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I've been studying algebra. In trying an exercise, I've come across the following issue.

I know that $\mathbb{Q}(a) = \mathbb{Q}(b)$, for some $a,b \notin \mathbb{Q}$. I want to conclude that $a$ and $b$ have the same minimal polynomial over $\mathbb{Q}$.

If $p(x)$ is the minimal polynomial of $a$ and $q(x)$ is the minimal polynomial of $b$ over $\mathbb{Q}$, then $ \mathbb{Q}[x]/(p(x)) \simeq \mathbb{Q}(a) = \mathbb{Q}(b) \simeq \mathbb{Q}[x]/(q(x)). $ It's clear that $p(x)$ and $q(x)$ are irreducible, monic, and have the same degree. However, I can't find a proof or a counterexample to determine whether $p(x)$ must equal $q(x)$.

Any advice? Thanks.

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    This is false. Take, for example, $a = \sqrt{2}, b = \sqrt{2} - 1$.2012-11-28

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To expand a little bit on Qiaochu's comment. Consider an extension $\mathbb Q(a)$ and note that for any $c \in \mathbb Q$ we have that $\mathbb Q(a)=\mathbb Q(a+c)$ but $a$ and $a+c$ can't have the same minimum polynomial for every $c \in \mathbb Q$. A polynomial only has a finite number of roots.

In the case that $a$ and $b$ have the same minimum polynomial, we call $a$ and $b$ algebraic conjugates. Note though that usually algebraic conjugates don't generate the same extension for instance $\sqrt[3]{2}$ and $\omega\sqrt[3]{2}$, where $\omega$ is a primitive third root of unity, are both solutions to $x^3-2$. But $\mathbb Q(\sqrt[3]{2}) \neq \mathbb Q(\omega\sqrt[3]{2})$, due to considering imaginary parts. What is true is that they're isomorphic.