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.

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Graduate student

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