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Let $D$ be an integral domain and $F=\mathrm{Frac}(D)$ be the field of fractions of $D$. We will look at $F$ as a set of equivalence classes from the equivalence relation $\sim$ on $D\times \left(D-\{0\}\right)$ by $(a,b)\sim(c,d)$ if and only if $ad=bc$ with the usual fraction addition and multiplication. We'll just denote $$\frac{p}{q} = \{ (a,b) \in D\times \left(D-\{0\}\right) \mid (a,b)\sim(p,q)\}=\{ (a,b) \in D\times \left(D-\{0\}\right) \mid aq=bp\}.$$

I am wondering if we have $\frac{p}{q} \in F$ under what conditions is it true that if $\frac{a}{b}=\frac{p}{q}$ then there exists some $h \in D-\{0\}$ where either $$\frac{a}{b}=\frac{hp}{hq} \text{ and $a=hp$ and $b=hq$ or } \frac{p}{q}=\frac{ha}{hb} \text{ and $p=ha$ and $q=hb$}.$$

I'd like to be able to do this to have a better idea of what the elements look like that are equal to $\frac{p}{q}$, to simplify what $\frac{p}{q}$ or $\frac{a}{b}$ looks like as a set.

Edit: Changed what I am looking for exactly I think.

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    $\begingroup$ one element worth thinking about would be $(a,a)$ where obviously $a \neq 0$. $\endgroup$ – James S. Cook Aug 31 '13 at 1:10
  • $\begingroup$ In the fraction field of e.g. ${\Bbb Z}[x,y]$, $x/x=y/y$ but there is no $h$ for which $x=hy$ or $y=xh$. (This is what James is talking about above.) I don't expect the condition that there is such an $h$ to in general be reducible into simpler or more familiar conditions. $\endgroup$ – anon Aug 31 '13 at 1:52
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    $\begingroup$ @anon Noetherian rings with this property are exactly the discrete valuation rings (or fields). $\endgroup$ – user714630 Aug 31 '13 at 21:56
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In this answer, I will prove the following result, where we allow ourselves to work in only noetherian rings when necessary:

Theorem. Let $D$ be a noetherian domain and let $F$ be its field of fractions. Suppose that when ever $a/b=c/d$ in $F$, there is an element $h$ of $D$ such that $a=hc$ and $b=hd$ or $c=ha$ and $d=hb$. Then, there exists an element $m$ of $D$ such that every element of $D$ takes the form $um^n$ where $n$ is a nonnegative integer and $u$ is an invertible element of $D$.

Let's start with a lemma. It is a restatement of this property in terms of ideals.

Lemma I. Let $D$ be an integral domain and let $F$ be its field of fractions. A necessary and sufficient condition for the principal ideals of $D$ to form a total order is $a/b=c/d$ in $F$ implies there exists $h\in D$ such that $ah=c,bh=d$ or $ch=a,dh=b$.

Proof. Start by assuming the principal ideals are totally ordered (by inclusion). Suppose $a/b=c/d$ in $F$, so that $ad=bc$ in $D$. Without loss of generality, we may assume $(a)\subseteq (c)$. That is, there exists $h\in D$ such that $a=hc$. Then $hcd=bc$. Cancel $c$ to get $b=hd$, as desired.

Now, we show the converse. If $a,b$ are nonzero elements of $D$, then $a/a=b/b$ implies either $a=hb$ or $b=ha$. That is, one of $(a),(b)$ is contained in the other. $\square$

The total order on the principal ideals is a strong property, we will call the property $T$ to save typing. For instance,

Proposition II. A noetherian ring with property $T$ is a principal ideal ring.

Proof. Let $I$ be an ideal, and note that $$I=\bigcup_{x\in I}(x).$$ The noetherian property implies there is a maximal element of $\{(x):x\in I\}$. Since that set is a chain, $I=(x)$ for some $x\in I$. $\square$

And moreover,

Proposition III. If $R$ is a ring satisfying $T$, then $R$ is a local ring; in other words, $R$ has a unique maximal ideal.

Proof. Let $M$ be a maximal ideal of $R$, where $R$ satisfies $T$. Suppose $x\in R\setminus M$, then $(x)\not\subseteq (m)$ for any $m\in M$. Thus, for all $m\in M$, $(m)\subseteq (x)$, and this implies $$M=\bigcup_{m\in M}(m)\subseteq (x).$$ Since $x\notin M$, $(x)=(1)$ and thus $x$ is invertible. It follows that the set of noninvertible elements of $R$ is an ideal, which is a condition for $R$ to be local. $\square$

Consider the case $D$ is a noetherian integral domain with $T$. Then $D$ is a principal ideal domain, and thus it is a unique factorization domain. The only irreducible element of $D$, however, is the generator $m$ of the maximal ideal; this is because, in principal ideal domains, irreducible elements generate maximal ideals. Thus, if $a\in D$, then $a$ is a unit multiple of a power of $m$. This completes the proof of the theorem.

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  • $\begingroup$ A correction to the proof of Lemma I. In the first paragraph, assume $c\ne0$ so that we can really cancel $c$. If $c=0$, then $a=0$, and either $(b)\subseteq(d)$ or $(d)\subseteq(b)$. Suppose the first, and find $h$ such that $b=hd$. Of course, $a=0=hc$ as well. $\endgroup$ – user714630 Sep 4 '13 at 22:07
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Since $aq=bp$ we have $\dfrac{ap}{aq}=\dfrac{pa}{pb}$. I belive that is the nearest one can get to the "cancellation" that you wanted.

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  • $\begingroup$ I don't think this is exactly what I wanted so I will see if I can come up with a better way to say what I am trying to do. What you say here though proves the existence of the $h$. $\endgroup$ – Frudrururu Aug 31 '13 at 1:45
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    $\begingroup$ Well, not a "single" $h$, but a different $h$ for each fraction. $\endgroup$ – André Nicolas Aug 31 '13 at 1:51
  • $\begingroup$ Yes indeed thanks my oversight $\endgroup$ – Frudrururu Aug 31 '13 at 2:10

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