In this post, I'll remind you how the field of quotients is constructed and how it works (but I won't prove that it works, since that's done in every textbook ever), and then we will use similar ideas to create a structure called the Partial Ring of Quotients, and find the key to creating unity!
(Added later: Some have told me this is also called the localization of a ring. My approach differs from Wikipedia's in notation, since I made this post based on an exercise in Fraleigh's algebra. I also aim to explain a bit more thoroughly than Wikipedia does!)
The Field of Quotients
The field of quotients' operations are defined on a set of equivalence classes of all ordered pairs of D. This sounds complicated, but it's not really! In $\mathbb{Q}$, for instance, we know that $\frac{1}{2} = \frac{3}{6}$. Those could be represented as ordered pairs, under some equivalence relation $\sim$ -- we'd like to generalize the idea that $(1, 2) \sim (3,6)$ to an arbitrary integral domain D. So, if D is an integral domain, we choose to say that $(n_1, d_1) \sim (n_2, d_2) \Leftrightarrow n_1 * d_2 = n_2 * d_1$, where $n_1, n_2, d_1, d_2 \in D$ (with $d_1, d_2 \neq 0$). Is this equivalence relation arbitrary--no! This is just like "cross-multiplying" to check if two fractions are equal, and disallowing fractions with a denominator of 0. This equivalence relation relies on $d_1$ and $d_2$ also not being zero divisors, and on the commutativity of $D$. Proof that it is an equivalence relation can be found in any introductory algebra text (or might even be relegated to the exercises).
The equivalence classes $[(n,d)]$ (read: numerator, denominator) form the elements of $F$, our field of quotients. The elements are equivalence classes because we want to make sure we make it clear that, for example, even though $(1, 2)$ and $(3, 6)$ may look different, they are actually representatives of the same exact element of $\mathbb{Q}$. You can add two equivalence classes in $F$ via $[(a,b)] + [(c,d)] = [(ad+bc, bd)]$ and multiply the equivalence classes by $[(a,b)]*[(c,d)]=[(ac, bd)]$. That these operations are well-defined regardless of representative ordered pairs, commutative, satisfy all the field axioms, etc., is, again, a topic textbooks go over immensely. But, you should be able to verify that if you represent two fractions in $\mathbb{Q}$ as ordered pairs, these operations work the same way as addition and multiplication in $\mathbb{Q}$.
In the field of quotients (which, to be fair, I haven't justified is actually a field), we now have a multiplicative inverse for every element not equivalent to 0, for if $[(a, b)]$ is an element of $F$ such that $a \neq 0$, the element $[(b,a)]$ is clearly also in $F$ and forms its multiplicative inverse -- just like taking the reciprocal. So even though we don't get multiplicative inverses in the integers, every element of $\mathbb{Q}$ does have a multiplicative inverse.
Can we think of $F$ as containing $D$? Of course. Simply define a map from $d \in D$ to $[(d, 1)]\in F$. It is a simple task to check that this gives an isomorphism from $D$ to a subset of $F$.
So above is basically the process you can use to turn any integral domain into a field that contains that integral domain -- extending it to permit division by any nonzero element.
So what about those rngs, and what's the partial ring of quotients?
But hope is not lost! As long as R has at least one element which is not a zero divisor, we can create a ring called $Q(R,T)$, the partial ring of quotients of $R$ with respect to a fixed set $T$. With it, we will create unity, (which may be the only type of unity the mathematician can hope to create, but one can dream) and we can create multiplicative inverses for at least some elements of our rng. But what is this set $T$, and why do we need it? It must be multiplicatively closed, and made up of non-zero divisors (you'll see why later).
If R has any element which is not a 0 divisor, then there is a multiplicatively-closed set of non-zero-divisors in R.
See, multiplying any non-zero-divisor $a$ by another non-zero-divisor $b$ (which may just be $a$ again) in our rng must yield a third non-zero divisor. Otherwise if $(ab)c = 0$ for nonzero c, then, using associativity, $a(bc)=0$, which since $a$ is not a 0 divisor, implies that $bc=0$, which is a contradiction -- $b$ was not a zero divisor.Thus the set of all non-zero-divisors in any ring or rng is multiplicatively closed; let's call this set T. (In fact, any multiplicatively closed set that contains no zero divisors will suffice throughout the argument, but this simply proves that if there are any non-zero-divisors in a rng, such a set must exist!)
We can now define the Partial Ring of Quotients $Q(R, T)$ on the set of equivalence classes of $R\times T$ the same way as we did for the field of quotients, with the same exact addition and multiplication operations. (Note that, as is convention, $R\times T$ denotes all ordered pairs $(r, t)$ with the first element in $R$ and the second element in $T$).
At this point, I will bother justifying that we still have an equivalence relation when describing the elements of $Q(R,T)$, since the potential existence of 0 divisors should make you uneasy about any claim regarding multiplication!
The relation $\sim$ defined as $(r_1, t_1) \sim (r_2, t_2) \Leftrightarrow r_1 t_2 = t_1 r_2$ is an equivalence relation (and thus, the elements of Q(R,T) are well defined).
Clearly, $\sim$ is reflexive, since R is commutative, so for any element $(r, t)\in R\times T$ we have $r*t = t*r$ and $(r,t)\sim(r,t)$.
That it is symmetric is also simple: If $(r_1,t_1) \sim (r_2, t_2)$ then $r_1 t_2 = t_1 r_2$. A quick rearrangement using R's commutativity yields that $r_2 t_1 = t_2 r_1$ so $(r_2, t_2) \sim (r_1, t_1)$
The transitivity is where we need that the elements of T are not zero divisors. (This argument is parallel, by the way, to the original argument that the elements of F the field of quotients are well defined) If $(a, b) \sim (c, d)$ and $(c, d) \sim (e, f)$, then we can write the equations:
$$ ad = bc \text{ and } cf = de $$
Multiplying both sides of the first equation by f, and both sides of the second equation by b, we get:
$$ adf = bcf \text{ and } bcf = bde $$
Rearranging and subtracting both sides of the equation from each other, we now get:
$$ \begin{align}
adf - bde &= bcf - bcf \\
daf - dbe &= 0 \quad \text{using commutativity here} \\
d(af - be) &= 0 \quad \text{left distribution works in all rings}
\end{align}$$
Since $d$ is an element of T, and not a 0 divisor, we must conclude that $(af-be) = 0$. Thus $af = be$ and $(a, b) \sim (e, f)$.
Though this argument is not materially different than the argument that the field of quotients' elements are well-defined, the last line illustrates the primary difference between the partial ring of quotients and the field of quotients: It is absolutely essential that the denominators are not zero divisors, or the last step of the argument would not work. This is why we can only allow the denominators to be members of a set which contains no zero divisors (but does not yet explain why that set must be closed under multiplication).
The proof that addition and multiplication are well-defined (at least in the sense that regardless of representatives, you get the same equivalence class) is not materially different from the proof for the field of quotients, except again it does hinge on the fact that T has no zero divisors. I will not repeat it here, since the crux of the differences is illustrated above.
On the way to ensuring $Q(R,T)$ is a ring, we need that $<Q(R,T), +>$ is a commutative group:
One thing that is not immediately obvious is that we need to make sure that $[(r_1, t_1)] + [(r_2, t_2)] = [(r_1 t_2 + t_1 r_2, t_1 t_2)]$ is actually an element of Q(R, T) -- this is one place where it is crucial that T is multiplicatively closed. Because of this, $t_1 t_2$ is an element of $T$, so the operation $+$ is closed in Q(R,T). If we did not require that T were multiplicatively closed, "+" might yield an element outside of T!
The argument for the associativity and commutativity of $+$ is identical to the argument for F, so I omit it here.
Using the additive identity $0 \in R$, the element $[(0, t)] \in Q(R,T)$ (for any representative $t$) acts as the additive identity. For $$[(r_1, t_1)] + [(0, t)] = [(r_1 t + 0 t_1, t_1 t)] = [(r_1 t, t_1 t)] = [(r_1, t_1)]$$ because $r_1 t t_1 = r_1 t_1 t$ by commutativity. Note that regardless of representative from $T$, $[(0, t_1)] = [(0, t_2)]$ since $0 t_2 = 0 = 0 t_1$.
Lastly, the existence of additive inverses in $R$ guarantees it in $Q(R,T)$ as well. For $[(r, t)] \in Q(R,T)$, there exists $[(-r, t)] \in Q(R, T)$ which satisfies the properties of the additive inverse. $$[(r, t)] + [(-r, t)] = [(rt + (-r)t, t^2)] = [(rt + -(rt), t^2)] = [(0, t^2)]$$ which is the additive identity in $Q(R,T)$.It doesn't take much more work to conclude that $Q(R,T)$ is a ring:
Multiplicative closure is, again, our biggest problem here, and it again hinges on T being multiplicatively closed. $[(r_1, t_1)] * [(r_2, t_2)] = [(r_1 r_2, t_1 t_2)]$ must give an element of $Q(R,T)$ because both R and T are closed under multiplication.
Multiplicative associativity is trivial (read: simply symbol-pushing) and hinges on its associativity in R. The argument is again the same as for the field of quotients.
$$[(a,b)]*([(c,d)]*[(e,f)]) = [(a,b)]*[(ce, df)] = [(a(ce), b(df)] \\
\qquad = [((ac)e, (bd)f)]=[(ac, bd)]*[(e,f)] = ([(a, b)] * [(c,d)]) * [(e,f)]$$
Thus, $Q(R,T)$ is a ring.
By this point you're probably wondering what the point is.
First cool thing: even if R does not have unity, $Q(R, T)$ does: If $t^*$ is any element of T, the element $[(t^*, t^*)]$ acts as unity in $Q(R, T)$. It's our multiplicative identity!
First, it is trivial to verify that we get the same element in Q(R,T) regardless of representative from T: $(t_1, t_1) \sim (t_2, t_2)$ because $t_1 t_2 = t_2 t_1$ in $R$, which is commutative.
If $[(r, t)]$ is an element of $Q(R, T)$, then $[(t^*, t^*)] * [(r, t)] = [(t^* r, (t^*)^2]$. In checking the equivalence relation we will see that this is the same element as $[(r, t)]$. For $(r, t) \sim (t^* r, =(t^*)^2$ because $ r (t^*)^2 = t^*r t^*$ via commutativity in $R$.Second cool thing: We really can consider $Q(R,T)$ to "contain" all of $R$. Fix an element of $T$, called $t^*$. Then the map $\phi : R\rightarrow T$ defined by $\phi(r) = (r t^*, t^*)$ is an isomorphism of $R$ into $Q(R,T)$.
Think about this as just multiplying any element of $R$ by our version of 1 in $Q(R,T)$!
First of all $\phi$ is one-to-one. If $\phi(r_1) = \phi(r_2)$ then $[(r_1 t^*, t^*)] = [(r_2 t^*, t^*)]$. Following the defined equivalence relation, $r_1 (t^*)^2 = r_2 (t^*)^2$, and since $t^*$ is not a 0 divisor, we can cancel and determine that $r_1 = r_2$.
Additionally, $\phi$ preserves $*$:Third cool thing: We can now think of $Q(R,T)$ as containing a multiplicative inverse for every element of $T$!
$$\begin{align} \phi(r_1 r_2) &= [(r_1 r_2 t^*, t^*)] \\
&= [(r_1 r_2 (t^*) ^2 ), (t^*) ^2] \quad \text{multiplication by 1} \\
&= [(r_1 t^*, t^*)] * [(r_2 t^*, t^*)] \quad \text{rearranging, using commuatativity.} \\
&= \phi(r_1) * \phi(r_2)
\end{align}$$
And $\phi$ preserves $+$:
$$\begin{align} \phi(r_1 + r_2) &= [( (r_1 + r_2 )t^*, t^*)] \\
&= [(r_1 + r_2)(t^*) ^2 ), (t^*) ^2] \quad \text{multiplication by 1} \\
&= [(r_1(t^*)^2 + r_2(t^*) ^2 ), (t^*) ^2] \\
&= [(r_1 t^*, t^*)] + [(r_2 t^*, t^*)] \\
&= \phi(r_1) + \phi(r_2)
\end{align}$$
Thus $\phi$ is an isomorphism (note that it might not be onto, just into).
So, for $t \in T$ we know its image under $\phi$ is $[(t t^*, t^*)] \in Q(R,T)$. But since T is multiplicatively closed, $t t^* \in T$ so we can rightfully say that $[(t^*, t t^*)] \in Q(R,T)$. If we multiply, $$ [(t t^*, t^*)] * [(t^*, t t^*)] = [t (t^*)^2, t (t^*)^2]$$ which is, as mentioned before, unity in $Q(R,T)$.
Notice that this might not be the image of an element in $R$, so the elements of our multiplicatively-closed subset T have inverses in $Q(R,T)$ even if they didn't in $R$.
So, with that, that is pretty much all of the advanced exercises from the section on the partial ring of quotients, from an old version of Fraleigh's Algebra. This took forever to write up. It's particularly infuriating to write something up nicely when it's already neatly in your head and needs no new organization.
At some point, I might have the energy to write up an example of a partial ring of quotients, say, $Q(\mathbb{Z}_9, \lbrace 1, 2, 4, 5, 7, 8\rbrace)$. But for right now, let's leave it at that.
Nice post; enjoyed reading it. I think one thing worth looking into is that (unless I'm mistaken) you can push this idea of partial rings of quotients even further by letting $$T$$ contain zero divisors and saying $$(a,t)\sim(b,s)$$ if $$as-bt$$ is a zero divisor with inverse in $$T$$.
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