Table of Content
An integral domain is not necessarily a field, but how can we make it a field? To make an integral domain a field, we need to add the inverses \(1/a\) and also add elements of the form \(b/a\), into the set.
First define a set
\[S = \{(a,b):a,b\in D, b\not =0\},\]where \(D\) is an integral domain and \((a,b)\in S\) is regarded as a formal fraction \(a/b\). Then define a relation \(\sim\) on \(S\) by
\[(a,b)\sim (c,d) \iff ad=bc,\]which is an equivalence relation. Let \(a/b\) denote the equivalence calss containing \((a,b)\), ant let
\[F = \{a/b: (a,b)\in S\}\]be the set of all equivalence classes. We now define addition and multiplication on \(F\), and then show that \(F\) becomes a field under these operations.
For \((a/b, c/d)\) in \(F\times F\), the assignments
\[+:(a/b,c/d)\mapsto(ad+bc)/bd\]and
\[\times:(a/b,c/d)\mapsto ac/bd\]are well-defined functions from \(F\times F\to F\).
There are two approaches to show an opeartion is well-defined, and below we only prove \(+\) is well-defined; \(\times\) can be proved in a similar way.
Approach \(\rm I\) (elements in two sums are equivalent)
We are going to show that if \((a_1, b_1)\sim(a_2,b_2)\) and \((c_1, d_1)\sim(c_2, d_2)\), then
\[(a_1d_1+b_1c_1, b_1d_1) \sim (a_2d_2+b_2c_2, b_2,d_2).\]\((a_1d_1+b_1c_1, b_1d_1) \in (a_1d_1+b_1c_1)/b_1d_1\).
We have
\[\begin{align*} (a_1d_1+b_1c_1)b_2d_2 &= (a_1b_2)d_1d_2 + (c_1d_2)b_1b_2 \\ &= (a_2b_1)d_1d_2 + (c_2d_1)b_1b_2 \\ &= (a_2d_2 + c_2b_2)b_1d_1. \end{align*}\]Thus \((a_1d_1+b_1c_1, b_1d_1) \sim (a_2d_2+b_2c_2, b_2,d_2)\) and \(+:F\times F\to F\) is well-defined. ◼
回頭看看 \(\sim\) 在 \(S\) 上的定義！
Approach \(\rm II\) (two sums are equal)
We now are to show that if \((a_1, b_1)\sim(a_2,b_2)\) and \((c_1, d_1)\sim(c_2, d_2)\), then
\[a_1/b_1 + c_1/d_1 = a_2/b_2 + c_2/d_2.\]We have
\[\begin{align*} a_1/b_1 + c_1/d_1 &= (a_1d_1 + b_1c_1)/b_1d_1 \\ &= b_2d_2(a_1d_1 + b_1c_1)/b_2d_2b_1d_1 \\ &= [(a_1b_2)d_1d_2 + (c_1d_2)b_1b_2] / b_1b_2d_1d_2 \\ &= [(a_2b_1)d_1d_2 + (c_2d_1)b_1b_2] / b_1b_2d_1d_2 \\ &= (a_2d_2 + b_2c_2)/b_2d_2 \\ &= a_2/b_2 + c_2/d_2. \end{align*}\]Thus the well-definedness of \(+\) is proved. ◼
可用 \(\mathbb{Z}_n\) 上的運算做類比，比較容易思考！
在這個例子中，Approach \(\rm I\) 比較方便。
Next we should prove that \(\langle F,+,\times\rangle\) is indeed a field.
By definition, \(D\) is an integral domain, which is a communtative ring without zero divisors. We can then observe that \(F\) also satisfies all the properties of an integral domain. Thus the remaining task for us is to show that every element in \(F\) is a unit:
For all \(a/b \in F\), there exist \(b/a \in F\) such that
\[a/b \times b/a = ab/ab = 1_D/1_D = 1_F \in F.\]Thus every element in \(F\) is a unit, and we have proved that \(F\) is a field. ◼
A subdomain of \((R,+,\circ)\) is a subset \(S\) of \(R\) such that \((S,+_S,\circ_S)\) is an integral domain.
The set \(D' = \{a/1:a\in D \}\) is a subdomain of \(F\), which is canonically isomorphic to \(D\).
For every integral domain \(D\), there exists a field containing a subdomain isomorphic to \(D\). Furthermore, if a field \(K\) contains \(D\), then it contains a subfield isomorphic to \(F\) defined earlier. Thus, \(F\) can be regarded as the smallest field containing \(D\).