7
$\begingroup$

Is there a "special formula" one can follow to prove whether something is a Euclidean domain or not? I've been looking around, but I haven't seemed to be able to find one, so I was wondering whether I was blind, or there just isn't one.

I have $\mathbb Z[\sqrt-3]=\{x+y\sqrt-3|x,y\in\mathbb Z\}$. I think I can remember having read somewhere that it's a Euclidean domain, but I'm not sure. I also don't know how to prove it. Any hints?

  • 0
    See [here](http://math.stackexchange.com/q/778122/28900), for example, to get a more general definition.2017-04-05

3 Answers 3

3

Hint: $4=(2)(2)=(1-\sqrt{-3})(1+\sqrt{-3}).$

11

To my knowledge, given an arbitrary integral domain, there is no "general" method to figure out whether it is a Euclidean domain.

To expand a bit on JessicaB's comment, though, we can completely determine which of the rings of the form $\Bbb Z[\sqrt{-n}]:=\{a+b\sqrt{-n}:a,b\in\Bbb Z\}$ are Euclidean domains (where $n$ is some positive integer). Let me outline how one might do it.


Given a positive integer $n$, we define $\rho_n$ from $\Bbb Z[\sqrt{-n}]$ to the nonnegative integers by $\rho_n(a+b\sqrt{-n}):=a^2+b^2n.$ This function will tell us important things about the ring. Some useful facts to prove are:

(A) $\rho_n$ is a multiplicative function--that is, $\rho_n(x\cdot y)=\rho_n(x)\cdot\rho_n(y)$.

(B) $x\in\Bbb Z[\sqrt{-n}]$ is a unit of $\Bbb Z[\sqrt{-n}]$ if and only if $\rho_n(x)=1$, and $x=0$ if and only if $\rho_n(x)=0$.

(C) If $x,y\in\Bbb Z[\sqrt{-n}]$ are associates (that is, differ by multiplication by a unit), then $\rho_n(x)=\rho_n(y)$. (The converse doesn't hold, though. Consider $1\pm2\sqrt{-n}$.)

(D) If $x\in\Bbb Z[\sqrt{-n}]$ is nonzero and not a unit, then $x=x_1\cdots x_k$, where each $x_j$ is irreducible in $\Bbb Z[\sqrt{-n}]$. (That is, we have existence, though not necessarily uniqueness, of irreducible factorizations.)

(E) If $n\geq 2$, then $\sqrt{-n}$ is irreducible in $\Bbb Z[\sqrt{-n}]$.

(F) If $n\geq 3$, then $2$ is irreducible in $\Bbb Z[\sqrt{-n}]$.

Having these handy facts in our arsenal, it isn't too difficult to prove the following two results:

($1$) If $n=1$ or $n=2$, then $\Bbb Z[\sqrt{-n}]$ is a Euclidean domain, with Euclidean function $\rho_n$.

($2$) If $n\geq 3$ (whether $n$ is square-free or not), then $\Bbb Z[\sqrt{-n}]$ is not a UFD, so not a Euclidean domain. [As JessicaB pointed out, you need only show that $2$ is not prime in $\Bbb Z[\sqrt{-n}]$. You may want to do two cases, for $n$ odd and $n$ even.]

  • 0
    It occurs to me: did you mean "homomorphic image," rather than "holomorphic image"? If so, then$I$*can* answer your question (and just did, [here](http://math.stackexchange.com/questions/574260/problem-related-to-homomorphic-map-of-a-pid-r/574280#574280)).2013-11-20
4

To prove something is or is not a Euclidean domain, it seems useful to use the following chart:

Fields $\subset$ Euclidean Domain $\subset$ Principal Ideal Domain $\subset$ Unique Factorization Domain $\subset$ Domain

In particular, to prove something is a Euclidean domain, you may prove either it is a field (only if it actually is a field), or you may prove it is a Euclidean domain directly (See below for details).

To prove something is not a euclidean domain, you may prove that it is not one of the latter ones: i.e., prove there exists an ideal that is not principal, a factorization that is not unique, or zero divisors.

To prove something is a euclidean domain, by and large you must prove the existence of a division algorithm using the standard definition that $\forall x,y \in D \exists q,r$ satisfying $ x = qy + r$ and $\mathcal{N}(r) < \mathcal{N}(y)$, where $\mathcal{N}$ is a norm on your domain $D$.