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This is the problem statement:

The set of all polynomials of degree greater than 3 together with the zero polynomial in the vector space of $P$ of all polynomials with coefficients in $\mathbb{R}$.

It is not subspace because the given description corresponds to an infinitely generated polynomial space?

Is that correct? If not, how I can solve this?

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    $\begingroup$ Vector spaces can be infinitely generated, as can their subspaces. As the problem states, for example, the set of all polynomials with coefficients in $\mathbb{R}$ forms a vector space. It is an infinitely-generated subspace of the set of polynomials in two variables. $\endgroup$ Commented Apr 11, 2016 at 21:44

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It is not a subspace because it's not stable by addition: take $F(X)=1-X^4, G(X)=X^4$. Then $F(X)+G(X)=1$ has degree $0$.

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  • $\begingroup$ The ideal generated by $X^4$ doesn't contain any polynomials of degree $5$, though, does it? $\endgroup$
    – user296602
    Commented Apr 11, 2016 at 22:13
  • $\begingroup$ It contains all polynomials of order $\ge4$, plus $0$. For instance, $X^5=x\cdot X^4\in (x^4)$. $\endgroup$
    – Bernard
    Commented Apr 11, 2016 at 22:32
  • $\begingroup$ Right, sorry - I was thinking about the action of $\mathbb{R}$ not $\mathbb{R}[X]$. However, I still don't see how to get all polynomials of degree $4$, for example $X^4 + X$. $\endgroup$
    – user296602
    Commented Apr 11, 2016 at 22:34
  • $\begingroup$ It's my turn to be sorry. I meant polynomials of order $\ge 4$ (the order in the sense of formal power series, for instance). $\endgroup$
    – Bernard
    Commented Apr 11, 2016 at 22:37
  • $\begingroup$ Ah, ok, and this is equivalent to being divisible by $X^4$. But this won't be the space described in the question, which is just about degree $\ge 4$. $\endgroup$
    – user296602
    Commented Apr 11, 2016 at 22:42

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