What elements may I adjoin to $\mathbb{Q}[\sqrt{3}]$ in order to get to $\mathbb{Q}[\sqrt{7+\sqrt{3}}]$

Question

The field extension $\mathbb{Q}[\sqrt{7+\sqrt{3}}]/\mathbb{Q}$ has degree four and $\sqrt{7+\sqrt{3}}$ is a primitive element.

I'm interested in dividing this into two successive field extensions of degree 2, namely $\mathbb{Q}[\sqrt{7+\sqrt{3}}]/\mathbb{Q}[\sqrt{3}]$ and $\mathbb{Q}[\sqrt{3}]/\mathbb{Q}$.

I'd like to know if I can adjoin a different element than $\sqrt{7+\sqrt{3}}$ to $\mathbb{Q}[\sqrt{3}]$ in order to get to $\mathbb{Q}[\sqrt{7+\sqrt{3}}]$. After all, $\sqrt{7+\sqrt{3}}$ is already a primitive element of the full extension.

In the case of $\mathbb{Q}[\sqrt{2}+\sqrt{3}]/\mathbb{Q}$, for example, I can choose to either adjoin the two elements $\sqrt{2}$ and $\sqrt{3}$ or only the primitive element $\sqrt{2}+\sqrt{3}$. So what I am looking for is the analogue to the second square root here.

Answer 1

If $K$ is a field, and $t$ has no square root in $K$, then the extension $K\subset K[\sqrt{t}]$ has degree $2$.

It is straightforward to show that an element $a+b\sqrt{t} \in K[\sqrt{t}]$ generates the extension if and only if $b\neq 0$.

It follows that the generators of the extension $\mathbb{Q}[\sqrt{3}] \subset \mathbb{Q}[\sqrt{7 + \sqrt{3}}]$ are exactly the numbers of the form $a + b\sqrt{7+\sqrt{3}}$, with $a,b\in\mathbb{Q}[\sqrt{3}]$ and $b\neq 0$.

These all have degree $4$ over $\mathbb{Q}$, so the literal answer to the question is no. However, the generator $\sqrt{3}\sqrt{7+\sqrt{3}} = \sqrt{21 + 3\sqrt{3}}$ is an interesting choice that is equivalent to $\sqrt{7+\sqrt{3}}$, but not trivially so.

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To expand on Jyrki's comment, $\mathbb{Q}[\sqrt{7+\sqrt{3}}]$ is contained in the Galois extension $\mathbb{Q}\subset\mathbb{Q}[\sqrt{7+\sqrt{3}}, \sqrt{7-\sqrt{3}}]$, and corresponds to a non-normal subgroup of the Galois group $D_8$.

(Note: It's not hard to show that the Galois group is $D_8$, since every order $8$ subgroup of $S_4$ is isomorphic to $D_8$.)

But a non-normal subgroup of index $4$ in $D_8$ is contained in a unique subgroup of index $2$. It follows that $\mathbb{Q}[\sqrt{3}]$ is the only intermediate field extension of $\mathbb{Q}\subset\mathbb{Q}[\sqrt{7+\sqrt{3}}]$, so looking for anything like a "second square root" is futile.

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Maybe it's easier than all this. We cannot have $\mathbb{Q}[\sqrt{7+\sqrt{3}}] = \mathbb{Q}[\sqrt{3}, \sqrt{n}]$, because the latter is Galois while the former is not.

Answer 2

Unless I am mistaken the degree of $\Bbb Q( \sqrt{7+ \sqrt{3}})$ over $\Bbb Q$ is $4$, right?

Any intermediate field would be quadratic then, i.e. $\Bbb Q(\sqrt{n})$. One could probably prove at this point that $a + b \sqrt{n} + c \sqrt{3} + d \sqrt{3 n} $ cannot be equal to $\sqrt{7+\sqrt{3}}$.

Another method is to consider Galois automorphisms. If such an extension exists then the Galois group would be generated by $\sqrt{3} \rightarrow - \sqrt{3}$ and ditto for $\sqrt{n}$, and I guess the automorphisms of the given extension can be computed and the group is a different one $(\sqrt{3} \rightarrow - \sqrt{3}$ and $\sqrt{7+\sqrt{3}} \rightarrow - \sqrt{7+\sqrt{3}}$, which shows that the surds of the extensions must be fixed points of one of the Galois automorphisms), but it is late^^.

Answer 3

I think that the following argument shows that the type of element you are looking for does not exist.

Let $u=\sqrt{7+\sqrt{3}}$ and $v=\sqrt{7-\sqrt{3}}$. The zeros of the polynomial $$ p(x)=(x^2-7)^2-3\in\Bbb{Z}[x] $$ are $\pm u$ and $\pm v$, so $M=\Bbb{Q}(u,v)$ is a Galois extension of the rationals. It should be easy to show that $[M:\Bbb{Q}]=8$. It's late here, so I will skip that for now. The Galois group $G$ acts on the four roots faithfully, so we can identify $G$ as a subgroup of the symmetric group $S_4$. Therefore $G$ is a Sylow 2-subgroup of $S_4$, so it has to be $\cong D_4$. Tha mapping $\sigma:v\mapsto -v,u\mapsto u,$ is the only non-trivial automorphism of $M$ that fixes all the elements of $L=\Bbb{Q}(u)$.

If we identify $D_4$ as the group of symmetries of the square, we see that $\sigma$ must correspond to a reflection w.r.t. a diagonal. Therefore $\sigma$ is contained in a unique subgroup of order $4$ (the one generated by the reflections w.r.t. both diagonals). Therefore $\Bbb{Q}(u)$ has a unique quadratic subfield. The non-existence of the desired element follows from this (see my comments).

The group $D_4$ does have three subgroups of order $4$, so $M$ has three quadratic subfields. In addition to $\Bbb{Q}(\sqrt3)=\Bbb{Q}(u^2)=\Bbb{Q}(v^2)$ we have $\Bbb{Q}(uv)=\Bbb{Q}(\sqrt{46})$, and thus also $\Bbb{Q}(\sqrt{138})$. These involve $v$, so are not subfields of $\Bbb{Q}(u)$.