# 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 $x^4-2$ Sep 14, 2019 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. Sep 14, 2019 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. Sep 14, 2019 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 $$x^4-2$$).