Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

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ADVANCED ABSTRACT ALGEBRA
(m,a) (p,c) + (n, b) (p,c) = (mp, pa + mc + ac) + (np, pb + nc + bc)
= (mp + np, pa + mc + ac + pb + nc + bc)
Therefore
[(m,a) + (n,b)](p,c) = (m,a) (p,c) + (n,b) (p,c)
Similarly we can check it for right distributive law.
(vii)
(1,0) (m, a) = (m,a) = (m,a) (1, 0)
Hence (1, 0) = 1 is unity of S.
Hence S is a ring with unit element.
Consider the set
Since
and
T = {(0,a) | A ∈ R}
(0,a) + (0,b) = (0, a+b) ∈ T
0 = (0, 0) ∈ T
− (0,a) = (0, −a) ∈ T
(0,a)(0,b) = (0, ab) ∈ T,
therefore T is a subring of S.
We define a mapping
by
f : R → T
f(a) = (0,a) , a ∈ R
Then
f(a+b) = (0, a+b)
= (0, a) + (0, b)
= f(a) + f(b)
and
f(ab) = (0, ab)
= (0, a) (0, b)
= f(a) + f(b)
Thus f is a ring homomorphism. Also,
f(a) = f(b) (0, a) = (0, b)
a = b .
Therefore f is an isomorphism and hence R can be imbedded in S.
Theorem. Every integral domain can be imbedded in a field.
Proof. Let D be an integral domain and
S = {(a,b) | a, b ∈ D, b ≠ 0}