# Proof that $\mathbb{R}[x]$ is not a finite dimensional vector space

How can we prove that the vector space of polynomials in one variable, $\mathbb{R}[x]$ is not finite dimensional?

It has an infinite linearly independent set, $\{1,x,x^2,x^3,\dots\}$.

• The theorem that finitely generated space has no infinite linearly independent set seems to need some dimension theory. Here there is an infinite basis, and the argument in that case is immediate. – zyx Mar 21 '13 at 4:52

If so then the $1$-$1$ map $\rm\: f \to xf\:$ is onto so $\rm\:\exists f\!:\ xf = 1\:\Rightarrow\: 0=1\:$ by evaluating at $\rm\:x=0.$

• lovely solution :) – Ittay Weiss Mar 21 '13 at 4:08
• A gem. ,.,.,..,.,.15c..,.,., – zyx Mar 21 '13 at 4:16
• Or the onto map $f\longmapsto f'$ is $1-1$. Very nice, +1. – Julien Mar 21 '13 at 6:50

For all $n\ge 1$, $\mathbb R[x]$ contains the subspace $P_n$ of all polynomials of degree less than $n$. Thus the dimension of $\mathbb R[x]$ is not smaller than the dimension of $P_n$, which is $n$ (since for instance $P_n$ is isomorphic to $\mathbb R^{n}$). The result follows.

Suppose it is and let $v_1$ through $v_n$ be a basis. Let $d$ be the highest degree appearing in the $v_i$. Because addition of polynomials never increases degree, any linear combination of the $v_i$ will have degree less than or equal to $d$. But clearly $\mathbb R[x]$ contains polynomials of arbitrarily high degree. This contradiction proves the theorem.

There are two different interpretations of the problem.

1. The vector space with basis the symbols $<1>,<x>,<x^2>,...$ is infinite dimensional (there is also some multiplication law, but that does not change the vector space and its dimension). Well, we just wrote down an infinite basis, so that is that. The infinite dimensionality is true, pretty much by definition, for real coefficients, complex coefficients, over finite fields, rings, etc.

2. The vector space of polynomial functions on $\mathbb{R}$, functions that can be calculated by "evaluating at [any given real number] $x$" a single element of $\mathbb{R}[X]$, is infinite dimensional. This space is equivalent (isomorphic) to the space on the formal symbols for powers of $X$ , and consequently has the same infinite dimension, if (and only if) there is no polynomial that evaluates to $0$ at all arguments. That is true for any infinite field, such as $\mathbb{R}$, using the theorem that a polynomial of degree $n$ has at most $n$ roots. Over a finite field there are many polynomials representing the same function.

It is usually taken for granted that having an infinite basis means the space is not finite dimensional, but a proof that "infinite basis implies contradiction in the assumption of finite dimension" can be done by the degree argument.

Define $p = \inf \{ k \in \mathbb{N} | \ x^k \in \operatorname{sp}\{ 1,x,...,x^{k-1}\} \}$.

If $p< \infty$, then $x^p = \sum_{k=0}^{p-1} \alpha_k x^k$. Differentiating $p-1$ times gives $p! x = 0$, evaluating at $x=1$ yeilds a contradiction. Hence $p = \infty$.