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?

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    $\begingroup$ This is the usual construction that gives us the rationals from the integers, where we put $(x,a)$ equivalent to $(y,b)$ if $bx=ay$, and define addition, multiplication on equivalence classes by cheating and seeing what $\frac{x}{a}+\frac{y}{b}$ and $\frac{x}{a}\cdot\frac{y}{b}$ ought to be. Everything works in exactly the same way if second coordinates are restricted to a set of non zero-divisors closed under product. $\endgroup$ – André Nicolas Jun 23 '12 at 6:32
  • $\begingroup$ @AndréNicolas, can you clarify what you mean by "cheating"? Are you saying we go ahead and see what we should define, then write it from top to bottom? $\endgroup$ – 000 Jun 23 '12 at 6:35
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    $\begingroup$ Temporary notation to save typing, $[x,a]$ is the equivalence class of ordered pair $(x,a)$. Want to define $[x,a]+[y,b]$. Recall that informally $x/a+y/b=(bx+ay)/ab$, so define $[x,a]+[y,b]$ to be $[ax+by,ab]$. there is an npleasant thing that needs checking. We have defined addition of equivalence classes by using representatives. Need to check that if $[x',a']=[x,a]$ and $[y',b']=[y,b]$ we get the same answer, that is, that $(a'x'+b'y',a'b')$ and $(ax+by,ab)$ are equivalent. Multiplication is easier to define! Then need to verify get a ring. (Cont.) $\endgroup$ – André Nicolas Jun 23 '12 at 6:45
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    $\begingroup$ the calculations are routine, but there are quite a few. closure under operations is built-in, as is commutativity. Need to check associativity of the two operations, and distributivity, existence of additive inverse (easy, for $[x,a]$ use $[-x,a]$. The embedding of the ring maps $x$ to $[xb,b]$ where $b$ is arbitrary, or more naturally if $1$ is in our collection of allowed denominators, map $x$ to $[x,1]$. Nothing new, it is minor generalization of construction of the rationals from $\mathbb{Z}$. $\endgroup$ – André Nicolas Jun 23 '12 at 6:52
  • $\begingroup$ I see what you're saying. We kinda define things as we want and then make sure it works out as we wanted, it seems. Thanks for this way of looking at things. I had not fully realized it. $\endgroup$ – 000 Jun 23 '12 at 7:00

I'd say you're essentially on the right track, but I think you should think more carefully about how to rigorously define "$a/b$" (take a look at the page on localization if you need help). The "quotient ring" the question is asking about is called the total ring of fractions of $\mathfrak{R}$, which is a field if and only if $\mathfrak{R}$ is a domain.

Side note: I prefer the terminology "ring of fractions" (or "field of fractions") to "quotient ring", which to me is a term that refers to a quotient of a ring by an ideal.

  • $\begingroup$ Thanks a bunch. I definitely think that there is a significant difference in modern terminology and notation as compared to B.L.'s. There are also a few differences in subtle details, which makes it very interesting to read the more modern treatments. $\endgroup$ – 000 Jun 23 '12 at 6:41

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.

Voloch's title is probably a joke - since said presentation-based approach 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 [1].

[1] Ranicki, A.(ed). Noncommutative localization in algebra and topology. ICMS 2002


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