$. The key point is that being isomorphic as modules is a lot weaker than being isomorphic as rings -- it is just a question of what the linear algebra of the two things look like ! – babu_babu Mar 04 '21 at 09:31
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As $R$-modules, $R[X]$ has a simple basis $\{1, X, X^2, \ldots\}$ and $R[X,Y]$ has $\{X^n Y^m \mid n \ge 0, m \ge 0\}$ (where $X^0 = Y^0 =1$ of course etc.) as a basis.
Two $R$-modules with countable bases are isomorphic. Just fix your favourite bijection between the bases and extend $R$-linearly..
Henno Brandsma
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