Let $L/K$ be a field extension. For $a\in L$, $a$ algebraic over $K$, its conjugate elements are the other roots of its minimal polynomial. Then, their simple field extensions $K(b)$, $b$ conjugate of $a$, are isomorphic to $K(a)$.

When is $K(a)=K(b)$ for a conjugate $b$ of $a$? (E.g., $\mathbb{Q}(\sqrt{2}+\sqrt{3})=\mathbb{Q}(\sqrt{2}-\sqrt{3})$.)

  • $\begingroup$ $\forall b, b \in K(a)$ iff $K(a)/K$ is a normal extension $\endgroup$ – reuns Apr 10 '17 at 17:26

In the example you gave, the reason that $\mathbb Q(\sqrt 2+\sqrt 3)=\mathbb Q(\sqrt 2-\sqrt 3)$ is because $\mathbb Q(\pm\sqrt 2+\pm\sqrt 3)$ contains $\sqrt 2$ and $\sqrt 3$:

Notice that $(\pm \sqrt 2+\pm\sqrt 3)^2=2+3\pm 2\sqrt 6$, so $\sqrt 6$ is in the extension, but since $\sqrt 6=\sqrt 3\sqrt 2$, it follows that both $\sqrt 2 $ and $\sqrt 3$ are also in the extension.

This is essentially telling you that $K=\mathbb Q(\sqrt 2+\sqrt 3)=\mathbb Q(\sqrt 2,\sqrt 3)$, which is the splitting field for $(x^2-2)(x^2-3)$ (which is equivalent to saying that $K$ is a Galois extension), and the fact that $K$ is splitting (Galois) tells us that all the conjugates of any $\alpha\in K$ is also in $K$.

I should add, to answer your question, that what I am essentially saying is that an extension $F(\alpha)/F$ contains all conjugates of $\alpha$ if and only if it is a splitting field (resp. Galois).


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