Page 199 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Let us denote the equivalence class that contains (a,b) by [a,b]. Thus,
[a,b] = {(c,d) | c,d  R, d  0 and ad = bc }.

Let F be the set of all equivalence classes of K with respect to ~.
Let us define + and in F as follows. (It might help you to keep in mind the rules for adding and
multiplying rational numbers.)

[a,b] + [c,d] = [ad+bc,bd] and
[a,b].[c,d] = [ac,bd].
.
Do you think + and are binary operations on F?

Note b # 0 and d # 0 in the integral domain R imply bd # 0. So, the right-hand
sides of the equations given above are well defined equivalence classes. Thus, the sum and
product of two elements in F is again an element in F.
We must make sure that these operations are well defined.
So, let [a,b] = [a’,b’] and [c,d] = [c’,d’]. We have to show that [a,b] + [c,d] = [a’,b’] + [c’,d’],
i.e., [ad+bc,bd] = [a’d’+b’c’,b’d’].
Now, (ad+bc) b’d’ – (a’d’ + b’c’) bd
= ab’dd’, + cd’bb’ – a’bdd’ – cdbb’
= (ab’ – a’b)dd’ + (cd’ - c’d) bb’

= (0) dd’ + (0)bb’, since (a,b) - (a’, b’) and (c,d) ~ (c’,d’).
= 0
Hence, [ad+bc,bd] = [a’ d’ + b’c’,brd’], i.e., + is well defined.

Now, let us show that [a,b] . [c,d] = [a’,br] . [c’,d’],
i.e., [ac,bd] = [a’c’ b’d’].
Consider (ac) (b’d’) - (bd) (a’c’)
= ab’cd’ – ba’dc’ = ba’cd’ – ba’ cd’, since ab’ = bar and cd’ = dc’
= 0

Therefore, [ac,bd] = [a’c’,b’d’]. Hence,. is well defined.
We will now prove that F is a field.
(i) + is associative : For [a,b], [c,d], [u,v] E F,

([a,b] + [c,d]) + [u,v] = [ad+bc,bd] + [u,v]
= [(ad+bc)v + ubd, Wv]
= [adv + b(cv+ud), bdv]
= [a,b] + [cv+ud,dv]
= [a,b] + ([c,d]+ [u,v])

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