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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?

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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
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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

1 Answer 1

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.

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