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For a quadratic extension $\mathbb{Q}(\alpha)$ of $\mathbb{Q}$ where $\alpha$ is a root of $x^2 - d$ with $d > 0$, we have two real embeddings $$\sigma_1:\alpha \mapsto \sqrt{d} \quad \text{ and } \quad \sigma_2:\alpha \mapsto -\sqrt{d},$$ but the images $\sigma_1(\mathbb{Q}(\alpha))$ and $\sigma_2(\mathbb{Q}(\alpha))$ are seen to be equal to the same field $\mathbb{Q}(\sqrt{d})$.

More generally, suppose $K = \mathbb{Q}(\alpha)$ where $\alpha$ is a root of an irreducible polynomial $f \in \mathbb{Q}[x]$ of degree $n$.

Question: Given two real embeddings $\sigma_1, \sigma_2: K \to \mathbb{R}$, how to determine whether $\sigma_1(K) \stackrel{?}{=} \sigma_2(K)$?

More concretely, is there an algorithm to decide it based on approximations of $\sigma_1(\alpha), \sigma_2(\alpha) \in \mathbb{R}$?

Note that the $\sigma_i(\alpha)$ are just real roots of $f$.

For a concrete example, I am interested in some $f$ of degree 71 with 3 real roots (and Galois group $S_{71}$).

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We can use the Galois group action in case $\operatorname{Gal}(f) = S_n$.

There is no need to distinguish between real or complex embeddings.

If $n > 2$ and $\operatorname{Gal}(f) = S_n$ then the images of all embeddings are pairwise distinct.

Proof: Suppose $\beta, \gamma$ are two distinct roots of $f$, and let $\mu$ be a third. If $\mathbb{Q}(\gamma) \subset \mathbb{Q}(\beta)$ then $\gamma = P(\beta)$ for some polynomial $P \in \mathbb{Q}[x]$. View the equation $\gamma = P(\beta)$ in the splitting field of $f$ and apply the Galois action of a transposition interchanging $\gamma$ and $\mu$ (fixing $\beta$ and $\mathbb{Q}$, so $P(\beta)$ too) to obtain $\gamma = \mu$: absurdity. Q.E.D.

A corollary:

If $n > 2$ and $\operatorname{Gal}(f) = S_n$ then $\operatorname{Aut}(\mathbb{Q}(\alpha)/\mathbb{Q}) = \{ \operatorname{id} \}$ for every root $\alpha$ of $f$.

Proof: The only root of $f$ in $\mathbb{Q}(\alpha)$ is $\alpha$ itself, by the proof above.

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