Here's the exercise, as quoted from B.L. van der Waerden's Algebra,
Show that any commutative ring $\mathfrak{R}$ (with or without a zero divisor) can be embedded in a ''quotient ring" consisting of all quotients $a/b$, with $b$ not a divisor of zero. More generally, $b$ may range over any set $\mathfrak{M}$ of non-divisors of zero which is closed under multiplication (that is, $b_1$, $b_2$ is in $\mathfrak{M}$ when $b_1$ and $b_2$ are). The result is a quotient ring $\mathfrak{R}_{\mathfrak{M}}$.
I'm not sure if I am stuck or if I am overthinking. My answer goes like this:
Commutative rings without zero divisors are integral domains as defined in Algebra and, by removing all $a/b$ which are not in the the commutative ring $R$ from the field $R \hookrightarrow Q$ where $Q$ is the field of all quotients $a/b$, one shows that any commutative ring without zero divisors can be embedded in a quotient ring. (This is more rigorously outlined in Algebra itself.)
Now what has me confused is how this case differs from a case with zero divisors. Doesn't the exact same logic hold for a commutative ring with zero divisors? The only thing that I think zero divisors would interfere with is solving equations of the form $ax=b$ and $ya=b$ since there aren't inverses of zero divisors. However, we do not need to show that they can be embedded in a quotient field. We're only showing they can be embedded in a quotient ring. So, there's really no issue here and we apply the same approach as above.
Now, as for the last part, I think that all that needs to be said is that if the set $\mathfrak{M}$ wasn't closed under multiplication, neither would the "ring" be and hence it would not be a ring by definition. Therefore, $\mathfrak{M}$ has to be closed and any commutative ring where $b$ ranges over any set of non-divisors of zero can be embedded in a quotient ring.
I guess my real issue here is that I can't tell if I'm overthinking or underthinking. Does anyone care to elucidate this for me?
Best Answer
The fraction ring (localization) $\rm\,S^{-1} R\,$ is, conceptually, the universal way of adjoining inverses of $\rm\,S\,$ to $\rm\,R.\,$ The simplest way to construct it is $\rm\,S^{-1} R = R[x_i]/(s_i x_i - 1).\,$ This allows one to exploit the universal properties of quotient rings and polynomial rings to quickly construct and derive the basic properties of localizations (avoiding the many tedious verifications always "left for the reader" in the more commonly presented pair approach). For details of this folklore see e.g. the exposition in section 11.1 of Rotman's Advanced Modern Algebra, or Voloch, Rings of fractions the hard way.
Likely Voloch's title is a joke - since said presentation-based method is by far the easiest approach. In fact both Rotman's and Voloch's expositions can be simplified. Namely, the only nonobvious step in this approach is computing the kernel of $\rm\, R\to S^{-1} R,\,$ for which there is a nice trick:
$\quad \begin{eqnarray}\rm n = deg\, f\quad and\quad r &=&\rm (1\!-\!sx)\,f(x) &\Rightarrow&\ \rm f(0) = r\qquad\,\ \ \ via\ \ coef\ x^0 \\ \rm\Rightarrow\ (1\!+\!sx\!+\dots+\!(sx)^n)\, r &=&\rm (1\!-\!(sx)^{n+1})\, f(x) &\Rightarrow&\ \rm f(0)\,s^{n+1}\! = 0\quad via\ \ coef\ x^{n+1} \\ & & &\Rightarrow&\ \rm\quad r\ s^{n+1} = 0 \end{eqnarray}$
Therefore, if $\rm\,s\,$ is not a zero-divisor, then $\rm\,r = 0,\,$ so $\rm\, R\to S^{-1} R\,$ is an injection.
For cultural background, for an outstanding introduction to universal ideas see George Bergman's An Invitation to General Algebra and Universal Constructions.
You might also find illuminating Paul Cohn's historical article Localization in general rings, a historical survey - as well as other papers in that volume: Ranicki, A.(ed). Noncommutative localization in algebra and topology. ICMS 2002.