Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer
Thank you Matt E. – H.E Mar 16 '14 at 17:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.