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We know that polynomial rings over $\mathbb{Q}$ is a vector space over $\mathbb{Q}$. It has a well-known basis $1, x, x^2,\ldots$ but can we classify all bases?

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    Asaf just showed you a way to construct uncountably many different bases. Furthermore, it is straightforward to extend **any** finite independent set into a basis....2012-11-24

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To answer a question posted in the comments under the question: there are lots of other bases. If you take one polynomial of each degree, for degrees $0,1,2,3,\ldots$, that's always a basis (take degree $0$ to mean nonzero constant polynomials). You can find a set of four third-degree polynomials that spans the set of all polynomials of degree $\le 3$, so you don't need one of each degree. (For example, the difference between two of them could be $x^2$, and between two others could be $x$, etc.) A basis must have $\le n+1$ polynomials of degree $n$ or less (any more would make it linearly dependent since the dimension of that subspace is $n+1$.

A linearly independent set of polynomials is a basis if for infinitely many $n$ the number of members of the set having degree $\le n$ is $n+1$. I'm guessing that that sufficient condition is also a necessary condition. More later, maybe, . . . .