Form of the elements of a localization

Question

If I have a ring $R$, a multiplicatively closed subset $U\subset R$, and consider an element of the localization: $\frac{r}{r'} \in U^{-1}R$, can I then assume without loss of generality that $r'\in U$?

If I'm not mistaken the definition only gives me that there exist some $r''\in R$ and $u,u'\in U$ such that $\frac{r}{r'}=\frac{r''}{u}$, that is: $u'ur=u'r'r''$, or in case of $R$ being an integral domain, just that $ur=r'r''$.

If I can't assume that $r'$ is in $U$, what would be a counterexample where this would get me into trouble?

Edit: Sorry, I mixed the symbols up at first. Should be correct now.

Answer 1

Let $R$ be a integral domain and $S$ a closed multiplicative subset. Then one defines $$ S^{-1}R=\{(r,s):r\in R,s\in S\}/\sim $$ where $(r,s)\sim(r',s')$ iff $rs'=sr'$. This set happens to be a ring with the operations $$ (r,s)+(r',s'):=(rs'+r's,ss'),\hspace{.3cm}(r,s)(r',s'):=(rr',ss'). $$ As you can see, you get nothing but elements of $S$ in "the denominator" (the second coordinate).

If you have $R$ a ring that is not an integral domain, then the above relation $\sim$ no longer define an equivalence relation; you need to identify more elements. Now, $$ S^{-1}R:=\{(r,s):r\in R,s\in S\}/\sim' $$ where $(r,s)\sim'(r',s')$ iff it exists some $s''\in S:s''(rs')=s''(sr')$. The sum and product operations are defined exactly as before.

Still, as you can see, there is nothing but elements of $S$ in the denominator. The point is that the original set $S^{-1}R$ (viewed just a a set) contains just elements of the form $(r,s)$ for $r\in R$ and $s\in S$, and we are identifying some of them to obtain cosets (which we denote exactly as before but realising that $(r,s)=(rs',ss')$ for every $s\in S$) but this changes nothing cause any element you choose as the representative of your coset (element of ring) will belong to the original set $S^{-1}R$ (viewed just as a set).

Update: If the ideal $(0)$ is a prime ideal in $R$ (that is to say, if $R$ is an integral domain), then you might consider the localization at $(0)$, $S^{-1}R=:R_{(0)}$, where $S=R\setminus\{0\}$. This ring is called the field of fractions of $R$, usually denoted by $Q(R)$. Now, for every other multipicative set $S'\subset R$ you have $S'\subset S$ (given that $0\not\in S'$, but in that case $S'^{-1}R$ would be the trivial ring $\{0\}$, so let's obvious that case). Here you have a natural application $\varphi:S'^{-1}R\to S^{-1}R=Q(R)$ given by $\varphi(r,s)=(r,s)$, that is well defined because $\varphi(r,s)=\varphi(rs',ss')$ for $s'\in S'$. Actually this application is a ring homomorphism. Still, in $S^{-1}R$ you have more elements identified (cause $S$ is bigger than $S'$), and it might happen that for one element $a=\varphi(r,s)\in S^{-1}R$ there are some representative in $S^{-1}R$ that doesn't belong to $S'^{-1}R$. This is not odd. Even if we can think of $S'^{-1}R$ as a subring of $S^{-1}R$, the sets $S'^{-1}R$ and $S^{-1}R$ are different.

Answer 2

You should be clear what you mean by the fraction. In general, if one writes $\frac{r}{s}$ where $r,s\in R$ for some (commutative) ring $R$, then it is implicitly assumed that the denominator $s$ is in the unit group of the rings.

Starting with a commutative ring $R$ and a multiplicative subset $S$, one can construct the localization $S^{-1}R$ of $R$ with respect to $S$. This new ring comes together with a canonical map $R\to S^{-1}R$ which is not necessarily injective. However in general one denotes the image of an element $r\in R$ simply by $r$ (the context making clear in which ring this element is considered). The notation $\frac{r}{1}$ is also used and does make clear that one works in the localization.

The image of an element $s\in S$ in $S^{-1}R$ is by construction a unit. That is why it makes sense to write $\frac{x}{s}$ inside $S^{-1}R$ and in fact all elements of $S^{-1}R$ are of the form $\frac{r}{s} = \frac{r}{1}\cdot (\frac{s}{1})^{-1}$ with $r\in R$ and $s\in S$.

If you mean by $\frac{r}{s}$ the class of $(r,s)\in R\times S$ under the equivalence relation that defines the localization, then by definition(!) the element $s$ should be in $S$. But this notation is compatible with the more general situation described above (the element that would be written by $\frac{1}{s}$ is indeed the multiplicative inverse of $\frac{s}{1}$).

Answer 3

What he asks is the following:

Given $\frac{r}{s}$ in the total ring of fractions of $R$. If it happens to be that $\frac{r}{s} \in U^{-1}R$ for some multiplicative system $U$, can we deduce $s \in U$.

The answer is clearly NO.

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Let $R$ be a noetherian domain. By Krull's principal ideal theorem, any non-unit $r \in R$ is contained in a prime of height $1$, i.e. the intersection of all $S = R \setminus P$, where $P$ ranges through all primes of height $1$ is the set of units.

From this - given a statement like $\frac{r}{s} \in S^{-1}R \Longrightarrow s \in S$ - we would easily deduce $R = \bigcap\limits_{\operatorname{ht} P=1}R_P$: If $\frac{r}{s} \in R_P$ for all primes of height $1$, we would get that $s$ is contained in the interesction of all $R \setminus P$, i.e. in the set of units. But this equality happens to be false for some non-normal domains, i.e. such an argument is deadly wrong.

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This is actually more subtle as one would think. The false assumption, that one could argue like this, leads to false proofs of a correct statement, which are though accepted and upvoted here...Luckily, there exist correct proofs, too.