Answer to Question #164882 in Abstract Algebra for Ashish Kumar

Question

Let D be an integral domain. Then prove that there exists a 

field F that contains a subring isomorphic to D.

Answer 1

Solution:

Proof:

Let a, b ∈ D with b$\ne$ 0.

Since E is a field, $b^{-1}$ ∈ E.

Thus define the function φ : F → E as φ(a, b) = $ab^{−1}$ .

Here we show that φ is well defined, so suppose (a, b),(a'b') ∈ F such that (a, b) = (a'b'). (Note that b, b'$\ne$ 0.)

Then by the definition of equivalence classes in the field of quotients, we have ab' = ba' ⇒ $ab^{−1}$ = $a'b'^{−1}$ ⇒ φ((a, b)) = φ((a'b' )).

Hence φ is well-defined.

Next, we show that φ preserves addition and multiplication.

Let (a, b),(c, d) ∈ F (b, d '$\ne$0).

Then φ((a, b)+(c, d)) = φ((ad+bc, bd)) = (ad+bc)($d^{-1}b^{-1}$ ) = $add^{−1} b^{-1}+bcd^{−1} b^{−1} = ab^{−1}+cd^{−1}$ = φ((a, b))+φ((c, d)) and φ((a, b)(c, d)) = φ((ac, bd)) = $acd^{−1} b ^{−1} = ab^{−1} cd^{−1}$ = φ((a, b))φ((c, d))

Thus φ preserves the operations and is a homomorphism.

Note that φ : F → φ(F) is onto by definition.

Next, we show that φ is one-to-one. Suppose that φ((a, b)) = φ((c, d)). Then $ab^{−1} = cd^{−1}$ which implies ad = bc.

Hence (a, b) = (c, d) by the definition of equivalence classes in the field of quotients and therefore φ is one-to-one. Since φ is a bijection, φ is an isomorphism.

We know that φ(F) is a field by its construction.

φ(F) is an integral domain (D) plus the inverses of the elements of D. Therefore we have a subfield of E, specifically φ(F) that is isomorphic to F.

Hence, proved.