Do conjugate fields have the same maximal order?

Suppose that $\alpha,\beta\in K$ are conjugates elements, i.e. zeros of an irreducible polynomial over $\mathbb{Q}$. Then we know that the fields $K_{1}=\mathbb{Q}(\alpha)$ and $K_{2}=\mathbb{Q}(\beta)$ are conjugate fields that are also isomorphic.

Every number field $K$ contains a unique maximal order $\mathcal{O}_{K}$ (i.e. ring of algebraic integers of $K$). My question is:

Do isomorphic conjugate fields possess the same maximal order? In particular, do $K_{1}=\mathbb{Q}(\alpha)$ and $K_{2}=\mathbb{Q}(\beta)$ have the same unique maximal order?

Thank you.

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This is not the case. There is even an equivalence $$K_1 = K_2 \Leftrightarrow \mathcal O_{K_1} = \mathcal O_{K_2}$$

The reason is that $K_1$ is uniquely determined by $\mathcal O_{K_1}$. An algebraic way to see this is by $K_1 = Q(\mathcal O_{K_1})$ where the latter expression denotes the quotient field embedded in $K_1$. Also, $K_1 = \Bbb Q(\mathcal O_{K_1})$ is the field generated by $\mathcal O_{K_1}$ over $\Bbb Q$.

Nevertheless, both orders are isomorphic by any isomorphism determining $K_1 \cong K_2$.

If you would like to see an example, consider $K_1 = \Bbb Q (\sqrt[4]{2})$ and $K_2 = \Bbb Q (i\sqrt[4]{2})$ which are conjugated, as both are subfields of the splitting field of the irreducible polynomial $X^4-2$. However, $K_1 \cap K_2 = \Bbb Q$. If both had the same maximal order $\mathcal O$, then it would be in $\Bbb Z$, a contradiction.

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Thank you Benh. Great answer! – H.E Mar 16 '14 at 17:17
You are welcome! – benh Mar 16 '14 at 17:35

If you are asking if the maximal orders are literally the same, the answer is no, as benh explains.

However, the isomorphism between $K_1$ and $K_2$ will induces an isomorphism between their maximal orders. In particular, their maximal orders are isomorphic.

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Thank you Matt E. – H.E Mar 16 '14 at 17:18