# Proving we have a basis for $F[x]$

So $F$ is an arbitrary field, and $F[x]$ denotes the set of of formal polynomials with coefficients in $F$. And $A=\{f_i \mid i\geq 1\}$.

I need to show two things,

1. If $A$ is such that $deg (f_i) \neq deg (f_j)$ for $i\neq j$ then $A$ is linearly independent.

2. If $A$ also satisfies that $\{deg(f_i) \mid f_i\in A\} = \mathbb{N}$ then $A$ is a basis for $F[x]$.

So for 1, I'm using coordinates in the standard basis for $F[x]$ and showing that those are clearly independent...but appealing to coordinates isn't something I particularly enjoy, especially since in this case it seems pretty circular given that the standard basis for $F[x]$ is such an $A$.

And while 2 makes total sense intuitively, I'm not sure how to show that $A$ spans...at least not with the precise formalism that proving it entails.

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Hint for 2: induction. Show that all polynomials of degree $n$ are in the span of $A$. – Thomas Andrews Sep 28 '12 at 3:32

Hints:

(1) Suppose $\,A\,$ is linearly dependent, so there exists some $\,f_k\in A\,$ that is a (finite, of course) linear combination of other elements in $\,A\,$ . Now remember that multiplying a polynomial by a non-zero scalar does not change the polynomial's degree.

(2) Show that $\,A\,$ is a maximal linear independent set in $\,\Bbb F[x]\,$

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Hint $\rm\ (1)\$ A sum of distinct degree polynomials has degree that of the largest degree summand.

$\rm(2)\$ By $(1)$ they are independent, and a simple induction shows their span includes all $\rm\:x^n.\:$

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