1

Does every extension of degree 4 over a field $F$ contain a sub-extension of degree 2 over F? If yes, prove it. If not give a counterexample.

I just want to know if my procedure is right.

There is a theorem that says that if $F \subseteq F_1 \subseteq K$ are fields then $[K:F] = [K:F_1][F_1:F]$, therefore $[F_1:F]$ divides $[K:F]$, since in our case $[K:F] = 4$ and $[F_1:F] = 2$, then $2|4$ and our assertion is right.

Thanks

Jyrki Lahtonen
  • 133,153
Leonhard Leibniz
  • 1,258
  • 1
    If such a subfield $F_1$ exists, then your calculations are correct. But the whole question is whether or not such an $F_1$ exists, and you haven't proven that it does. – Brent Kerby Mar 17 '15 at 17:51
  • Yes, I am starting to study this. Does it has to do with the fundamental theorem? maybe Primitive element theorem? – Leonhard Leibniz Mar 17 '15 at 18:04
  • 2
    Related links: http://math.stackexchange.com/questions/141727/why-this-extension-doesnt-contain-a-subextension-of-degree-2 http://math.stackexchange.com/questions/294780/field-extension-of-composite-degree-has-a-non-trivial-sub-extension – Viktor Vaughn Mar 17 '15 at 18:27
  • 2
    If the fields truly are finite, as suggested by your choice of tags, then there always is such an intermediate field, because the related Galois groups are always cyclic. If $F$ is an infinite set, say $F=\Bbb{Q}$, then Brent Kerby's answer points the way (and the choice of tag was wrong). – Jyrki Lahtonen Mar 17 '15 at 21:29
  • @JyrkiLahtonen, ah yes, I didn't notice the (finite-fields) tag. – Brent Kerby Mar 17 '15 at 22:11
  • Leonhard, Are you insisting that the fields are finite or not? The answer depends on that!! @Brent I suspect that Leonhard is simply tagging incorrectly. Many beginners here use the tag finite-fields, when they ask about finite extensions of fields - not realizing that the adjective finite applies to field in the former and to extension in the latter case. This is the umpteenth time I face this problem :-) – Jyrki Lahtonen Mar 18 '15 at 13:50

1 Answers1

2

A counterexample can be given by taking $F=\mathbb Q$ and $K$ to be the extension field by adjoining a root of an irreducible polynomial of degree 4 whose splitting field has the alternating group $A_4$ as its Galois group. Then $F$ is the fixed field of a subgroup $H$ of order 3 of $A_4$, and the subfields of $K$ are in one-to-one correspondence with the subgroups of $A_4$ containing $H$; in particular, a subfield of degree 2 would correspond to a subgroup of $A_4$ of order 6, but $A_4$ has no such subgroup.

Brent Kerby
  • 5,539
  • 2
    What works equally well is to pick a Galois extension $K/F$ with Galois group $G\cong S_4$, and let $F_1$ be the fixed field of a copy of $S_3\le G$. A copy of $S_3$ is a maximal subgroup of $S_4$, so... +1, of course :-) – Jyrki Lahtonen Mar 17 '15 at 21:27