When are rational functions considered equal? Eg, $\frac{z-1}{z^2-1}$ vs $\frac{1}{z+1}$ I came across an example like this recently. Consider
$$ f(z) = \frac{z-1}{z^2 - 1}.$$
I want to determine the singularities of $f$.
Since $z^2-1 = (z-1)(z+1)$, algebraically $f(z) = \frac{1}{z+1}$, i.e. $f$ has a simple pole in $-1$. However, when checking the solution to this problem, $1$ was also listed as a singularity of $f$, albeit a removable one. This does make sense to me: When we consider
$g(z) = \frac{1}{z^2 - 1}$ and $h(z) = z - 1$, then $f(z) = g(z)h(z)$, and clearly, $g$ has simple poles in $1, -1$.
However, this raises the question of when two rational functions are really considered equal. By this example, it would seem like polynomial division is a transformation that only preserves equality modulo removable singularities. Is there a more formal description of this problem?
 A: We can construct the field $\Bbb C(z)$ of rational functions with complex coefficients in $z$ from the ring $\Bbb C[z]$ of polynomials in $z$ the same way we construct $\Bbb Q$ from $\Bbb Z$: As equivalence classes of pairs $(a,b)$ of polynomials where $b$ is not the zero polynomial and we declare the equivalence relation $(a,b)\sim (a'b')\iff ab'=a'b$ and introduce the notation $\frac ab$ for the equivalence class of the pair $(a,b)$. Note that this construction does not do anything at all like plugging in numerical values for $z$ and hence does not care whether $b$ has zeroes, or has zeroes in common with $a$.
It is worth noting that in spite of their name these rational functions are not functions per se, in fact already polynomials are mere algebraic constructs and to be distinguished from polynomial functions. This distinction is particularly important over finite fields: There are infinitely many polynomials in $\Bbb F_2[x]$, but only four functions $\Bbb F_2\to \Bbb F_2$. $x^2+x$ is not the zero polynomial, but the associated polynomial function is the zero function because it evaluates to $0$ at all points of $\Bbb F_2$.
After the above construction, we can interpret rational functions as functions $\Bbb C\cup\{\infty\}\to \Bbb C\cup \{\infty\}$, namely by (for given $z$) picking a representative of the equivalence class for which not both the numerator and the denominator are $=0$.
So in $\Bbb C(z)$, we can take for example $\frac{1}{z-1}$ and $\frac{z^2-1}1$ and multiply them as $\frac{1}{z-1}\cdot \frac{z^2-1}1=\frac{z+1}1$ and observe that this product gives $2$ when evaluated at $1$. To reach the same goal with functions, we can't multiply that easily. We first have to bring both functions to a common domain, which would by $\Bbb C\setminus \{1\}$ and then the product function would only be defined in this domain as well and could afterwards be extended into this removable singularity if desired,  but the extension is not the product proper.
