If $\vartheta_1,\,\vartheta_2,\,\ldots,\,\vartheta_n$ , are the algebraic conjugates of the algebraic number $\vartheta_1$ then the algebraic number fields $\mathbb{Q}(\vartheta_1),\,\mathbb{Q}(\vartheta_2),\,\ldots,\,\mathbb{Q}(\vartheta_n)$ , are the conjugate fields of $\mathbb{Q}(\vartheta_1)$

Notice that the conjugate fields of $\mathbb{Q}(\vartheta_1)$ are always isomorphic but not necessarily distinct.

All conjugate fields are equal, i.e. $\mathbb{Q}(\vartheta_1)= \mathbb{Q}(\vartheta_2)=\ldots=\mathbb{Q}(\vartheta_n)$ or equivalently $\vartheta_1,\ldots,\vartheta_n$ belong to $\mathbb{Q}(\vartheta_1)$ if and only if the extension $\mathbb{Q}(\vartheta_1)/\mathbb{Q}$ is a Galois extension of fields. The reason for this is that if $\vartheta_1$ is an algebraic number and $m(x)$ is the minimal polynomial of $\vartheta_1$ then the roots of $m(x)$ are precisely the algebraic conjugates of $\vartheta_1$

For example, let $\vartheta_1 = \sqrt{2}$ Then its only conjugate is $\vartheta_2=-\sqrt{2}$ and $\mathbb{Q}(\sqrt{2})$ is Galois and contains both $\vartheta_1$ and $\vartheta_2$ Similarly, let $p$ be a prime and let $\vartheta_1=\zeta$ be a primitive $p$ root of unity. Then the algebraic conjugates of $\zeta$ are $\zeta^2,\ldots,\zeta^{p-1}$ and so all conjugate fields are equal to $\mathbb{Q}(\zeta)$ and the extension $\mathbb{Q}(\zeta)/\mathbb{Q}$ is Galois. It is a cyclotomic extension of $\mathbb{Q}$

Now let $\vartheta_1=\sqrt[3]{2}$ and let $\zeta$ be a primitive $3$ root of unity (i.e. $\zeta$ is a root of $x^2+x+1$ so we can pick $\zeta=\frac{-1+\sqrt{-3}}{2}$ . Then the conjugates of $\vartheta_1$ are $\vartheta_1$ $\vartheta_2=\zeta\sqrt[3]{2}$ and $\vartheta_3=\zeta^2\sqrt[3]{2}$ The three conjugate fields $\mathbb{Q}(\vartheta_1)$ $\mathbb{Q}(\vartheta_2)$ and $\mathbb{Q}(\vartheta_3)$ are distinct in this case. The Galois closure of each of these fields is $\mathbb{Q}(\zeta,\sqrt[3]{2})$