Closed subsets of rational functions when the coefficients are endowed an Euclidean topology

Question

Suppose $\text{rat}(n)$ denotes the collection of proper rational functions such that: if $f \in \text{rat}(n)$, then \begin{align*} f(z) = \frac{ b_{n-1} z^{n-1} + \dots + b_0 }{z^n + a_{n-1}z^{n-1} + \dots + z_0}. \end{align*} In other words, the numerator is a polynomial of degree at most $n-1$ and the denominator is a monic polynomial of degree $n$. Endow the coefficients $(a_0, \dots, a_{n-1}, b_0, \dots, b_{n-1}) \in \mathbb R^{2n}$ the Euclidean topology. I read in a book without proof that the subset of $\text{rat}(n)$ (togologize the coefficients) that has no common factor (in the numerator and denominator) is an open subset. It states that having a common factor is an open condition. How do I prove this fact?

Answer 1

Let $f,g$ be polynomials in a variable $z$. They have a common factor iff there exists some $a$ such that $f(a)=g(a)=0$. This immediately tells you that having a common factor is a closed condition. Consequently, having no common factor is an open condition.

Answer 2

I realized the answer lies in the resultant of two polynomials. If $f, g$ has no common factor, then the resultant $\text{res}(f, g) = 0$. And apparently $\text{res}(f, g)$ is a polynomial in the coefficients, which is precisely a closed condition on the coefficients.