I heard that in a general Euclidean Domain, the quotient $q$ and remainder $r$ when an element $a$ is divided by an element $b$ by Division Algorithm, may not be unique. Is it true? If yes, can anyone give me examples where such a thing happens.
3 Answers
The ring of Gaussian integers $\mathbb{Z}[i]$ is a Euclidean domain with the complex norm in the role of the function measuring the size of elements. There we have plenty of examples. For example, take $a=3$ and $b=1+i$. Do we write the division step as $$ 3=(1-i)\cdot(1+i)+1, $$ i.e. $q=1-i$ and $r=1$, or $$ 3=(2-i)\cdot(1+i)-i, $$ i.e. $q=2-i$ and $r=-i$ or some other way (there are two more possibilities with $r=-1$ and $r=i$).
The problem is that the size function does not in general order the elements. Even in $\mathbb{Z}$ we get the following kind of non-uniqueness ($a=3, b=2$) $$ 3=2\cdot2-1=1\cdot2+1 $$ unless we specify that the remainder $r$ must be non-negative. In a more general Euclidean domain there is no such thing as a set of positive elements, so we have no choice but to give up on that requirement.
You do get uniqueness of $q$ and $r$ in the important case of a polynomial ring over a field. The degree of a polynomial is a very well-behaved function in this sense.
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It is almost never unique. If $u$ is any unit, then usually $a$ and $u \cdot a$ have the same degree. Thus there are far more than $d$ elements of degree $<d$. Even in $\mathbb{Z}$, there are $2d-1$ elements of degree $<d$.
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1+1 for the key observation that the number of small elements exceeds the number of cosets of the ideal generated by the divisor (in the cases where these numbers are finite). A much better explanation than my mumbling about singling out positive remainders. – Jyrki Lahtonen Apr 28 '13 at 09:43
Not much is demanded of the size function $\lambda$ for a Euclidean domain. In particular, we may have $\lambda(x+y)>\max\{\lambda(x),\lambda(y)\}$. This is crucial to the question of uniqueness in the divisional algorithm, which can be rephrased as asking whether some equation \begin{equation}\tilde r = \varphi g+r \end{equation} has a nontrivial solution $\varphi$, given $g\ne 0$ and $\max\{\lambda(\tilde r),\lambda(r)\}<\lambda(g)$.
For polynomial rings over integral domains, we always have $\deg(\tilde r-r)\le\max\{\deg(\tilde r),\deg(r)\}$, and $\deg(\varphi g) \ge \deg(g)$, so if the division algorithm yields some solution, then it must be unique.
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