Page 191 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Thus, for a, system (R,+,.) to be a field it must satisfy the ring axioms R to R as well as the
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following axioms.
(i) is commutative,
(ii) R has identity (which we denote by 1) and 1  0, and
(iii) every non-zero element x in R has a multiplicative inverse, which we denote by x .
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Just as a matter of information we would like to tell you that a ring that satisfies only (ii) and (iii)
above, is called a division ring or a skew field or a non-commutative field. Such rings are very
important in the study of algebra, but we will not be discussing them in this course.
Let us go back to fields now. The notion of a field evolved daring the 19th century through the
research of the German mathematicians Richard Dedekind and Leopold Kronecker in algebraic
number theory. Dedekind used the German word Körper, which means field, for this concept.
This is why you will often find that a field is denoted by K.

As you may have realised, two of the best known examples of fields are R and C. These were the
fields that Dedekind considered. Yet another example of a field is the following ring.


Example: Show that Q  2Q  {a  2b|a,b Q} is field.
Solution: From Unit 14 you know that F  Q  2Q is a commutative ring with identity 1 + 2.

Now, let a  2b be a non-zero element of F. Then either a  0 or b  0. Now, using the
rationalisation process, we see that

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

 a  2b  = 1  1  2b  a fib
2
a  2b (a  2b)(a  2b) a  2b 2
1 ( b)

= 2 2  2 2 2  F
a  2b a  2b
(Note that a – 2b  0, since 2 is not rational and either a  0 or b  0.)
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Thus, every non-zero element has a multiplicative inverse. Therefore, Q  2Q is a field.
Can you think of an example of a ring that is not a field? Does every non-zero integer have a
multiplicative inverse in Z? No. Thus, Z is not a field.
By now you have seen several examples of fields. Have you observed that all of them happen to
be integral domains also? This is not a coincidence. In fact, we have the following result.

Theorem 5: Every field is an integral domain.
Proof: Let F be a field. Then F  {0} and 1  B. We need to see if F has zero divisors. So let a and
b be elements of F such that ab = 0 and a  0. As a  0 and P is a field, a exists.
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Hence, b = I . b = (a – la) b = ad (ab) = a 0. Hence, if a  0 and ab = 0, we get b = 0, i.e., F has no
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zero divisors. Thus, F is a domain.
Theorem 5 may immediately prompt you to ask if every domain is a field. You have already
seen that Z is a domain but not a field. But if we restrict ourselves to finite domains, we find that
they are fields.

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