I am looking at this article here:
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.27.4955&rep=rep1&type=pdf
and trying to understand Theorem 3 on Page 13. They have the following figure for the fundamental theorem:
followed by these theorem statements,
Theorem 3: Let $K:F$ be a Galois Extension, and set $G = Aut(K/F)$. The group $G$ is known as the Galois group. There is a $1-1$ (inclusion reversing) correspondence between intermediate subfields $E$ of $K$ and subgroups $H$ of $G$, with the following properties (summarized in Fig 0.2):
- $[K:E] = |H|$, and $[E:F] = |G|/|H|$.
- $K:E$ is always Galois, with $Aut(K/E) = H$.
- $E:F$ is Galois if, and only if, $H$ is a normal subgroup of G. If this is the case, then $Aut(E/F)$ is the quotient group $G/H$.
I get the idea of Galois correspondence and the fundamental theorem. However, my question is about the existence of such intermediate fields especially when $F=Q$, the field of rationals. If $K$ is a Galois Extension of $F (=Q)$, wouldn't $K$ be the splitting field of any irreducible polynomial in $Q$, in which case, how can there even be an intermediate normal extension $E$, unless $E=K$?
Sorry if the question is too elementary, I am wrapping my heads around this whole Galois theory, which is driving me nuts :-(
If $K$ is Galois over $F$, then any polynomial irreducible over $F$ that has a zero in $K$ splits over $K$. But there are polynomials irreducible over $F$ that don't have a zero in $K$ (e.g., $x^2-3$ is irreducible over the rationals but has no zero in ${\bf Q}(\sqrt2)$, which is Galois over the rationals), and there may be polynomials irreducible over $F$ that have a zero in $K$ and split over some proper subfield of $K$ (as in the example reuns gives, where $x^2-2$ splits in a proper subfield of the splitting field of $x^4-2$).