Let D be an integral domain. Then there exists a field F (called the field of quotients of D) that contains a subring isomorphic to D. PROOF Let S = {(a, b) | a, b E D, b ± 0}. We define an equivalence rela- tion on S by (a, b) = (c, d) if ad = bc. Now, let F be the set of equivalence classes of S under the relation = and denote the equivalence class that contains (x, y) by xly. We define addition and multiplication on F by alb + cld = (ad + bc)/(bd) and alb - cld = (ac)(bd). (Notice that here we need the fact that D is an integral domain to ensure that multiplication is closed; that is, bd + 0 whenever b + 0 and d + 0.) Since there are many representations of any particular element of F (just as in the rationals, we have 1/2 = 3/6 = 4/8), we must show that these two operations are well-defined. To do this, suppose that alb = a'lb' and cld = c'ld', so that ab' = a'b and cd' = c'd. It then follows that (ad + bc)b'd' = adb'd' + bcb'd' = (ab')dd' + (cd')bb' = (a'b)dd' + (c'd)bb' = a'd'bd + b'c'bd = (a'd' + b'c')bd. Thus, by definition, we have (ad + bc)/(bd) = (a'd' + b'c')/(b'd'), and, therefore, addition is well-defined. We leave the verification that multiplication is well-defined as an exercise (Exercise 55). That F is a field is straightforward. Let 1 denote the unity of D. Then 0/1 is the additive identity of F. The additive inverse of alb is - alb; the multipli- cative inverse of a nonzero element alb is bla. The remaining field prop- erties can be checked easily.
Let D be an integral domain. Then there exists a field F (called the field of quotients of D) that contains a subring isomorphic to D. PROOF Let S = {(a, b) | a, b E D, b ± 0}. We define an equivalence rela- tion on S by (a, b) = (c, d) if ad = bc. Now, let F be the set of equivalence classes of S under the relation = and denote the equivalence class that contains (x, y) by xly. We define addition and multiplication on F by alb + cld = (ad + bc)/(bd) and alb - cld = (ac)(bd). (Notice that here we need the fact that D is an integral domain to ensure that multiplication is closed; that is, bd + 0 whenever b + 0 and d + 0.) Since there are many representations of any particular element of F (just as in the rationals, we have 1/2 = 3/6 = 4/8), we must show that these two operations are well-defined. To do this, suppose that alb = a'lb' and cld = c'ld', so that ab' = a'b and cd' = c'd. It then follows that (ad + bc)b'd' = adb'd' + bcb'd' = (ab')dd' + (cd')bb' = (a'b)dd' + (c'd)bb' = a'd'bd + b'c'bd = (a'd' + b'c')bd. Thus, by definition, we have (ad + bc)/(bd) = (a'd' + b'c')/(b'd'), and, therefore, addition is well-defined. We leave the verification that multiplication is well-defined as an exercise (Exercise 55). That F is a field is straightforward. Let 1 denote the unity of D. Then 0/1 is the additive identity of F. The additive inverse of alb is - alb; the multipli- cative inverse of a nonzero element alb is bla. The remaining field prop- erties can be checked easily.
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter7: Real And Complex Numbers
Section7.1: The Field Of Real Numbers
Problem 22E: Prove that if F is an ordered field with F+ as its set of positive elements, then F+nen+, where e...
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Show that the mapping Φ : D --> F in the proof of Theorem is a
ring homomorphism.
Transcribed Image Text:Let D be an integral domain. Then there exists a field F (called the
field of quotients of D) that contains a subring isomorphic to D.
PROOF Let S = {(a, b) | a, b E D, b ± 0}. We define an equivalence rela-
tion on S by (a, b) = (c, d) if ad = bc. Now, let F be the set of equivalence
classes of S under the relation = and denote the equivalence class that
contains (x, y) by xly. We define addition and multiplication on F by
alb + cld = (ad + bc)/(bd)
and
alb - cld = (ac)(bd).
(Notice that here we need the fact that D is an integral domain to ensure
that multiplication is closed; that is, bd + 0 whenever b + 0 and d + 0.)
Since there are many representations of any particular element of F
(just as in the rationals, we have 1/2 = 3/6 = 4/8), we must show that
these two operations are well-defined. To do this, suppose that alb = a'lb'
and cld = c'ld', so that ab' = a'b and cd' = c'd. It then follows that
(ad + bc)b'd' = adb'd' + bcb'd' = (ab')dd' + (cd')bb'
= (a'b)dd' + (c'd)bb' = a'd'bd + b'c'bd
= (a'd' + b'c')bd.
Thus, by definition, we have
(ad + bc)/(bd) = (a'd' + b'c')/(b'd'),
and, therefore, addition is well-defined. We leave the verification that
multiplication is well-defined as an exercise (Exercise 55). That F is a
field is straightforward. Let 1 denote the unity of D. Then 0/1 is the
additive identity of F. The additive inverse of alb is - alb; the multipli-
cative inverse of a nonzero element alb is bla. The remaining field prop-
erties can be checked easily.
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