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I have several questions concerning the proof. I don't think I quite understand the details and motivation of the proof. Here is the proof given by our professor.

The space of polynomials $F[x]$ is not finite-dimensional.

Proof. Suppose $$F[x] = \operatorname{span}\{f_1,f_2,\dots,f_n\}$$

Let us choose a positive integer $N$ such that $N > deg (f_i)$ for all $i = 1,\dots,n$. As $\{f_1,f_2,\dots,f_n\}$ spans $F[x]$ we can find scalars $a_1, a_2, \dots,a_n$ such that $x^N = a_1f_1 + a_2f_2 + \cdots + a_nf_n$. Then the polynomial

$$G(x) = x^N - a_1f_1 - a_2f_2 - \cdots - a_nf_n \equiv 0$$

is a polynomial of degree $N$ which is identically zero. This is a contradiction since $G(x)$ cannot have more than $N$ roots.

Questions

  • Why is $G(x)$ identically zero and what does it mean for it to be identically zero?
  • After obtaining that $G(x) = x^N - a_1f_1 - a_2f_2 - \cdots - a_nf_n \equiv 0$, why must we have more than $N$ roots?
  • What is the motivation behind choosing $N$ such that $N > deg (f_i)$ for all $i = 1,\dots,n?$

  • What are the implications if we chose $N$ less than or equal to $f_i$ such that $f_i$ has the greatest degree? How will the proof fail in this case?

  • If there are any other important details and key insights that are worthy to be pointed out please let me know so that I can better my understanding of the argument presented.

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3 Answers

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To say that a polynomial $p(x)$ is identically zero is to say that $p(x)=0$ for all values of $x$. Here the coefficients $a_1,\dots,a_n$ were chosen so that $$x^N=a_1f_1(x)+a_2f_2(x)+\dots+a_nf_n(x)$$ for all values of $x$, and $G(x)$ was defined by $$G(x)=x^N-a_1f_1(x)-a_2f_2(x)-\dots-a_nf_n(x)\;,$$ so for each $x$ we have $$G(x)=x^N-\Big(a_1f_1(x)+a_2f_2(x)+\dots+a_nf_n(x)\Big)=0\;.$$ This is exactly what’s meant when we say that $G(x)$ is identically zero and write $G(x)\equiv 0$.

In particular, $G(x)=0$ for every value of $x$, so every possible value of $x$ is a root of $G(x)$. Assuming that your field of scalars is infinite, that means that $G(x)$ has infinitely many roots and therefore certainly has more than $N$ roots.

The reason for choosing $N$ to be greater than the degree of any of the $f_i$ is to ensure that the polynomial $$a_1f_1(x)+a_2f_2(x)+\dots+a_nf_n(x)$$ has degree less than $N$, so that when it is subtracted from $x^N$, the resulting polynomial $G(x)$ still has an $x^N$ term, i.e., is still of degree $N$. Otherwise $G(x)$ might actually be the zero polynomial: all of the terms might cancel out. The whole point is to get a polynomial that on the one hand cannot be the zero polynomial, because its leading term is $x^N$, but on the other hand must be the zero polynomial, because it’s the difference of two equal polynomials.

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I'd be a bit careful with saying things like: "To say that a polynomial $p(x)$ is identically zero is to say that $p(x) = 0$ for all values of $x$." Because that seems to suggest that a polynomial is a function defined on the ring it's over. Consider $f = X^2 + X$ over $\mathbb Z/2\mathbb Z$. $f(0) = f(1) = 0$, but $f$ is certainly not the zero polynomial. –  kahen Feb 29 '12 at 11:08
    
@kahen: I debated with myself, but finally decided that it was probably a useful paraphrase here, since the context appears to be a first course in linear algebra. –  Brian M. Scott Feb 29 '12 at 12:48
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The idea of the proof is this. A basis for the set of all polynomials is $$1,x,x^2,x^3,\ldots,x^n,\ldots$$ If you have a finite basis, there will be some polynomials that you will be missing, in particular polynomials with high degree. For example, if you take the span of $$x^3+105x^4+x^7, 1+x^8$$ then for sure you'll be missing any polynomial with degree larger than $8$. The rest of the proof is just taking this idea and making it formal. You choose $N$ to be larger than all the degrees since you know that in that case $x^N$ isn't in the span. You prove that by claiming that you can't have the equality $$x^N = A(x^3+105x^4+x^7) + B(1+x^8)$$ in $F[x]$. If there were an equality, then the difference between the two sides will be the zero polynomial. That's what we mean by identically zero. We know that the difference isn't identically zero since, well, it's not the zero polynomial, but rather a degree $N$ polynomial.

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I think the proof in your question is incorrect. Let's assume that $F$ is a field and $x$ an indeterminate. A correct proof might (I believe) go as follows:

Suppose $$ F[x]=\operatorname{span}\{f_1,f_2,\dots,f_n\}. $$

Let us choose a positive integer $N$ such that $N > \deg (f_i)$ for all $i=1,\dots,n$. As $\{f_1,f_2,\dots,f_n\}$ spans $F[x]$ we can find scalars $a_1, a_2, \dots,a_n$ such that $$ x^N=a_1f_1+a_2f_2+\cdots+a_nf_n. $$ This yields the contradiction $$ 0= x^N - a_1f_1 - a_2f_2 - \cdots - a_nf_n\neq0. $$

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