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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?

Blue
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tolUene
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    $z^2+1\neq (z-1)(z+1)=z^2-1$ – Ritam_Dasgupta Jul 19 '21 at 19:37
  • @Ritam_Dasgupta Oops, yeah. Fixed that. – tolUene Jul 19 '21 at 19:38
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    You could define it either way. 1) Rational functions are equal if they are equal as functions. 2) Rational functions are equal if they are equal up to removable singularities. Which makes more sense will surely depend on your application. – Charles Hudgins Jul 19 '21 at 19:40
  • Two functions are identically equal iff they have the same domains and take the same values at each input. – Vishu Jul 19 '21 at 19:41
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    And we require that the domain of a function be determined "without any simplification"? I.e. polynomial division changes the domain? – tolUene Jul 19 '21 at 19:42
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    To be more concrete, whether you consider $z/z$ equivalent to $1$ depends on how you define things. $z/z$ is not defined at $z = 0$ since $0/0$ is not defined. Meanwhile $z \mapsto 1$ is defined everywhere. As functions with the most natural domain, then, they are not equal. As algebraic objects, we may say they are equal using the equivalence relation $a/b \sim c/d$ iff $ad = bc$. In this case, $z/z \sim 1$ because $z = z$. For another definition still, we could say rational functions are equal if they agree at all but finitely many points, in which case, again, z/z = 1$. – Charles Hudgins Jul 19 '21 at 19:59
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    As always in mathematics, which definition you use depends on what you're trying to do, but you should make it clear which definition you are using if it isn't clear from context. – Charles Hudgins Jul 19 '21 at 20:00
  • I generally teach my undergraduate non-STEM students that rational functions are equal to their simplification restricted to the original domain. – Alan Jul 19 '21 at 20:21

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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.

Hagen von Eitzen
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    [mathjax: seems you forgot a slash in front of infty. ] – coffeemath Jul 19 '21 at 19:54
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    If I understand you correctly, in this example, $\frac{z-1}{z-1} \sim 1$, since $(z-1)1 = 1(z-1)$. Then, $\frac{1}{z+1} = 1 \cdot \frac{1}{z+1} \sim \frac{z-1}{(z-1)(z+1)} = \frac{z-1}{z^2-1}$. Doesn't this mean that this constructions does indeed care whether $a, b$ share zeroes, i.e. divisors? – tolUene Jul 19 '21 at 23:24