proof verification - Proving that $[mathbb{Q}(sqrt{sqrt{p+q}+sqrt{q}},sqrt{sqrt{p+q}-sqrt{q}}):mathbb{Q}]=8$.

Some days ago I posted a question in MSE in order to correct a solution to the problem of Prove that $[\mathbb{Q}(\sqrt{4+\sqrt{5}},\sqrt{4-\sqrt{5}}):\mathbb{Q}]=8$.

After posting this another question, I found a general argument for this type of extensions. I think that the ideas at the solution of Bill Dubuque in this question could be used to solve the following problem:

Let $p$ and $q$ be distinct positive prime numbers such that $p+q$ is a perfect square. Then $[\mathbb{Q}(\sqrt{\sqrt{p+q}+\sqrt{q}},\sqrt{\sqrt{p+q}-\sqrt{q}}):\mathbb{Q}]=8.$

My attempt of solution:

Let $\alpha_1 = \sqrt{\sqrt{p+q}+\sqrt{q}}$ and $\alpha_2=\sqrt{\sqrt{p+q}-\sqrt{q}}).$ Let $\mathbb{K}=\mathbb{Q}(\alpha_1,\alpha_2)$.

First observe that

$$\alpha_1^2 = \sqrt{p+q}+\sqrt{q},$$ and $$\alpha_1 \alpha_2 = \sqrt{p}.$$

Let $\mathbb{L}=\mathbb{Q}(\alpha_1^2,\alpha_1 \alpha_2)=\mathbb{Q}(\sqrt{q},\sqrt{p}).$ We have that $[\mathbb{L}:\mathbb{Q}]=4,$ hence $\mathbb{L}$ is a 2-dimensional vector space over $\mathbb{Q}(\sqrt{q}),$ with basis $\{1,\sqrt{p}\}$. We will prove now that $\alpha_1 \not\in \mathbb{L}:$

Suppose that $\alpha_1 \in \mathbb{L}$ (this imply directly that $\alpha_2 \in \mathbb{L}$ too), then exists unique $a,b \in \mathbb{Q}(\sqrt{q})$ with

$$\alpha_1 = a + b\sqrt{p}.$$ Hence, $$\sqrt{p+q}+\sqrt{q} = a^2 + p b^2 + 2ab\sqrt{p},$$ or equivalently: $$2ab\sqrt{p} = \sqrt{p+q}+\sqrt{q} - a^2 - p a^2.$$

Since the right member of the equality is in $\mathbb{Q}(\sqrt{q}),$ must be $a=0$ or $b=0$.

• If $a=0$ then $\alpha_1 = b\sqrt{p}=b\alpha_1 \alpha_2,$ hence $1=b\alpha_2$ and we conclude that $\alpha_2^{-1}=b \in \mathbb{Q}(\sqrt{q}).$




• If $b=0$ then $\alpha_1=a \in \mathbb{Q}(\sqrt{q}).$

Both cases gets a contradiction since $\sqrt{\sqrt{p+q}\pm\sqrt{q}}\not\in\mathbb{Q}(\sqrt{q}).$ If we suppose that

$$\sqrt{\sqrt{p+q}\pm\sqrt{q}}\in\mathbb{Q}(\sqrt{q}),$$ then exists unique $a,b \in \mathbb{Q}$ such that $$\sqrt{\sqrt{p+q}\pm\sqrt{q}}=a+b\sqrt{q}.$$ Hence
$$\sqrt{p+q}\pm\sqrt{q} = a^2 + qb^2+2ab\sqrt{q},$$ and must be $ab=\pm1/2$ and $\sqrt{p+q} = a^2 + qb^2.$ Solving for $a$ we get that $a$ is a root of the polynomial $$4x^4-4\sqrt{p+q}x^2+q.$$ Hence $a$ have one of the following four values: $$\pm\sqrt{\frac{\sqrt{p+q}}{2}\pm\frac{\sqrt{p}}{2}},$$ but any of these values is a rational, if not, $$\bigg(\pm\sqrt{\frac{\sqrt{p+q}}{2}\pm\frac{\sqrt{p}}{2}}\bigg)^2=\frac{\sqrt{p+q}}{2}\pm\frac{\sqrt{p}}{2} \in \mathbb{Q}.$$

With this we conclude the proof and get the original claim.

End.

The problem I posted some days ago is a special case with $p = 11$ and $q = 5$.

Is this approach correct? I'm interested in reading Galois-type solutions since I think they are more "beautiful". Which are the pair of distinct positive primes whose sum is a perfect square? I see the pairs $(11,5)$, $(23,2)$ and $(31,5)$ for example.

Thaks to everyone.

Answer

But this is again a quick application of the "kummerian argument" which I used in my answer to your question of a few days ago. Introduce $k=\mathbf Q(\sqrt p, \sqrt q)$, which is a biquadratic field because $pq$ cannot be a square in $\mathbf Q$ (by unique factorization in $\mathbf Z$)
. Consider then the extensions $k(\sqrt {\sqrt {p+q} \pm \sqrt q})$, where $p+q$ is a perfect square. Since $\sqrt {\sqrt {p+q}+\sqrt q} .\sqrt {\sqrt {p+q}-\sqrt q}=p$ is a square in $k^*$, the kummerian argument above $k$ shows that the extensions $k(\sqrt {\sqrt {p+q} \pm \sqrt q})$ are the same field, say $K$. Applying again Kummer over $\mathbf Q(\sqrt q)$ as base field, we see that $K=k=\mathbf Q(\sqrt q)(\sqrt p)$ iff $p(\sqrt {p+q}\pm \sqrt q)$ are squares in $\mathbf Q(\sqrt q)$; multiplying the two relations, we get that $p^3$ is a square in $\mathbf Q(\sqrt q)$ : impossible. Hence $[K:k]=2$ and $[K:\mathbf Q]=8$.

Remark: In the kind of questions you are dealing with, the kummerian approach is more natural in the sense that it appeals only to the multiplicative structure of the fields involved, whereas a blunt direct approach mixes the multiplicative and additive structures.