Degree of field extension question. Last lecture it was written " $[\mathbb{Q}[X]:\mathbb{Q}]=\infty$. The elements $1,x,x^2,\ldots$ are linearly independent (but not a basis)."
This confused me.. why isn't this a basis? Given any polynomial with rational coefficients you can clearly write it as a linear combination of a finite number of powers of $x$..? What would be a basis, if not this?
 A: The notation $L/K$ means that $L$ is a field extension of $K$: so both $L$ and $K$ are fields and $K$ is a subfield of $L$ (so $K\subseteq L$). If $L/K$ then we use the notation $[L:K]$ to denote $\dim_KL$, the dimension of $L$ considered as a vector space over $K$, which we call the degree of the extension.
The notation $\Bbb Q[X]$ stands for the ring of polynomials in the variable $X$ with coefficients from $\Bbb Q$; this ring is not a field. For example, the element $X\in\Bbb Q[X]$ does not have an inverse. However it is true that $\Bbb Q[X]$ is an infinite-dimensional vector space over $\Bbb Q$, i.e. $\dim_{\Bbb Q}\Bbb Q[X]=\infty$ (in particular the dimension is $\aleph_0$; there are different sizes of infinities, called cardinal numbers, and an infinite-dimensional vector space has as its dimension an infinite cardinal number). Since $\Bbb Q[X]$ is not a field, we do not write $[\Bbb Q[X]:\Bbb Q]$, and furthermore $\{1,X,X^2,\cdots\}$ is a $\Bbb Q$-basis for $\Bbb Q[X]$.
The notation $\Bbb Q(X)$ stands for the field of rational functions in the variable $X$ with coefficients coming from $\Bbb Q$. To use abstract algebra lingo, $\Bbb Q(X)$ is the "fraction field" of $\Bbb Q[X]$. It is not difficult to check that the sum and product and reciprocals of rational functions are rational functions (if we except division by zero). Here, too, $\Bbb Q(X)$ is $\aleph_0$-dimensional over $\Bbb Q$, so $[\Bbb Q(X):\Bbb Q]=\aleph_0$. However the set ${\cal B}=\{1,X,X^2,\cdots\}$ is not a $\Bbb Q$-basis for $\Bbb Q(X)$; the $\Bbb Q$-span of ${\cal B}$ is $\Bbb Q[X]$ which is a proper subset of $\Bbb Q(X)$ (because for example $X^{-1}\in\Bbb Q(X)$ but $X^{-1}\not\in\Bbb Q[X]$).
