Are rational functions a vector space?

Question

Let $\mathscr P_n[x]$ be the space of real polynomials in $x\in\mathbb R$ of degree at most $n\in\mathbb N$, and $$\mathscr P[x] := \lim_{n\to\infty} \mathscr P_n[x]$$ the set of all real polynomials in $x$.

Then consider the set $V$ whose elements $v(x)$ are defined by $$v(x) = \frac{p(x)}{q(x)} \qquad \text{for} \quad p(x), q(x) \in \mathscr P [x],$$ with $q(x) \ne 0$ (the zero function). Is this set a vector space under addition?

My naïve answer would be yes, with a basis of elements of the form $$b(x) = \frac{(x-\hat x^1)(x-\hat x^2)\cdots(x-\hat x^m)} {(x-\hat x_1)(x-\hat x_2)\cdots(x-\hat x_n)} \qquad \hat x^i, \hat x_j \in \mathbb C$$ such that the coefficients of the numerator and denominator are real, but I'm not sure whether I'm neglecting something. It certainly seems to me that both linear independence and spanning should work.

Bonus: If it is a vector space, does it have a name?

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There is a superficially similar question to mine here, but that one is asking for fixed $q(x)$ only.

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Edits: courtesy of comments, have changed slightly the definitions of $q(x)$ and $b(x)$.

Answer 1

Putting together everything from the comments:

1. Given an arbitrary real polynomial quotient $\frac{P(x)}{Q(x)}$ $\left(P(x), Q(x) \in \mathscr P[x]\right)$, you can uniquely decompose it into basis functions by

   1. applying the method of long division, to obtain $$\frac{P(x)}{Q(x)} = R(x) + \frac{\bar P(x)}{Q(x)},$$ where $R(x), \bar P(x)\in \mathscr P[x]$ with $\operatorname{deg} \bar P < \operatorname {deg} Q,$ then

   2. exploiting the unique decomposition via partial fractions of $\frac{\bar P(x)}{Q(x)}$ to write $$\frac{\bar P(x)}{Q(x)} = \sum_{i=1}^r \left[ \frac{c_{i_{1}}}{x - b_{i}} + \dots + \frac{c_{i_{l_i}}}{(x - b_{i})^{l _ {i}}}\right] + \sum _ { j=1}^s \left [ \frac{d _ {j _ {1} } x + e _ {j _ {1} } }{x ^ {2} + p _ {j} x + q _ {j} } + \dots + \frac{d _ {j _ { t _ j } } x + e _ {j _ { t _ j } } }{( x ^ {2} + p _ {j} x + q _ {j} ) ^ {t _ {j} } } \right ] ,$$ for some coefficients $c, d, e \in\mathbb R$ to be determined.

   This holds for any $Q(x)$, which has been here maximally factorised in the reals to the form $$Q(x) = b _ {0} ( x - b _ {1} ) ^ {l _ {1} } \cdots ( x - b _ {r} ) ^ {l _ {r} } ( x ^ {2} + p _ {1} x + q _ {1} ) ^ {t _ {1} } \cdots ( x ^ {2} + p _ {s} x + q _ {s} ) ^ {t _ {s} } .$$

   This decomposition (exists, unique) means that we have a set of (spanning, independent) basis vectors $$1, x^u, \frac{1}{(x-b)^l}, \frac{1}{(x^2+px+q)^t}$$ with $u,l,t \in \mathbb N_*$ and $b, p, q \in \mathbb R_*$ (and $p^2 - 4q <0$).

   Hence, it is a vector space.

2. Apparently the vector space is called the rational functions in $x$ on the reals and denoted $\mathbb R (x)$, but I can't find any reference for that. (@Robert Israel)

3. More than a vector space, $\mathbb R (x)$ is actually a field, because it has multiplicative structure (see the answer by @Jos).

Thank you to everyone who contributed to sorting this out.

Answer 2

Since the vector space requirements are met you can say they form a vector space. But you can make the stronger claim that they are a field, since you can multiply them with each other, and they have (except for the zero element) a multiplicative inverse, the latter by just exchanging $p$ and $q$.