# Abstract Nonsense

## Euclidean Domains (Pt. I)

Point of Post: In this post we set the stage, in the typical manner, for the decreasing sequence of ‘nice integral domains’ starting from the nicest (non-field), Euclidean domains.

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Motivation

Now that we have ample information about general rings, it would seem to behoove us to start looking of classes of better behaved rings (similar to looking at Hausdorff spaces for topological spaces). In a somewhat surprising, but standard, approach we start in the opposite direction from where one might expect. In other words, we shall start with the nicest class of rings we shall consider and work our way down to the least nice, but most general. So, what is this class of nice rings for which we start? Well, anyone who has done any kind of number theory knows that the inability of division in $\mathbb{Z}$ is not as hard an obstacle for one to overcome as may be initially thought (of course, this inability to divide is what makes number theory a rich subject). Indeed, this is the case because while $\mathbb{Z}$ doesn’t have division, it has a division algorithm. Given any two integers $a,b$ we know that we can write $a=qb+r$ for some integer $q$ and some number $0\leqslant b<|r|$. The operative part, at least for us, is that we know we can make the remainder be ‘small’ (compared to $b$) since this allows us to perform minimality arguments (i.e. every subgroup of $\mathbb{Z}$ is generated by the element of least norm). This is the idea we’d like to abstractify. Namely, we’d like to consider classes of integral domains where we have a notion of ‘size’ which allows us to create a ‘division algorithm’ where the remainder has small ‘size’. In particular, this will entail us having an integral domain $R$ for which there is a ‘size function’ $d:R-\{0\}\to\mathbb{N}\cup\{0\}$ such that for every $a,b\in R$ there exists $q\in R$ such that $a=qb+r$ where $0\leqslant d(r).

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Euclidean Domains

A norm on an integral domain $R$ is a function $N:R\to\mathbb{N}\cup\{0\}$ with the property that $N(0)=0$. Norms are, as goes without saying, not very interesting. Some more interesting qualities a norm can have is that it may be multiplicative or positive $N(a)>0$ for $a\ne 0$. It’s clear actually, that only integral domains (or more precisely, rings without zero divisors) can have positive multiplicative norms, for if $ab=0$ then $N(ab)=0$ and so $N(a)N(b)=0$, but if $a,b\ne 0$ then $N(a),N(b)>0$ and so $N(a)N(b)>0$ contradictory to what was previously proven. Conversely, if $R$ is an integral domain then defining $N(r)=1-\delta_{r,0}$ gives a positive multiplicative norm on $R$. Thus, one could phrase that integral domains are precisely those commutative unital rings which admit positive multiplicative rings.

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Norms aren’t what ultimately interests us right now though, not even positive multiplicative norms. No, we are interested in a special type of norm called a Euclidean valuation. In particular, a norm $d:R\to\mathbb{N}\cup\{0\}$ on an integral domain $R$ is called a Euclidean valuation if given any $a,b\in R$ with $b\ne 0$ there exits $q,r\in R$ with $a=bq+r$ such that either $r=0$ or $d(r). Such a $q,r$ are called a quotient and remainder respectively. It is standard practice to call the value $d(a)$ the degree of $a$.

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So, why is this function useful? Well, the first thing it allows us to do is perform minimality arguments related to division. To illustrate consider the following proof that every ideal of the Euclidean domain $R$ is principal (generated by a single element of $R$):

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Theorem: Let $R$ be a Euclidean domain with Euclidean valuation$d$, then if $\mathfrak{a}$ is an ideal of $R$ then $\mathfrak{a}=(a)$ where $a$ is any element of $\mathfrak{a}$ of minimal degree. A Euclidean Domain is an integral domain $R$ for which there exists a Euclidean valuation $d$ on $R$.

Proof: First of all, it’s clear by the well-ordering principle that there exists elements of $\mathfrak{a}$ of minimal degree. Suppose that $a$ is such an element. Clearly $(a)\subseteq\mathfrak{a}$ and so it suffices to prove the reverse inclusion. To do this let $b\in\mathfrak{a}$ be arbitrary. We know then, by definition, that there exists $q,r\in R$ with $b=qa+r$  with $r=0$ or $d(r). That said, since $r=b-qa\in \mathfrak{a}$ and $a$ has minimal degree in $\mathfrak{a}$ we have that $d(r) cannot be the case, and so $r=0$. Thus, $b=qa$, or $b\in(a)$. Since $b$ was arbitrary we have that $\mathfrak{a}\subseteq(a)$ and the conclusion follows. $\blacksquare$

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Ok, fine, so we can see that an integral domain being Euclidean has its perks (this ‘principal ideal property’ is actually just one small fact), but for the notion of Euclidean domains to be useful, they must be common enough that we’d care. In other words, what are some examples of Euclidean domains? Well, probably the most obvious examples besides fields (where one can just define $d(r)=1$ for all $r\ne 0$)  is $\mathbb{Z}$ with $d(x)=|x|$. The fact that $\mathbb{Z}$ is, in fact, an integral domain was proven to you before, but it was just called the ‘division algorithm’. Other important examples include the quadratic integer ring $\mathbb{Z}[i]=\left\{a+bi:a,b\in\mathbb{Z}\right\}$ with $d(a+bi)=a^2+b^2$. This is, in fact, not hard to show that $k[x]$ is a Euclidean domain where $k$ is some field $d(p)=\deg(p)$. These three examples will constitute a lot of our time in the future, and we shall prove that they are Euclidean domains when their time comes. I should point out, that although the focus of the definition of Euclidean domains is on the $d$ function, probably the hardest part, which is implicitly implied in the definition is showing first off that every element can be written in the form $a=bq+r$.

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So, what are some non-examples of Euclidean domains? Well, while a lot of the theory that we shall see for Euclidean domains will be developed in later posts I discussed the fact that all ideals in Euclidean domains are principal because it gives a good way of proving that a given integral domain is not Euclidean. Indeed, it’s easy to see that the ring $\mathbb{Z}[x]$ is not Euclidean. Why? Well, I claim that the ideal $(p,x)$ for any prime $p$ is not principal in $\mathbb{Z}[x]$. Why though? Well, suppose for a second that it was, that $(p,x)=(q(x))$ for some polynomial $q(x)$. Since $p\in (q(x))$ this implies that $p=r(x)q(x)$ for some $r(x)\in\mathbb{Z}[x]$, and so, in particular $q(x)$ is constant. But, this then implies that $r(x)$ is also constant, and since $q(x)=q$ divides $p$ this implies that $q=\pm 1$ or $q=\pm p$. Well, clearly $q\ne \pm1$ otherwise $(q(x))=\mathbb{Z}[x]$ which would contradict $(q(x))=(p,x)$. Thus, we may conclude that $q(x)=\pm p$ but, evidently $(\pm p)\subsetneq(p,x)$ which is again a contradiction. Thus, $(p,x)$ is not principal in $\mathbb{Z}[x]$ and so $\mathbb{Z}[x]$ cannot be Euclidean (note, this means that there does NOT EXIST a $d$ which makes $\mathbb{Z}[x]$ into a Euclidean domain–not that it isn’t a Euclidean domain for a specific $d$ function)

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

October 13, 2011 -

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