Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $A$ least not with the precise formalism that proving it entails.

share|cite|improve this question
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
up vote 3 down vote accepted


(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 $\,\Bbb F[x]\,$

share|cite|improve this answer

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 $\rm\:x^n.\:$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.