Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A$ be an integral domain. If $S=A-\left\{0\right\}$, then $S^{-1}A$ is the field of fractions of $A$.

What is the problem if we actually take $S=A$? From what i see, in that case $0/0=1/1$ and the zero of the ring $S^{-1}A$ will be equal to its identity, thus $S^{-1}A=0$. Is there any other issue of non well-posedness?

share|cite|improve this question
I mean, that's a pretty big issue. What else are you looking for? – Qiaochu Yuan May 4 '12 at 16:04
What else is there? :-) – Manos May 4 '12 at 16:10
What do you mean ? Usually the localization is defined for a subset $S$ such that the product of two elements $S$ is in $S$ and $S$ does not contain 0. You can, if you wish, admit 0 in $S$ but then you obtain the ring with one element, as you almost proved it. – Lierre May 4 '12 at 16:34
I don't understand the question. If you take $S = A$ you get the zero ring and you do not get the field of fractions. What else is there to say? – Qiaochu Yuan May 4 '12 at 16:47
@QiaochuYuan: You said "it's a pretty big issue". Since this is a mathematics forum, i assume that you mean what you say. Irony does not help. – Manos May 4 '12 at 17:15
up vote 2 down vote accepted

Nothing horrible happens... except that you get the $0$ ring.

Recall that if $S$ is a nonempty multiplicative subset of a commutative ring (not necessarily with $1$) $A$, then $S^{-1}A$ is the ring whose underlying set is the quotient $$\left.\left\{\frac{a}{s} \, \right|\, a\in A, s\in S\right\}\Bigm/\sim$$ where $\frac{a}{s}\sim \frac{b}{s'}$ if and only if there exists $s''\in S$ such that $s''(as'-bs) = 0$ in $A$; and with addition and multiplication defined by $$\begin{align*} \frac{a}{s} + \frac{b}{t} &= \frac{at+bs}{st}\\ \frac{a}{s}\times\frac{b}{t} &= \frac{ab}{st}. \end{align*}$$ These operations are well defined, make $S^{-1}A$ into a ring with unity (the unity being the class of $\frac{s}{s}$ for any $s\in S$), and there is a natural homomorphism $\varphi\colon A\to S^{-1}A$ given by $\varphi(a) = \frac{as}{s}$, where $s\in S$ is an arbitrary element (this is also well-defined).

There is absolutely no problem if you include $0$ in your multiplicative subset $S$... but if you do, all you get is the zero ring. Because if $0\in S$, then for all $a,b\in A$, $s,s'\in S$, we have $\frac{a}{s}\sim\frac{b}{s'}$, since $0(as'-bs) = 0$ holds. Thus, $S^{-1}A$ has a single element, and so is the $0$ ring.

Because this occurs if you have $0\in S$, many authors exclude the case in which $0\in S$; not because the universe explodes if you put $0\in S$ or anything like that, but merely because as soon as $0\in S$, you just descend into the triviality of the zero ring. This is true with any commutative ring $A$, whether or not it is a domain.

You can extend the notion of "ring of [left/right] fractions" along similar lines with noncommutative rings, but only in some cases; one large class was studied by Ore, and you can find a lot of results on this in Lam's Lectures on Rings and Modules, Chapter 4. Again, one usually excludes the case of $0\in S$ to avoid trivialities.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.