# Subfields of a quartic extension of $\mathbb{Q}$

While going through the lecture notes of Pete L. Clark on Field Theory I found an interesting exercise.

If $$K$$ is a field extension of $$\mathbb{Q}$$ of degree $$4$$ then either

• there is no intermediate subfield $$F$$ with $$\mathbb{Q} \subset F\subset K$$ or
• there is exactly one such intermediate field $$F$$ or
• there are three such intermediate fields.

An example of second possibility is $$K=\mathbb{Q} (\sqrt[4]{2})$$ with $$F=\mathbb{Q} (\sqrt{2})$$. For the third case we can take $$K=\mathbb {Q} (\sqrt{2}, \sqrt{3})$$ with $$F$$ being any of $$\mathbb {Q} (\sqrt{2}),\mathbb {Q}(\sqrt{3}),\mathbb{Q} (\sqrt{6})$$.

I couldn't think of an obvious example of the first case where no intermediate field exists.

Also it is not possible that there are only two such intermediate fields because if we have two such intermediate fields, say $$\mathbb{Q} (\sqrt{a} ), \mathbb {Q} (\sqrt{b})$$ with $$a, b\in\mathbb {Q}$$ then $$\mathbb {Q} (\sqrt{ab})$$ is another such subfield.

My guess is that if there are more than three such subfields then the degree of $$K$$ containing all these must be at least $$8$$. I don't know if this is true and I have not yet proved it so far and need some help to complete this proof (or perhaps this simply does not work).

I also think that there is another approach which sort of handles all the cases in a unifying manner.

And I also need an example for first case where there is no intermediate subfield (first case).

• how about $\mathbb Q(\varphi)$ where $\varphi^4+\varphi+1=0$? Jul 21, 2021 at 9:51
• For the first case you first need a quartic $f(x)\in\Bbb{Q}[x]$ with Galois group $G$ either $S_4$ or $A_4$. If $\alpha$ is a zero of $f$ then $\Bbb{Q}(\alpha)/\Bbb{Q}$ has no intermediate subfields. This is because the stabilizer of $\alpha$ in $G$ is a maximal subgroup. I think we have covered that case earlier. Give me a minute. Jul 21, 2021 at 10:25
• I explained the argument in this old answer. Jul 21, 2021 at 10:26
• Geoff Robinson gives a nice different argument specific to the quartic case here. I seem to have handled this twice :-( Jul 21, 2021 at 10:32
• An older variant answered by Pete L. Clark himself. Jul 21, 2021 at 10:40

In what follows $$C_n$$ is the cyclic group of order $$n$$ (also denoted as $$\mathbb {Z} _n$$ or $$\mathbb {Z} /n\mathbb {Z}$$), $$D_{2n}$$ denotes the dihedral group of order $$2n$$ (with $$D_4$$ also known as Klein four group). $$S_n$$ denotes the group of all permutations on $$n$$ elements and $$A_n$$ denotes the group of all even permutations on $$n$$ elements.

We are given that $$K$$ is a field extension of $$\mathbb {Q}$$ of degree $$4$$. Hence we can assume that $$K=\mathbb {Q} (a)$$ where $$a$$ is a root of an irreducible polynomial $$f(x) \in \mathbb {Q} [x]$$ of degree $$4$$.

And now we can deal with various cases depending on number of roots of $$f$$ contained in $$K=\mathbb {Q} (a)$$.

If $$K$$ contains all the roots of $$f(x)$$ then $$K$$ is a Galois extension of $$\mathbb {Q}$$ and we have the Galois group $$\text{Gal} (K/\mathbb {Q} )$$ as either $$C_4$$ (eg $$f(x) =(x^5-1)/(x-1)$$) or as $$D_4$$ (eg $$f(x) = x^4-10x^2+1$$). In the first case we have just one subgroup namely $$C_2$$. And $$D_4$$ has three subgroups each of which looks like $$C_2$$.

Thus we have either one subfield or three subfields between $$\mathbb {Q}$$ and $$K$$.

Next we deal with the case when $$K$$ contains two roots of $$f(x)$$. Then the remaining two roots are root of a quadratic polynomial $$g(x) \in K[x]$$ and hence we can adjoin one of them to $$K$$ to get the splitting field $$L$$ of $$f(x)$$ with $$[L:K] =2$$.

Next $$[L:\mathbb{Q}] =8$$ and we know that $$\text{Gal} (L/\mathbb {Q})$$ is of order $$8$$ and contains a transposition (which permutes two roots of $$g(x)$$) and further the group acts transitively on roots of $$f(x)$$ (as $$f(x)$$ is irreducible). Thus the Galois group of $$f(x)$$ is $$D_8$$.

The subgroup corresponding to extension $$K\subset L$$ is $$\text{Gal} (L/K)$$ and it contains a single transposition apart from identity. A subfield $$F$$ of $$K$$ corresponds to a subgroup of $$\text{Gal} (L/\mathbb {Q}) \simeq D_8$$ containing the subgroup $$\text{Gal} (L/K)$$. There is exactly one such subgroup and it is isomorphic to Klein-4 group. And thus we have just one subfield $$F$$. A simple example of this case is the polynomial $$f(x) =x^4-2$$ (the polynomial $$x^4+2$$ also works and has same splitting field).

Finally we consider the most complicated case where $$f(x)$$ has only one root in $$K$$. The remaining three roots are the roots of a cubic $$g(x) \in K[x]$$ and $$g(x)$$ is irreducible (none of its roots lie in $$K$$). Adjoining one root of $$g(x)$$ to $$K$$ we get another field extension $$L$$ with $$[L:K] =3$$. It may be possible that $$L$$ contains all roots of $$g(x)$$ in which case $$L$$ is the splitting field of $$f(x)$$ with $$[L:\mathbb{Q}] =12$$ and the Galois group of $$f(x)$$ is isomorphic to $$A_4$$.

Now $$\text{Gal} (L/K)$$ corresponds to a subgroup of $$A_4$$ of order $$3$$ and there is just one such subgroup which is isomorphic to $$A_3$$ (equivalently to $$C_3$$).

A subfield $$F$$ of $$K$$ corresponds to a proper subgroup of $$\text{Gal} (L/\mathbb{Q})$$ which contains $$\text{Gal} (L/K)$$ properly. Thus its order must be $$6$$. But $$A_4$$ has no subgroup of order $$6$$ and hence such a field like $$F$$ does not exist. An example of such a case is $$f(x) = x^4+8x+12$$ (taken from notes by Keith Conrad).

We still have to deal with the case when $$L$$ does not contain all the roots of $$g(x)$$. Then $$L$$ contains just one root of $$g(x)$$ and remaining two roots are the roots of a quadratic polynomial $$h(x) \in L[x]$$. Adjoining one root of $$h(x)$$ to $$L$$ we get the splitting field $$M$$ of $$f(x)$$ with $$[M:L] =2$$ and $$[M:\mathbb{Q}] =24$$ so that the Galois group of $$f(x)$$ is isomorphic to $$S_4$$.

The field $$K$$ corresponds to a subgroup of order $$6$$ of $$S_4$$ and there is only one such subgroup which is isomorphic to $$S_3$$.

A subfield $$F$$ of $$K$$ corresponds to a proper subgroup of $$S_4$$ which contains $$S_3$$ properly. Hence its order must be $$12$$. The only subgroup of order $$12$$ is $$A_4$$ and it consists of even permutations only and hence it can't contain $$S_3$$. Thus we don't have the desired subgroup and hence no such desired subfield $$F$$ exists. An example for this case is given by $$f(x) =x^4+x+1$$ (this is also mentioned in comments to question, Keith Conrad mentions $$x^4-x-1$$ as another example in his notes).

• A local resource for the biquadratic polynomials. Observe that, unlike for Dummit & Foote, for many authors (Jacobson and others) $D_n$ is the group of symmetries of a regular $n$-gon. So what you called $D_4$ is the Klein four group, and what you called $D_8$ is $D_4$. Not criticizing your choices, just to make sure that there won't be misunderstandings. Personally I'm on record for following Jacobson's convention (I was raised that way). Somewhere here Keith Conrad described a conversation he had with one of D&F. He asked about this! Jul 24, 2021 at 8:47
• Thanks @JyrkiLahtonen for the feedback on notation. I will update the post with the convention used. Jul 24, 2021 at 9:46
• Do what you think is best. Many users have learned stuff from D&F also. This is just one of those things algebraists cannot agree on. Jul 24, 2021 at 10:19
• @JyrkiLahtonen the place where I wrote about my conversation with Foote is math.stackexchange.com/questions/2560348/….
– KCd
Aug 26 at 17:40