Questions tagged [euclidean-domain]

Use for questions related to commutative rings that can be endowed with a Euclidean function, which allows a suitable generalization of the Euclidean division of the integers.

A Euclidean domain (also called a Euclidean ring) is a commutative ring that can be endowed with a Euclidean function, which allows a suitable generalization of the Euclidean division of the integers. This generalized Euclidean algorithm can be put to many of the same uses as Euclid's original algorithm in the ring of integers: in any Euclidean domain, one can apply the Euclidean algorithm to compute the greatest common divisor of any two elements. In particular, the greatest common divisor of any two elements exists and can be written as a linear combination of them (Bézout's identity). Moreover, every ideal in a Euclidean domain is principal, which implies a suitable generalization of the fundamental theorem of arithmetic: every Euclidean domain is a unique factorization domain.

It is important to compare the class of Euclidean domains with the larger class of principal ideal domains (PIDs). An arbitrary PID has much the same "structural properties" of a Euclidean domain (or, indeed, even of the ring of integers), but when an explicit algorithm for Euclidean division is known, one may use the Euclidean and extended Euclidean algorithms to compute greatest common divisors and the quantities in Bézout's identity. In particular, the existence of efficient algorithms for Euclidean division of integers and of polynomials in one variable over a field is of basic importance in computer algebra.

So, given an integral domain R, it is often very useful to know whether R has a Euclidean function: if so, that implies that R is a PID. If there is no "obvious" Euclidean function, however, then determining whether R is a PID is generally a much easier problem than determining whether it is a Euclidean domain.

Euclidean domains appear in the following chain of class inclusions:

commutative rings ⊃ integral domains ⊃ integrally closed domainsGCD domains ⊃ unique factorization domains ⊃ principal ideal domains ⊃ Euclidean domainsfieldsfinite fields

260 questions
46
votes
6 answers

Prove that the Gaussian Integer's ring is a Euclidean domain

I'm having some trouble proving that the Gaussian Integer's ring ($\mathbb{Z}[ i ]$) is an Euclidean domain. Here is what i've got so far. To be a Euclidean domain means that there is a defined application (often called norm) that verifies this two…
30
votes
5 answers

How is the logarithm of an integer analogous to the degree of a polynomial?

I've recently been reading Serge Lang's Math Talks for Undergraduates, specifically a section about the abc conjecture. Lang starts by stating and proving the Mason-Stothers Theorem: Let $f,g \in \mathbf{C}[t]$ be nonconstant and relatively prime.…
22
votes
5 answers

The ring $\Bbb Z\left [\frac{-1+\sqrt{-19}}{2}\right ]$ is not a Euclidean domain

Let $\alpha = \frac{1+\sqrt{-19}}{2}$. Let $A = \mathbb Z[\alpha]$. Let's assume that we know that its invertibles are $\{1,-1\}$. During an exercise we proved that: Lemma: If $(D,g)$ is a Euclidean domain such that its invertibles are $\{1,-1\}$,…
13
votes
5 answers

GCD in arbitrary domain

Is there a domain where computing GCD of two elements is not trivial (i.e. Euclid's algorithm will not work)? AFAIK we can always use the Euclid's algorithm in a Euclidean Domain.
11
votes
3 answers

Dividing one polynomial by another

How is this done? For example, how would one simplify the following? $$\frac{x^3-12x^2+0x-42}{x^2-2x+1}$$ I can do it with long division, but it never makes intuitive sense to me. Either an explanation of the long division algorithm or a new way…
10
votes
4 answers

$\mathbb{Z}[\sqrt{11}]$ is norm-euclidean

I'm trying to show that $\mathbb{Z}[\sqrt{11}]$ is Euclidean with respect to the function $a+b\sqrt{11} \mapsto|N(a+b\sqrt{11})| = | a^2 -11b^2|$ By multiplicativity, it suffices to show that $\forall x \in \mathbb{Q}(\sqrt{11}) \exists n \in…
9
votes
2 answers

Euclidean domain $\mathbb{Z}[\sqrt{d}]$

I am trying to generalized, for which integral values of $d$, $\mathbb{Z}[\sqrt{d}] = \{ a + b\sqrt{d} \vert a,b\in\mathbb{Z}\}$ is an Euclidean domain? I am interested specially in positive integral values of $d$.
9
votes
1 answer

Wlog a Euclidean domain size function satisfies $ V(a) \le V(ab)$

In the definition of domain, we first define a degree function $\vartheta: R^\times \rightarrow \mathbb{N}$ with such two constraints: (1) $\vartheta(f)\leq \vartheta(fg)$ for all $f,g\in R^\times$. (2) for all $f,g\in R$ with $f\in R^\times$, there…
9
votes
3 answers

Let $ I $ be an ideal in $\mathbb Z [i]$. Show that $\mathbb Z[i] /I $ is finite.

Let $I$ be an ideal in $\mathbb{Z}[i]$. I want to show that $\mathbb{Z}[i]/I$ is finite. I start with $\mathbb{Z}[i]/I$ is isomorphic to $\mathbb{Z}$. $\mathbb{Z}$ is ID then $I$ is prime.
8
votes
0 answers

Natural Euclidean Function Not Satisfying the $d$-inequality

Let me provide some background before I begin (although I feel as though it's hardly needed): Let $R$ be an integral domain. I call a function $d:R\setminus \{0\}\to\mathbb{N}\cup\{0\}$ a Euclidean function if for every $a,b\in R$, $a\ne0$,…
8
votes
2 answers

Why is $v(a) \leq v(ab)$ superfluous in Euclidean domain definition?

Can anyone make me understand in simple language why the second condition for being an Euclidean domain is superfluous ? Why $v(a) \leq v(ab)$ is not needed? How we can deduce from the first one?
8
votes
2 answers

Showing $(3 + \sqrt{3})$ is not a prime ideal in $\mathbb{Z}[\sqrt{3}]$

Let $I = (3+\sqrt{3})$ Looking at the field norm we note that $N(3 + \sqrt{3}) = 6$. We also know that $\mathbb{Z}[\sqrt{3}]$ is a Euclidean Domain. We want to find some $\alpha, \beta \in \mathbb{Z}[\sqrt{3}]$ s.t. $\alpha \cdot \beta = 3 +…
7
votes
1 answer

Degree of the kernel of a module map $R^n\rightarrow R^n$ for an Euclidean domain $R$

Let $R$ be an Euclidean domain with the degree function $d$. Let $A\in R^{n\times n}$ be an $n\times n$-matrix with entries in $R$ such that det$(A)=0$. As a module map $A:R^n\rightarrow R^n$, there always exists a kernel element $v\in R^n$ since…
7
votes
1 answer

When is $\mathbb Z[\zeta_n]$ a Euclidean Domain?

After having accidentally duplicated this question, I thought I'd follow up with a related question. In an answer to the linked question, Zev Chonoles quotes the first page of Chapter 11 of Washington's Introduction to Cyclotomic Fields which…
6
votes
0 answers

$\Bbb Z\left [\frac{-1+\sqrt{-19}}{2}\right ]$ is not a Euclidean domain

Definition: A universal side divisor, is an element $s\in R\setminus R^\times$ such that for every $x\in R$ either $s\mid x$, or there is some unit $u\in R^\times$ such that $s\mid x+u$. Fact: A Euclidean domain $R$ has universal side…
1
2 3
17 18