Related to my last question:
In Silverman and Tate, Rational Points on Elliptic Curves, there is the following proposition.
Proposition 6.5. Let $E$ be an elliptic curve given by a Weierstrass equation $$E:y^2=x^3+ax^2+bx+c, \quad a,b,c \in \mathbb{Q}.$$ (a) Let $P=(x_1,y_1)$ be a point of order dividing $n$. Then $x_1$ and $y_1$ are algebraic over $\mathbb{Q}$.
(b) Let $$E[n]=\{(x_1,y_1) , \cdot \cdot \cdot , (x_m,y_m) , \mathcal{O} \}$$be the complete set of points of $E(\mathbb{C})$ of order dividing $n$. Let $$K=\mathbb{Q}(x_1,y_1 , \cdot \cdot \cdot , x_m,y_m )$$ be the field generated by the coordinates of all of the points in $E[n]$. Then $K$ is a Galois extension of $\mathbb{Q}$.
So, (a) proves that the extension $K=\mathbb{Q}(x_1,y_1 , \cdot \cdot \cdot , x_m,y_m )$ is an algebraic extension.
Wikipedia says
A Galois extension is an algebraic field extension $E/F$ that is normal and separable; or equivalently, $E/F$ is algebraic, and the field fixed by the automorphism group Aut$(E/F)$ is precisely the base field $F$.
Since all embeddings of extensions of $\mathbb{Q}$ into $\mathbb{C}$ fix $\mathbb{Q}$, can we immediately conclude that $K=\mathbb{Q}(x_1,y_1 , \cdot \cdot \cdot , x_m,y_m )$ is a Galois extension of $\mathbb{Q}$? (The proof in the text shows that all field homomorphisms $\sigma: K \rightarrow \mathbb{C}$ are automorphisms of $K$, and concludes this shows that $K$ is a Galois extension of $\mathbb{Q}$.)
Yes the multiplication by $n$ map $(a,b)\mapsto [n](a,b)$ is given by a pair of rational functions in $a,b$ with coefficients in $\Bbb{Q}$ thus for $\sigma \in Gal(\overline{\Bbb{Q}}/\Bbb{Q})$ then $$[n](x_j,y_j) =O\implies \text{ the denominator vanishes} \implies [n](\sigma(x_j),\sigma(y_j))=O$$ ie. $(\sigma(x_j),\sigma(y_j))\in K^2$ and $\sigma(K)\subset K$ ie. $K/\Bbb{Q}$ is normal, in characteristic $0$ it is separable thus it is Galois. Moreover $\sigma$ commutes with the addition of $E$ so that $Gal(K/\Bbb{Q})$ is a subgroup of $Aut(E[n])$.