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Let $L/K$ be a field extension. Suppose $f(x)$ is the minimal polynomial of $\alpha$ and $\alpha'$ over $K$ and suppose $g(x)$ is the minimal polynomial of $\beta$ and $\beta'$ over $K$. If $L = (\alpha, \beta)$, is it true that the rules $\sigma(\alpha) = \alpha'$ and $\sigma(\beta) = \beta'$ define a $K$-automorphism of $L$?

For example: $\mathbb{Q} \subset \mathbb{Q}(\sqrt{7}, \sqrt{13})$. I wish to define an automorphism $\sigma$ satisfying $\sigma(\sqrt{7}) = - \sqrt{7}$ and $\sigma(\sqrt{13}) = \sqrt{13}$. That is, I put $\sigma(a + b \sqrt{7} + c\sqrt{13} + d \sqrt{91}) = a - b \sqrt{7} + c \sqrt{13} - d \sqrt{91}$. Will this always extend to a ring morphism?

If this is not true, then how do I define an automorphism in practice? Do I always have to check that $\sigma$ is multiplicative?

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  • $\begingroup$ I think it isn't always possible to define an automorphism in this manner, because the roots of $g$ and $f$ could be algebraically dependent, which would "force" the automorphism to map $\beta$ to a certain element if we set $\sigma(\alpha)=\alpha'$ $\endgroup$ Commented Nov 16 at 21:40

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This is not possible, take $\alpha'=\alpha$, $\beta=\alpha$ and $\beta'\neq\beta$, such an automorphism $\sigma$ should satisfy the identities $\sigma(\alpha)=\alpha'=\alpha$ and $\sigma(\alpha)=\sigma(\beta)=\beta'$ so that $\alpha=\beta'$, which is not the case.

Maybe this answer won't suit you, so let's suppose $\alpha\neq\beta$. This is still not possible in general. Take $\alpha=\sqrt[3]{2},\alpha'=j\sqrt[3]{2},\beta=j,\beta'=j^2$ and $K=\mathbb{Q}$, in this case ${\rm Gal}(L/K)\simeq\mathfrak{S}_3$. There are $2$ types of automorphisms of $L/K$ :

  • the $3$-cycles, from which only $\sqrt[3]{2}\mapsto j\sqrt[3]{2}\mapsto j^2\sqrt[3]{2}\mapsto\sqrt[3]{2}$ sends $\alpha$ to $\alpha'$, but in this case $\sigma(j)=j$,
  • the transpositions, but these fix $\mathbb{Q}(\alpha)$ so we can't have $\sigma(\alpha)=\alpha'$,

so there is no such $\sigma$.

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  • $\begingroup$ Nice! So I guess the best way to approach this is to use primitive element theorem? It seems quite impractical to check $\sigma(ab) = \sigma(a) \sigma(\beta)$ every time. $\endgroup$ Commented Nov 16 at 23:30
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    $\begingroup$ The primitive element theorem works, but in practice it is hard to find a primitive element (though it is VERY useful in theory). You can also use inductive arguments, for instance if $M/K$ and $L/K$ are both Galois, the restriction map $$\sigma\in{\rm Gal}(M/K)\mapsto\sigma_{|L}\in{\rm Gal}(L/K)$$ is surjective. $\endgroup$ Commented Nov 16 at 23:52

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