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 $S_{71}$).


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.