# Abstract Nonsense

## Euclidean Domains (Pt. II)

Point of Post: This is a continuation of this post.

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The $d$-inequality

There is an axiom that we have omitted from our definition of Euclidean domain but is sometimes included in textbooks (for example [3] as listed below). To make clear what this is, we define a function $f:R-\{0\}\to\mathbb{N}\cup\{0\}$ for an integral domain $R$ to have the $d$-inequality if $f(a)\leqslant f(ab)$ for all $a,b\in R$. Then, people often define a Euclidean domain to be an integral domain $R$ such that there exists a Euclidean valuation $d$ on $R$ which satisfies the $d$-inequality. Why? This why is actually two why’s rolled into one–why is this desirable, and why the discrepancy? Well, let’s answer these questions in the order asked. So, why is the $d$-inequality desirable? Well, in the back of everyone’s mind, we think of units as things that ‘act like $1$‘.  Unfortunately, this intuition does not help us find units, or even more fundamentally, decide whether a given ring element is a unit. It would be nice if we had some way of quantitatively deciding, given a ring element, if its a unit. Well, this is precisely the case if we are given a Euclidean valuation satisfying the $d$-inequality. Namely, we’ll prove that if $d$ is a Euclidean domain satisfying the $d$-inequality then $a$ being a unit is equivalent to $d(a)=d(1)$.

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Lemma: Let $(R,d)$ be a Euclidean domain, with $d$ satisfying the $d$-inequality. Then, $d(a)\geqslant d(1)$ for all $a\in R-\{0\}$, $d(ab)=d(a)$ for all $b\in R^\times$ and $a\in R-\{0\}$, and $d(ab)>d(a)$ if $b\notin R^\times$.

Proof: This first assertion follows immediately since $d(a)=d(1\cdot a)\geqslant d(1)$. To prove the second assertion we merely note that $d(ab)\geqslant d(a)$ by the $d$-inequality, and since $d(a)=d(abb^{-1})\geqslant d(ab)$ and so $d(a)=d(ab)$. To prove the last fact we assume that $d(ab)=d(a)$. We know then that $a=qab+r$ with either $r=0$ or $d(ab)>d(r)$. Now, we may rewrite this as $a(1-qb)=r$ and since $qb\ne 1$ (since $b$ is not a unit) we know for sure that $r\ne 0$. Thus, we have that $d(r), but $d(r)=d(a(1-qb))\geqslant d(a)$ which is a contradiction. $\blacksquare$

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Putting these three together we get the following results:

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Theorem: Let $(R,d)$ be a Euclidean domain with $d$ satisfying the $d$-inequality. Then, the units of $R$ are precisely the elements of $R$ of minimal degree. In particular, $d(1)$ has minimal degree in $R$ and $a\in R^\times$ if and only if $d(a)=d(1)$. Moreover, if $a,b\notin R^\times$ then $d(ab)>\max\{d(a),d(b)\}$.

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For example, if we assume for a second that the rings I stated earlier are integral domains with the $d$ I stated, then it’s easy to see that they all satisfy the $d$-inequality. In particular, the only units of $F[x]$ are those with $d(p(x))=d(1)=0$, and so the non-zero constants. The only units of $\mathbb{Z}[i]$ are those with $d(a+bi)=d(1)=1$ and so easily verified to be $a+bi\in\{\pm 1,\pm i\}$.

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Ok, so now it’s clear that having a Euclidean valuation which satisfies the $d$-inequality is a desirable thing. So, the question is, why do we not state that all our integral domains have it! Well, ostensibly you’d think that one camp of mathematical expositors said “The $d$-inequality puts too great of restriction on the rings we want to study, so let’s not include it. All the important results are independent of it anyways.” and the other camp is going “The $d$-inequality is an indispensable tool when dealing with integral domains! Sure, we have to forget some rings, but they’re the weird, degenerate cases anyways…it’s worth it!” Well, in fact, this is not the case. Indeed, the class of integral domains which are Euclidean domains and those for which there exists a Euclidean function satisfying the $d$-inequality are the same! To be more specific, clearly any ring possessing a Euclidean function satisfying the $d$-inequality will be an integral domain, and conversely given a Euclidean domain $(R,d)$ we shall see that there is a natural way of creating a new function $d'$ on $R$, built from $d$, which will be a Euclidean function satisfying the $d$-inequality. So, how should we go about building this function $d'$? Well, the above theorem about functions satisfying the $d$-inequality gives us a hint. Namely, we know that $d'(1)$ needs to be the minimal value of $d'$ on $R$. Indeed, what we should have is that the more things that some $x\in R$ divides the smaller it should be, obviously culminating in the fact that a unit (which divides everything) should have minimal degree. This suggests that our function $d'$ should depend on how many ring elements its argument divides, and depend inversely on this (the more things it divides the smaller it should be). Since we would like to build $d'$ from $d$ (since we need $d'$ to satisfy the division algorithm property) it seems sensible then to define $\displaystyle d'(x)=\min_{y\ne 0}d(xy)$. This turns out to be the right path to take, but why? Well, it’s clear that $d'$ has the one qualifying attribute we’d want, it satisfies the $d$-inequality. To see this we merely note that if $x\mid y$ then every thing divisible by $y$ is divisible by $x$, or said differently $\left\{xz:z\ne0\right\}\supseteq\left\{yz:z\ne0\right\}$. Since our function is taking the infimum over the image of these sets under $d$ this clearly implies that $d'(x)\geqslant d'(y)$. Ok, good! So we have a function which is now satisfies the $d$-inequality…but why is it a Euclidean function? Well, the first thing we might try to do is, given $a,b\in R$, we can try to just do the normal “there exists $q,r$ such that $b=qa+r$ and $r=0$ or $d(r)“. Of course, the problem with this is that we can’t assert a priori that $d'(r) for non-zero $r$. There is a trick though we can do. by definition $d'(a)=d(ax)$ for some $x\in R-\{0\}$ and we know that there exists $q,r\in R$ with $b=qax+r$. Let then $q'=qx$, so that $b=q'a+r$. If $r\ne 0$ then we know that $d(r). But, in general we know that $d'(r)\leqslant d(r)$ so that $d'(r)\leqslant d(ax)=d'(a)$. The conclusion then follows. Putting this all together we get:

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Theorem: Let $(R,d)$ be an integral domain. Then, if $d':R-\{0\}\to\mathbb{N}\cup\{0\}$ is defined by the rule $\displaystyle d'(x)=\min_{y\ne 0}d(xy)$ then $d'$ satisfies the $d$-inequality and $\left(R,d'\right)$ is a Euclidean domain.

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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.