Page 198 - DMTH403_ABSTRACT_ALGEBRA

Unit 19: The Field of Quotient Euclidean Domains

Notes
Example: Show that 2Z is a maximal ideal of Z , whereas (0,4,8) is not.
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Solution: You know that Z = Z/12Z and 2Z = 2Z/12Z. We see that Z / 2Z = (Z/12Z)/
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(2Z/12Z) = Z , which is a field. Therefore, 2Z  {0,2,4,6,8,10} is maximal in Z .
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Now {0, 4, 8} = 4Z  2Z  Z .
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Therefore, {0, 4, 8} is not maximal in Z .
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We first introduced you to a special ideal of a ring, called a prime ideal. Its speciality lies in the
fact that the quotient ring corresponding to it is an integral domain.
Then we discussed a special kind of prime ideal, i.e., a maximal ideal.
19.2 Field of Quotients
a
Consider Z and Q. You know that every element of Q is of the form , where a  Z and b  Z*.
b
a
Actually, we can also denote by the ordered pair (a, b)  Z × Z*. Now, in Q we know that
b
a c
b  d = - iff ad = bc. Let us put a similar relation on the elements of Z × Z*

Now, we also know that the operations on Q are given by
a c
a c ad bc and . a c a c Q.

, 
b  d  bd b d  b d  b d
Keeping these in mind we can define operations on Z × Z*. Then we can suitably define an
equivalence relation on Z × Z* to get a field isomorphic to Q.
We can generalise this procedure to obtain a field from any integral domain. So, take an integral
domain R. Let K be the following set of ordered pairs:
K= {(a,b) ) a , b  R and b  0)
We define a relation ~ in K by
(a, b) ~ (c, d) if ad = bc.

We claim that ~ is an equivalence relation. Let us see if this is so.
(i) (a, b) ~ (a, b)  (a, b)  K, since R is commutative. Thus, ~ is reflexive.

(ii) Let (a, b), (c, d)  K such that (a, b) ~ (c, d). Then ad = bc, i.e., cb = da. Therefore, (c, d) ~
(a, b). Thus, ~ is symmetric.
(iii) Finally, let (a,b), (c,d), ( u, v)  K such that (a,b) – (c,d) and (c,d) ~ (u,v ). Then ad = bc and
cv = du. Therefore, (ad) v = (bc)v = bdu, i.e., avd =bud. Thus, by the cancellation law for
multiplication (which is valid for a domain), we get av = bu, i.e., (a,b) – (u,v). Thus, – is
transitive.
Hence, ~ is an equivalence relation.

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