2
$\begingroup$

My problem: Proof that the vector space of polynomials over $\Bbb F$: $P(x, \Bbb F)$ isn't finetely generated.

The vector space isn't finetely generated if there is no finite set that would generate it. But how to prove it?

If have some polynomial from that set:

$p(x) = a_0 + a_1x + a_2x^2 + ... +a_nx^n $ than its basis would be $\{1, x, x^2, ...\} $and this basis is infinite so the dimension of my given space is infinite, hence my vector space isn't finitely generated?

  • 0
    You have to _try_ something. Don't just post the problem. – 2017-02-08
  • 0
    To explore the problem, take a few random polynomials, and see what polynomials are in the subspace they span. – 2017-02-08
  • 0
    But it is just so intuitive, if the polynomial has no finite degree its basis is infinite and its vector space isn't finitely generated. – 2017-02-08
  • 1
    Every polynomial has finite degree. – 2017-02-08
  • 1
    As a $F$-algebra, clearly $F[x]$ is finitely generated, but as a $F$-vector space $F[x^0,x^1,x^2,\ldots]$ isn't – 2017-02-08
  • 0
    @Leif: You wrote $$p(x) = a_0 + a_1x + a_2x^2 + ...$$ But a polynomial can't have infinitely many terms. It has to stop _somewhere_. – 2017-02-08
  • 0
    More precisely, a polynomial, in standard form, can't have infinitely many _nonzero_ terms. – 2017-02-08

2 Answers 2

1

Let $V$ be the vector space of all polynomials in the variable $x$, over the field $F$. let $S$ be a finite subset of $V$, and let $W$ be the space spanned by $S$. To show that $V$ is not finitely generated, it suffices to show that $W$ is a proper subset of $V$. What can be said about the degrees of the elements of $W$?

  • 0
    so $W = \langle S \rangle $ and $W$ is a subspace if it is closed to addition and multiplication by a scalar, but no idea what to do with that degree. – 2017-02-08
  • 0
    Try an actual set $S$ with a few polynomials. Didn't I already suggest this? – 2017-02-08
  • 0
    So every polynomial in that given space can be uniquely written as a finite linear combination of powers of $x$. So the powers of $x$ form a basis of the vector space of all polynomials, and there are infinitely many of them. – 2017-02-08
  • 0
    That's DonAntonio's version. No problem with that argument. – 2017-02-08
  • 0
    But you still don't see the alternate argument that I was alluding to? – 2017-02-08
  • 0
    I am trying right now, I want to understand it fully, it isn't a homework just a practice. – 2017-02-08
  • 0
    Good! Try a set S with say, two polynomials. – 2017-02-08
  • 0
    So $S$ is finite but I can get infinite numbers of polynomials by linear combinations so I actually can not say how many polynomials are in my given space. – 2017-02-08
  • 0
    Which set $S$ did you try? What are the degrees of the polynomials in your set $S$? What degrees are possible for elements of $\text{<}S\text{>}$? – 2017-02-08
  • 0
    Can I have a set like this? $S = \{x^2 +1, x-1\}$. And by their linear combination I can get for example: $x^2 + x$. But my degree would always be 2 or less. The dimension would be 3. – 2017-02-08
  • 0
    The dimension of is not relevant. The key result is that has no elements of degree greater than 2, hence it's not the full space. – 2017-02-08
  • 0
    Ok, I think I understand it, thank you. – 2017-02-08
  • 0
    Just to be sure, I can say that I have some finite basis in my given space $V$: $N = \{1, x, x^2 ... x^n\}$ and I have some vector $x^{n+1}$ in that space. Because $N$ is the basis of $V$ I should be able to write $x^{n+1}$ as the linear combination of the elements in $N$, but I can't so that is the contradiction that $N$ is finite. But even though $\{1, x, x^2 ...\}$ is an infinite basis I can create every possible polynomial from it and those polynomials have the finite degree, like $x^2 + x$ for example. – 2017-02-08
  • 0
    You can't assume that the elements of the finite set $S$ are just powers of $x$. Just assume that $S$ is a finite set of polynomials. Then regardless of the choice of elements of $S$, the set $S$ has some element of largest degree, $n$ say. It follows that $\text{<}S\text{>}$ does not contain any polynomials of degree greater than $n$, hence $\text{<}S\text{>}$ is not the full space. – 2017-02-08
1

Hint:

$$\left\{\,1,\,x,\,x^2,\,\ldots\,\right\}$$