So the question asks: Consider the vector space $\Bbb R[X]$ of all polynomials with real coefficients, and let $r$ be a fixed real number. Prove that the set $I(r) = \{f(X) ∈ \Bbb R[X] \mid f(r) = 0\}$ is an infinite-dimensional vector subspace of $R[X]$.

So so far I have:

Suppose $I(r) = \{f(X) ∈ \Bbb R[X] \mid f(r) = 0\}$ is a finite-dimensional vector subspace with $f_1,\ldots,f_n$ as the basis.

Let $N$ be the maximum of the degrees of the polynomials $f_1,\ldots,f_n$.

Then all linear combinations of $f_1,\ldots,f_n$ are in $I(r)$, the space of polynomials of degree $≤ N$.

Then any polynomial of higher degree, such as $f(x) = x^{N+1}$ will not be in the span of $f_1,\ldots,f_n$, which contradicts the facts that the vector space $R[X]$ contains all polynomials with real coefficients.

So prove by contradiction that $I(r) = \{f(X) ∈ \Bbb R[X] \mid f(r) = 0\}$ is an infinite-dimensional vector subspace of $R[X]$.

I feel that I "stated" too much but did not write enough "math stuff". Does this proof look alright?

  • $\begingroup$ You were on the right track, but somehow veered off a bit at the end. Your intial claim isn't that $f_{1}, \ldots, f_{n}$ span $R[X]$, but that they span the subspace $I_{r}$. Your idea works well though; just show that you can find a polynomial of higher degree which vanishes at $r$. $\endgroup$ Feb 21, 2016 at 2:11

1 Answer 1


Your proof is very fine, you need not worry about it (but you probably meant $(x-r)^{N+1}$ there). A quicker proof, though, would be to notice that $\{(x-r)^n\}_{n \geq 1}$ is linearly independent. If $I(r)$ contains an infinite linearly independent set, then $I(r)$ can't have finite dimension.


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