I am struggling with an example my textbook gave of a set that is not a vector space. It is stated that $V$, the set of all polynomials of degree exactly $3$ is not a vector space. The reason the textbook gives is that this set does not contain a zero vector. However is $f(x) = x^3$ not a polynomial in the set, thus leading to $ f(0) = 0 $ being a zero vector? I understand that this set would still not be a vector space due to a condition like this: $ [1 + 4x^2 + x^3] + [2-x+x^2-x^3] = 3-x+5x^2 $, with the sum not being in the set $V$. However I am still confused about the zero vector example my textbook gives.

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    $\begingroup$ No because it doesn't contain the zero function. $\endgroup$ May 12, 2020 at 6:17
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    $\begingroup$ Ah do you mean to say that $ f(x) = 0 $ is in fact of degree zero and thus not contained in the set $V$? $\endgroup$
    – Ludwig
    May 12, 2020 at 6:18
  • $\begingroup$ Yes. Not only that $V$ is not closed under polynomial addition; for example $x^3+x^2,-x^3\in V$ but their sum doesn't. $\endgroup$
    – user10575
    May 12, 2020 at 6:21
  • $\begingroup$ Tangential note: be careful with the degree of the zero function. Most authors will define the degree of the function defined by $f(x) = 0$ to be some kind of infinity. The reason for this is that degrees add with multiplication: $\deg(fg) = \deg(f) + \deg(g)$. But the zero function kills anything, thus $\deg(0) = \deg(0\cdot f) = \deg(0) + \deg(f)$. If we define $\det(0) = 0$, then $\deg(f) = 0$ for all $f$, which is nonsense (or, at least, not useful). Defining $\deg(0) = \infty$ solves this problem. $\endgroup$
    – Xander Henderson
    May 12, 2020 at 16:47

2 Answers 2


To offer a different perspective and to be a bit nit-picky:

The zero vector would be the zero polynomial, which is just $0$. The reason why I am pointing this out is because you don't specify the field $F$ over which you are working. In general, polynomials are not the same as polynomial functions, meaning that you "cannot plug in values into polynomials and say that the polynomial $f$ is the same as the collection $(f(x))_{x \in F}$."

For an example, assume that $F$ is the finite field with two elements. Then $f = x^2 + x$ is a polynomial different from $0$. However, if you plug in both elements $0$ and $1$, you get $f(0) = f(1) = 0$.

If you are working over an infinite field, e.g. the real numbers, then what you do is okay and not at all problematic. In this situation, we can identify polynomials with polynomial functions. (One can show that one can identify polynomials and polynomial functions if and only if the field one is working with is infinite.)


The 'zero' in this set would have to be the zero polynomial. Notice $f(x)=x^3$ is only $0$ at $x=0$. So the zero here is the zero function $f(x)= 0$. You want the set of odd degree polynomials to be $\{a_3 x^3 + a_2 x^2 + a_1x+a_0\}$, where not both $a_3$ and $a_1$ are zero so that the polynomial is odd in the traditional sense.

  • $\begingroup$ $x^3 + x^2$ is not odd in the traditional sense. (Or I don't understand what the traditional sense is...) $\endgroup$ May 12, 2020 at 6:26
  • $\begingroup$ @Marktmeister I'm not sure what you mean by the traditional sense. This is an odd degree polynomial and notice not both of $a_3$ and $a_1$ are zero for this polynomial. Perhaps what you are thinking is even/odd function (your example is neither). Or perhaps you misread the not both $a_3$ and $a_1$ are zero? $\endgroup$ May 12, 2020 at 6:30
  • $\begingroup$ I thought that it means that it is odd as an function. :-) Okay, thanks. $\endgroup$ May 12, 2020 at 6:31
  • $\begingroup$ Your first three sentences seem spot on, but then you go off in a direction which doesn't make sense to me. The question is about polynomials of degree exactly $3$, so I am not sure that I understand the relevance of $a_1$ in this context, nor of odd degree polynomials. Moreover, if $a_3=0$ and $a_2\ne 0$, then the polynomial is of even degree ($2$), independent of $a_1$. I don't understand what you are trying to say with that part of your answer. $\endgroup$
    – Xander Henderson
    May 12, 2020 at 16:51

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