$\begingroup$

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 $S_{71}$).