# Intermediate Normal Extensions and Galois Correspondence

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):

1. $$[K:E] = |H|$$, and $$[E:F] = |G|/|H|$$.
2. $$K:E$$ is always Galois, with $$Aut(K/E) = H$$.
3. $$E:F$$ 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 :-(

• $\Bbb{Q}(2^{1/2})$ the splitting field of $x^2-2$ is a subfield of $\Bbb{Q}(2^{1/4},i)$ the splitting field of $x^4-2$ – reuns Sep 14 at 0:18
• Ok, thanks! I get it now I guess. The definition of normal extension includes the possibility of having no roots in the extended field and thus there could be different polynomials through which extension could be made. However, I guess if we stick to one polynomial, there could be only one normal extension that would correspond to the identity automorphism. – user2167741 Sep 14 at 11:36
• In the above comment, where I said "there could be only one normal extension that would correspond to the identity automorphism." assumes the extension is the splitting field and not the irreducible base field. – user2167741 Sep 14 at 11:45

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 $$x^4-2$$).