0

I would like to understand the mathematics behind the standard long division algorithm, specifically for the ring of $\mathbb{Z}$. I'm looking for an algebraic or set theoretic explanation. Can anyone explain or point me to an explanation?

I asked here in which I was told the Wiki algorithm works over any field except for $\mathbb{Z}$. I was then referred to the non-monic algorithm proof in this post. I tried using the theorem and followed the proof, but I did not get an expected result. For instance using post:

$G = 1b^1 + 8b^0, F = 9b^0$, the leading coeff $a = 9$ then $i \ge max\{0, 1+deg G- degF\} = 2$. Substitution:

$a^i G = QF + R = 9^2(1b^1 + 8b^0) = Q9b^0 + R$, where $Q = 1(b^1+8b^0)$ and at this point it appears something went wrong.

Appreciate your guidance

Nick
  • 1,071
  • 2
    $\mathbb Z$ is a ring but not a field – J. W. Tanner Feb 08 '21 at 00:15
  • Are you trying to divide polynomials with integers coefficients or integer numbers? If the former then note that both polynomials have to be monic (i.e. leading coefficient equal to $1$; or to have at least invertible leading coeffcients) for this to work. Your second "polynomial" $F$ is not monic. – mrtaurho Feb 08 '21 at 00:24
  • I am looking to divide non-monic polynomials, which am assuming is what the high school standard integer long division algorithm does. – Nick Feb 08 '21 at 00:50

1 Answers1

1

First things first: $\Bbb Z$ is not a field (it's lacking multiplicative inverse for all non-zero elements distinct from $\pm1$). Polynomial long division works over any field $K$ as the polynomial ring $K[x]$ is a euclidean domain in this case. As I mentioned, $\Bbb Z$ is not a field and hence $\Bbb Z[x]$ not necessarily an euclidean domain (it's in fact even less; but that's not important right now).

We can, however, divide polynomials in arbitrary polynomial rings $R[x]$ whenever the leading coefficient of the divident is $1$ (or, more generally, invertible; this was mentioned in the post Bill Dubuque linked to you). In this case the usual algorithm works just fine. If the we have a field instead then every non-zero element is invertible which basically ensures that polynomial long division works out.

On to your specific case: you misread the precise statement of the divison algorithm on multiple occasions. As you have quoted

$$i\color{red}{\ge}\max\{1,1+\deg G-\deg F\}=2$$

which, in particular, includes the case $i>2$. You have only tested the case $i=2$. Moreover, the polynomials $Q,R$ are not of arbitrary degree; in fact $\deg R<\deg F$. But $\deg F=0$ and hence $\deg R<0$ which (by convention) implies $R\equiv0$ and we are only left with the possibility of $a^iG=QF$. This in turn does not leave that much choice for $Q$ other than $a^{i-1}G$ as $F=a$.

I am, honestly speaking, not sure what exactly you are asking.

mrtaurho
  • 16,103
  • @Nick As far as I can tell you're interesed in computations with integers numbers while the algorithm is only concerned with polynomials. Even though one can expand an integer base $10$ and it looks like a polynomial then it appears that it's arithmetical properties are quite different. The polynomial $X+8$ is only trivially "divisible" by $9$ as we lose the information that, for example, $9+1=X$ (more or less). I'm sorry, I don't think I can help you here. – mrtaurho Feb 08 '21 at 01:48
  • Your result follows the non-monic algorithm. I don't think this linked algorithm follows the standard integer long division people do by hand for computations because the steps appear different. Do you know the algorithm I'm searching for? – Nick Feb 08 '21 at 03:47
  • @Nick No, I don't think I can help you there. – mrtaurho Feb 08 '21 at 11:39
  • 2
    @Nick The non-monic version is exactly the same as the common high-school version except that we may need to scale the dividend by the lead-coeff of the divisor at each division step. – Bill Dubuque Feb 08 '21 at 21:02
  • @Nick I expanded the linked answer a bit. Maybe now it will be clearer to you. – Bill Dubuque Feb 09 '21 at 01:11