Let be a maximal ideal in If is a commutative ring, then must also be a commutative ring. Clearly, acts as an identity for We must also show that every nonzero element in has an inverse. If is a nonzero element in then Define to be the set We will show that is an ideal in The set is nonempty since is in If and are two elements in then
is in
Also, for any
it is true that
hence,
is closed under multiplication and satisfies the necessary conditions to be an ideal. Therefore, by
Proposition 16.10 and the definition of an ideal,
is an ideal properly containing
Since
is a maximal ideal,
consequently, by the definition of
there must be an
in
and an element
in
such that
Therefore,
Conversely, suppose that is an ideal and is a field. Since is a field, it must contain at least two elements: and Hence, is a proper ideal of Let be any ideal properly containing We need to show that Choose in but not in Since is a nonzero element in a field, there exists an element in such that Consequently, there exists an element such that and is in Therefore, for all Consequently,