2
$\begingroup$

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 $S_{71}$).

$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.