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?
Polynomial rings over rational numbers
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0For @Rahman: What do you mean by classifying all bases? – 2012-11-24
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1Note that every basis has $\aleph_0$ members, but that means that there are $2^{\aleph_0}$ different bases. To see this note that for every $A\subseteq\mathbb N$ the set $\{x^n\mid n\in A\}\cup\{x^n+1\mid n\notin A\}$ is also a basis, and every $A$ generates a different basis. – 2012-11-24
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0I wonder whether this vector space has another maximal independent set except for standard basis. – 2012-11-24
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0Asaf 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
1 Answers
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, . . . .