I am looking at this article here:


and trying to understand Theorem 3 on Page 13. They have the following figure for the fundamental theorem:

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 :-(

  • 2
    $\begingroup$ $\Bbb{Q}(2^{1/2})$ the splitting field of $x^2-2$ is a subfield of $\Bbb{Q}(2^{1/4},i)$ the splitting field of $x^4-2$ $\endgroup$
    – reuns
    Sep 14, 2019 at 0:18
  • $\begingroup$ 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. $\endgroup$ Sep 14, 2019 at 11:36
  • $\begingroup$ 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. $\endgroup$ Sep 14, 2019 at 11:45

1 Answer 1


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 $x^4-2$).

  • $\begingroup$ Yes, I get it now. I think we need to use more than one polynomial to find such an example and the definition of normal extension does provide that possibility. $\endgroup$ Sep 14, 2019 at 11:38

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