Lemma: Let $b,c \in \mathbb{Q}-\{0\}$. Then $\mathbb{Q}(\sqrt2,\sqrt3) =\mathbb{Q}(b\sqrt2+c\sqrt3+\sqrt6)$.

Question

I am attempting to prove that $\mathbb{Q}(\sqrt2,\sqrt3)=\mathbb{Q}(α)$ iff $α=a+b\sqrt2+c\sqrt3+d\sqrt6$ where $a,b,c,d\in \mathbb{Q}$ and two or more of $b,c,d$ are nonzero. The forward direction is easy, but I am having some trouble with the backward direction.

Let $α=a+b\sqrt2+c\sqrt3+d\sqrt6$, where $a,b,c,d\in \mathbb{Q}$ and at least two of $b,c,d$ are nonzero. I see that, without loss of generality, we can assume $a=0$. Now, the problem splits in two cases.

Either exactly two of $b,c,d$ are nonzero OR all three of $b,c,d$ are nonzero. I think I can tackle the former case on my own, but I could use some help on the latter.

Assume $b,c,d$ are all nonzero. WLOG, we can take $d=1$ so $α=b\sqrt2+c\sqrt3+\sqrt6$. I see that $\mathbb{Q}(\sqrt2,\sqrt3) \supseteq \mathbb{Q}(α)$, but I need some help proving $\subseteq$. An elementary, straightforward, complete proof without gaps is what I'm looking for.

I desire to prove the following Lemma:

Let $b,c \in \mathbb{Q}-\{0\}$. Then $\mathbb{Q}(\sqrt2,\sqrt3)=\mathbb{Q}(b\sqrt2+c\sqrt3+\sqrt6)$.

Answer 1

One inclusion is obvious, namely $\mathbb{Q}(b\sqrt{2}+c\sqrt{3}+\sqrt{6})\subseteq\mathbb{Q}(\sqrt{2},\sqrt{3})$. Note that the latter field has $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ as basis over $\mathbb{Q}$.

Let $r=b\sqrt{2}+c\sqrt{3}+\sqrt{6}$. If we prove $r$ can't have degree $2$ over $\mathbb{Q}$, we're done.

We have $r^2=2b^2+3c^2+6+2bc\sqrt{6}+4b\sqrt{3}+3c\sqrt{2}$; now $r^2+pr+q=0$, with $p,q\in\mathbb{Q}$, becomes $$ (2b^2+3c^2+6+q)+(3c+bp)\sqrt{2}+(4b+pc)\sqrt{3}+(2bc+p)\sqrt{6}=0 $$ Thus $p=-2bc$. Hence from $4b+pc=0$ we derive $$ 4b-2bc^2=0 $$ As $b\ne0$, we get $c^2=2$. A contradiction.

Answer 2

Let $\beta = a\sqrt{2} + b\sqrt{3}$. First we'll prove $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\beta)$. The inclusion $\mathbb{Q}(\beta) \subseteq\mathbb{Q}(\sqrt{2},\sqrt{3})$ is rather obvious. On the other side now consider $h(x) = \left(\sqrt{3} + \frac ab \sqrt{2} - \frac ab x\right)^2 - 3 \in \mathbb{Q}(\beta)[x]$. Now the roots of this polynomial are $\sqrt{2}$ and $\sqrt{2} + \frac {2a}b\sqrt{3}$. On the other side $\sqrt{2}$ satisfies $x^2 -2 \in \mathbb{Q}(\beta)[x]$ and as $\sqrt{2} + \frac {2b}a\sqrt{3}$ isn't a root of it we have that:

$$\min(\sqrt{2},\mathbb{Q}(\beta)) \mid \gcd(h,x^2-2) = x-\sqrt{2}$$

in $\mathbb{C}$ (In fact any extension containing the roots of $h$ and $x^2-2$ would do the trick). This yields that $\min(\sqrt{2},\mathbb{Q}(\beta)) = x - \sqrt{2}$, which implies that $\sqrt{2} \in \mathbb{Q}(\beta)$. From here the inclusion $\mathbb{Q}(\sqrt{2},\sqrt{3}) \subseteq \mathbb{Q}(\beta)$ follows.

Now it remains to prove that $\mathbb{Q}(\beta+\sqrt{6}) = \mathbb{Q}(\beta,\sqrt{6})=\mathbb{Q}(\beta)$, which can be done in a similar manner. The last equality follows as $\beta^2 = 2a^2 + 3b^2 + 2ab\sqrt{6}$, so $\sqrt{6} = \frac{\beta^2 - 2a^2 - 3b^2}{2ab} \in \mathbb{Q}(\beta)$. Hence the proof.

Answer 3

Let $u=a+ b \sqrt{2} + c\sqrt{3} + d\sqrt{2}\sqrt{3}$. Consider $u=u_1$, $u_2$, $u_3$, $u_4$ the numbers obtained by changing the signs of the radicals $\sqrt{2}$, $\sqrt{3}$ in all possible ways. Assume that numbers $u_1$, $\ldots$, $u_4$. are all distinct. Let $v = \alpha + \beta \sqrt{2} + \gamma \sqrt{3} + \delta \sqrt{2} \sqrt{3}$. We are looking for a polynomial of degree at most $4$ such that $P(u_i) = v_i$, $i=1\ldots, 4$. $P$ is the Lagrange interpolation polynomial $$P(X) = \sum_{i=1}^4 v_i \cdot \frac{\prod_{j\ne i}(X-u_j)}{\prod_{j\ne i}(u_i-u_j)}$$

Because of the symmetry of choices we have $P\in \mathbb{Q}[X]$. Moreover, $P(u_i)=v_i$, $i=1,\ldots, 4$. ( Lagrange's method, also used by Galois)

Now, the condition $u_i$ all distinct is equivalent to "at least two of the $b$, $c$, $d$ are non-zero".

In fact, Lagrange (and Galois) used a slightly different expression, that is easier for calculations. Considering $Q(X) = \prod_{i=1}^4(X-u_i)$, let's write $$R(X)= \sum_i v_i\cdot \frac{Q(X)}{X-u_i}$$. Again, $R\in \mathbb{Q}[X]$, and we have $$R(u_i)= v_i \cdot Q'(u_i)\\ v_i = \frac{R(u_i)}{Q'(u_i)}$$

Consider the example $$u_1 = \sqrt{2}+\sqrt{3}+\sqrt{2}\sqrt{3}\\ u_2 = -\sqrt{2}+\sqrt{3}-\sqrt{2}\sqrt{3}\\ u_3=\sqrt{2}-\sqrt{3}-\sqrt{2}\sqrt{3}\\ u_4=-\sqrt{2}-\sqrt{3}+\sqrt{2}\sqrt{3}$$ and $v=v_1=\sqrt{2}$. We calculate $$Q(X)=X^4 - 22 X^2 - 48 X - 23\\ \sum\frac{v_i}{X-u_i}=\frac{8 (X^2 + 6 X + 7)}{X^4 - 22 X^2 - 48 X - 23}\\ R(X)= 8 (X^2 + 6 X + 7)\\ R(u_1)=v_1 \cdot Q'(u_1)\\ v=\frac{R(u)}{Q'(u)}=\frac{8(u^2+6 u + 7)}{4(u^3-11u-12)}$$ that is $$\sqrt{2}=\frac{2(8u^2+6u+7)}{u^3-11u-12}_{u=\sqrt{2}+\sqrt{3}+\sqrt{6}}$$