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The Wikipedia page for partial fraction expansion mentions that it can be generalised to "regular" fractions, i.e. fractions of integers: https://en.wikipedia.org/wiki/Partial_fraction_decomposition#Fractions_of_integers

There is a webpage here which discusses such a generalization, and remarks that it is not unique:

https://planetmath.org/partialfractions

The question I have is: why is the partial fraction expansion (generalized to fractions of integers, i.e. rationals) not unique? What is the condition that an integral domain needs to satisfy for the expansion to be unique?

Theo H
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  • Conversely, why do you think it should be unique? EG Addition certainly isn't unique. – Calvin Lin Mar 10 '23 at 23:03
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    The "partial fraction" for integers involves fractions whose denominators are powers of primes, and where the numerator of a fraction with denominator $p^n$ is an integer $a$ with $0\leq |a|\lt p$. – Arturo Magidin Mar 10 '23 at 23:07
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    The decomposition should not "jump" directly to a fraction with denominator $p^n$. As with rational functions, you should use several fractions. Instead of a single one with denominator, say, $p^3$, you should write it with $\frac{a_1}{p}+\frac{a_2}{p^2}+\frac{a_3}{p^3}$. – Arturo Magidin Mar 10 '23 at 23:12
  • @CalvinLin To clarify, I am not expecting it to be unique for the integers, but clearly it is unique in the case of polynomials. – Theo H Mar 10 '23 at 23:12
  • iirc uniqueness does hold if you restrict to nonnegative numerators & remainders. Proofs are surely not that difficult. – Bill Dubuque Mar 10 '23 at 23:24
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    Likely it boils down to the fact that if $D$ is euclidean domain with euclidean algorithm having unique quotient and remainder, then either $D$ is field $F$ or $D=F[x],,$ see here. – Bill Dubuque Mar 10 '23 at 23:37

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For relatively prime $m$ and $n$, we can split up an fraction with denominator $mn$ into fractions with denominators $m$ and $n$: $mx+ny = 1$ for some integers $m$ and $n$, so $$ \frac{1}{mn} = \frac{y}{m} + \frac{x}{n}. $$ The ambiguity in the choice of numerators on the right side is reflected in the ambiguity in the choice of integers $x$ and $y$ making $mx+ny = 1$: if $x_0$ and $y_0$ are one solution then all others are $x=x_0+nk$ and $y=y_0-nk$ for $k\in \mathbf Z$.

Example. To break up $1/15$, we need to solve $3x+5y=1$. One choice is $x=2$ and $y=-1$, so $$ \frac{1}{15} = \frac{-1}{3} + \frac{2}{5}. $$ The general choices for $x$ and $y$ are $x=2+5k$ and $y=-1-3k$, so $$ \frac{1}{15} = \frac{-1-3k}{3} + \frac{2+5k}{5}. $$ In particular, it is impossible to make such a decomposition of $1/15$ with both numerators nonnegative. (That’s also obvious since $1/3$ and $1/5$ both exceed $1/15$.)

The same result applies with a denominator that is a product of relatively prime nonconstant polynomials $f(t)$ and $g(t)$: we can write $uf + vg = 1$ for some polynomials $u(t)$ and $v(t)$, so $$ \frac{1}{fg} = \frac{v}{f} + \frac{u}{g}. $$

Polynomials are different from integers in an important way: there are unique minimal choices of $u$ and $v$: there are unique $u$ and $v$ such that $\deg u < \deg g$ and $\deg v < \deg f$. Moreover, that uniqueness also holds if we want to write a polynomial $h$ as $uf+vg$ when $\deg h < \deg(fg)$. That leads to $$ \frac{h}{fg} = \frac{v}{f} + \frac{u}{g}, $$ which is why the partial fraction decomposition of a proper ratio (numerator of smaller degree than denominator) leads to a sum of parts that involves only proper ratios. There is no analogue of this for rational numbers because there is no canonical way to solve $mx+ny=1$ in integers. More specifically, the set of all possible remainders under polynomial division by a fixed polynomial is an additive group: $\{r : \deg r < d\} \cup \{0\}$ is an additive group. There is nothing like that in the integers. If we did not insist on low-degree numerators in a partial fraction decomposition of rational functions then the partial fraction decomposition of rational functions would not be unique!

Something special about Euclidean domains of the form $F[t]$ for a field $F$ is that it has unique quotients and remainders for its Euclidean function. That is not true for the integers or the Gaussian integers with their Euclidean functions $|n|$ and ${\rm N}(\alpha)$. This sounds surprising for integers, where we are taught that there is a unique quotient and remainder! But be careful: that happens when we require $a=bq+r$ with $0\leq r < |b|$, which does not involve the Euclidean function on $r$. If we use $|r|$ in place of $r$ to measure the size of $r$ then $r$ is not unique since it could be negative: $10 = 4\cdot 2 + 2 = 4\cdot 3-2$. The only Euclidean domains other fields with unique quotients and remainders are the polynomial rings $F[t]$. Bill has helpfully provided a link that will direct you to proofs of this in one of his comments above.

KCd
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