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For some of you this question might seem excessively easy. More importantly, it might be a duplicate, but the answers go over my head. The most promising "question that may already have your answer" was Degree 4 extension of $\mathbb {Q}$ with no intermediate field. But I haven't been able to understand this concept of "Galois over $\textbf{Q}$."

I'm also looking at the "similar questions." In his answer to What kinds of algebraic integers are of degree $4$?, Gerry Myerson mentions the numbers $~2x~=~\sqrt{\alpha-\beta}~-~\sqrt{\beta-\alpha+\dfrac2{\sqrt{\alpha-\beta}}}~,~$ $~\alpha=\sqrt[3]{\dfrac12+\dfrac{\sqrt{849}}{18}}~$ and $~\beta=4~\sqrt[3]{\dfrac2{3(9+\sqrt{849})}}~.~$ Those are presumably units in a ring that does have intermediate rings, but I have only wild guesses as to what those might be, e.g., $\mathcal{O}_{\textbf{Q}(\sqrt{849})}$.

Okay, so, given a squarefree integer $d \in \textbf{Z}$, it follows that the ring $\mathcal{O}$ of algebraic integers of $\textbf{Q}(\root 4 \of d)$ has $\mathcal{O}_{\textbf{Q}(\sqrt{d})}$ for an intermediate ring.

Then I thought what if $d$ is nontrivially divisible by a cube but not by a fourth power? For example, $d = 8$. But then doesn't that just devolve to $\mathcal{O}_{\textbf{Q}(\root 4 \of 2)}$?

More generally, though I might really be going out on a limb here, is $n$ is a positive composite integer, does it follow that all rings of degree $n$ have intermediate rings?

David R.
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    I don't know quite what your terms mean. Does "integer ring" mean "ring of integers of a number field"? And what does "intermediate ring" mean? Does it just mean "nontrivial subring"? Would you consider $\mathbb{Z}[2i]$ to be an intermediate ring of $\mathbb{Z}[i]$? – Qiaochu Yuan Oct 30 '18 at 19:33
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    @Qiaochu Maybe he means intermediate fields as defined in http://www.lmfdb.org/knowledge/show/nf.intermediate_fields ? Yes, no, maybe, Dave? – Robert Soupe Oct 30 '18 at 20:58
  • @RobertSoupe Yeah, that. – David R. Nov 02 '18 at 21:33
  • Okay, so to clarify, your question has nothing to do with rings of integers, and is about number fields? In that case the first question you linked to does in fact answer your question, although it's not particularly explicit. – Qiaochu Yuan Nov 02 '18 at 21:53
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    If this is really about intermediate fields, then Galois theory is your friend. If it is about rings, then, as Qiaochu Yuan pointed out, you always have non-maximal orders of number fields. As abelian groups those are still free of rank four. They are finite index (additive) subgroups of $\mathcal{O}_K$ that also happen to be rings. – Jyrki Lahtonen Nov 03 '18 at 05:07
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    Essentially the same piece of group theory + Galois theory shows that there are examples of fields $K$, $[K:\Bbb{Q}]=n$ such that there are no intermediate fields between $K$ and the rationals. To apply that we do need to solve the inverse Galois problem for the symmetric group, but that cn be done. – Jyrki Lahtonen Nov 03 '18 at 05:15
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    Also, if 1) your question is about fields, but 2) you had not heard of Galois theory, then I A) want to commend you for coming up with this question, B) wish that the moderators can cancel the bounty, because in that case this is too much of a duplicate. The general idea is that if you have a finite group $G$ that has a maximal subgroup $H$ of index $n$, and then can realize $G$ as the Galois group $G=Gal(L/\Bbb{Q})$, then the intermediate field $K$ consisting of the fixed points of the automorphisms in the subgroup $H$ will automatically be a degree $n$ extension with no intermediate fields. – Jyrki Lahtonen Nov 03 '18 at 05:21
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    You see, basic Galois correspondence would associate such an intermediate field with a group strictly in between $G$ and $H$, and those were assumed not to exist (and that can easily be shown to be the case with $G=S_n$, $H$= a point stabilizer $\simeq S_{n-1}$). – Jyrki Lahtonen Nov 03 '18 at 05:23

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