Question regarding galois correspondence in ${\mathbb R} /{\mathbb Q}$

Question

${\rm Aut }\left({\mathbb R}/{\mathbb Q}\right)$ is trivial, which implies there is no intermediate field between the field of rationals and field of real numbers, but we know there are infinitely many intermediate fields of $\mathbb R$ containing $\mathbb Q$.

Why I am getting an contradictions? Please explain where I went wrong.

Answer 1

As the other answers point out, the problem is that $\mathbb R/\mathbb Q$ is not a Galois extension. Their arguments are, however, incorrect: being finite is not necessary for an extension $L/K$ to be Galois, and there is a rich theory of infinite Galois extensions.

A Galois extension is an algebraic extension that is normal and separable. Separability is fine here, but the other two conditions do not hold:

1. $\mathbb R$ is not an algebraic extension of $\mathbb Q$. In particular, $\mathbb R$ contains transcendental extensions such as $\mathbb Q(\pi)$
2. Even if we replaced $\mathbb R$ with $\mathbb R\cap\overline{\mathbb Q}$, the extension would still not be normal. For example, it contains the real root of $X^3-2$ but none of the complex roots.

For these two reasons, one cannot apply Galois theory to the extension $\mathbb{R/Q}$.

Answer 2

Your first statement is wrong. The fact that $Aut(\Bbb{R}/\Bbb{Q})$ is trivial does not imply that there are no intermediate fields (for example take $\Bbb{Q}(\sqrt2))$.

You are probably applying Galois Fundamental Theorem wrong. In the hypothesis, the Field Extension has to be a Galois extension. In particular, it has to be a normal extension, but $\Bbb{R}/\Bbb{Q}$ is not normal, as $x^3-p$ (for example, with $p$ prime, is irreducible by Eisenstein), has a root in $\Bbb{R}$ but does not split on $\Bbb{R}$.