Why is $\mathbb Q[x] \ncong \mathbb Q[x,y]$?

Question

Consider $\mathbb Q[x]$ and $\mathbb Q[x,y]$ as $\mathbb Q$-modules. And now consider $\mathbb Q[x] \otimes_\mathbb Q \mathbb Q[x]$ It is easy to show that $$\mathbb Q[x] \otimes_\mathbb Q \mathbb Q[x] \cong \mathbb Q[x,y]$$ Using the map $ x \otimes x \mapsto xy$ However, I came up with this (probably flawed) proof: $$ \begin{align}\mathbb Q[x] \otimes_\mathbb Q \mathbb Q[x] &= \bigoplus_{n=1}^\infty \mathbb Q \otimes_\mathbb Q \bigoplus_{n=1}^\infty \mathbb Q \\ &= \bigoplus_{n=1}^\infty \left( \bigoplus_{n=1}^\infty\mathbb Q\right) \\ & = \bigoplus_{n=1}^\infty \mathbb Q \\&= \mathbb Q[x] \end{align} $$ The second to last line has to be wrong, can anybody help me see why? Is there a difference between summing infinitely many modules "twice" and "once". Are there cardinality issues?

Also, $\mathbb Q[x,y]$ is a free module, right?

Thanks

Answer 1

You can make a direct isomorphism between $\Bbb Q[x,y]$ and $\Bbb Q[x]$ by mapping any monomial $rx^my^n\in\Bbb Q[x,y]$ to $rx^{f(x,y)}\in\Bbb Q[x]$, where $f$ is your favourite bijection between $\Bbb N\times\Bbb N$ and $\Bbb N$.

Answer 2

Notice that $\mathbb{Q}[x]\cong\mathbb{Q}[x,y]$ as $\mathbb{Q}$-vector spaces because they have the same dimension. I don't think there's any problem with your proof. (But they are not isomorphic as algebras).