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If
 are the algebraic conjugates of the algebraic number
 , then the algebraic number fields
 are the conjugate fields of
 .
Notice that the conjugate fields of
 are always isomorphic but not necessarily distinct.
All conjugate fields are equal, i.e.
 , or equivalently
 belong to
 , if and only if the extension
 is a Galois extension of fields. The reason for this is that if
 is an algebraic number and  is the minimal polynomial of
 then the roots of  are precisely the algebraic conjugates of
 .
For example, let
 . Then its only conjugate is
 and
 is Galois and contains both
 and
 . Similarly, let  be a prime and let
 be a primitive  th root of unity. Then the algebraic conjugates of  are
 and so all conjugate fields are equal to
 and the extension
 is Galois. It is a cyclotomic extension of
 .
Now let
![$ \vartheta_1=\sqrt[3]{2}$](http://images.planetmath.org:8080/cache/objects/9487/l2h/img28.png "$ \vartheta_1=\sqrt[3]{2}$") and let  be a primitive  rd root of unity (i.e.  is a root of  , so we can pick
 ). Then the conjugates of
 are
 ,
![$ \vartheta_2=\zeta\sqrt[3]{2}$](http://images.planetmath.org:8080/cache/objects/9487/l2h/img36.png "$ \vartheta_2=\zeta\sqrt[3]{2}$") , and
![$ \vartheta_3=\zeta^2\sqrt[3]{2}$](http://images.planetmath.org:8080/cache/objects/9487/l2h/img37.png "$ \vartheta_3=\zeta^2\sqrt[3]{2}$") . The three conjugate fields
 ,
 , and
 are distinct in this case. The Galois closure of each of these fields is
![$ \mathbb{Q}(\zeta,\sqrt[3]{2})$](http://images.planetmath.org:8080/cache/objects/9487/l2h/img41.png "$ \mathbb{Q}(\zeta,\sqrt[3]{2})$") .
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