### Localization of a ring

So what exactly is the localization of a ring?

It is creating a field-imitation (and NOT necessarily a field with multiplicative inverses) for every ring. It includes the creation of multiplicative inverse- imitations (for elements that are not zero-divisors).

How does it do this? It apes the steps taken to create a field of fractions from an integral domain. A good description can be found on pg. 60 of Watkins’ “Topics in Commutative Ring Theory”, First Edition.

I would like to concentrate on one small point. Why is it that $\frac{a}{b}=\frac{c}{d}\implies t(ad-bc)=0$ where $b,d,t$ belong to the same multiplicative system? This description does not exactly ape the condition for integral domains: $\frac{a}{b}=\frac{c}{d}\implies (ad-bc)=0$. Simply put, why the extra $t$??

Because we have no means of isolating a regular element that is a common factor!! In every case, including that of integral domains, we get something like $h(ad-bc)=0$, where $h\in R$. Then we say $h$ is not a zero divisor. Hence, $ad-bc=0$. Here, we can’t say anything about $h$. Hence, the condition here is that $\frac{a}{b}=\frac{c}{d}\implies h(ad-bc)=0$.

Now where does the “multiplicative system” part come in?? That part seems fairly arbitrary, doesn’t it? How does it help prove transitivity? Because you’re going to get something of the form $h(af-be)$ anyway! Hence, if you didn’t add the $t$, how would you justify the $h$? Simple?

But why “multiplicative system”? Why not a simple $\frac{a}{b}=\frac{c}{d}\implies t(ad-bc)=0$ for some $t\in R$? Mainly because this allows us to localize $R$ with respect to different multiplicative systems within the ring. It helps us to generalize. We can choose the denominators and $t's$ (as in $t(ad-bc)$) from subsets of the ring, or the whole ring itself. It is the properties of these subsets that we will explore in the coming paragraphs.

Say we localize with respect to multiplicative system  $E$ of ring $R$. Then we choose all denominators from $E$. We need to choose denominators from the same multiplicative system because the way fraction multiplication is defined: we want multiplicative closure.

Why choose $t$ in $t(ad-bc)$ from the same multiplicative system as the denominators too? Why choose $t$ from a multiplicative system at all? We have to choose $t$ from the same multiplicative system because soon we start having products of $t's$, and we don’t want them to be from outside of the set we’re choosing the $t's$ from. This multiplicative system is the same as that of the denominators because denominators are also multiplied to the $t's$. Hence, we want algebraic closure.

I’ve always found the construction of the quotient field of a domain very arduous and time-consuming. Most people get lost somewhere in the proof. I am going to try and make it more transparent.

What are we trying to do? We’re trying to convert a ring into a field (please forgive my language). How are we doing it? We’re creating fractions out of numbers. Hence, consider the set of two-tuples $(a,b)$, where $a,b\in R$. Here, $R$ is the ring.

Why do we need $R$ to be an integral domain? Let us break this down

1. We (sort of) know that $(a,b)=\frac{a}{b}$. Now consider $\frac{a}{b}\times\frac{c}{d}=\frac{ac}{bd}$. By definition, $b,c\neq 0$. However, what if $bc=0$? Hence we can’t have zero-divisors.

2. We are making equivalence classes of elements, where $(a,b)\equiv (c,d)$ if $ad=bc$. This relationship is transitive only if the elements of the ring are commutative. Hence, we need commutativity. Construct a counter-example with $D_{2n}$ (which is non-commutative).

3. We also need $1\in R$, as we need a multiplicative unit in the field. A field always contains the multiplicative identity.

A ring with the three properties listed above is an integral domain. Hence, we need an integral domain.

Now just define addition and multiplication, and ensure that the operations are well-defined.

The thing that is tricky about this proof (for beginners) is that the construction of equivalence sets and verifying well-defined-ness is not intuitive. People just generally want to define addition and multiplication in the obvious way, and be done with it. As long as one remembers that checking whether the operations are well-define or not is important, one is fine. Note that if $a\equiv b$ and $c\equiv d$, and $a+c\not\equiv b+d$, then the operation is nonsensical. To make it clearer for the beginner, this is equivalent to saying $x+y\neq x+y$.