Proving the lemma $\sqrt2\notin\mathbb{Q}(\sqrt3,\sqrt5)$

Question

I am attempting to solve part (c) of the following problem from Artin's algebra textbook. I have already solved a, b, and d.

Determine the irreducible polynomial for $α=\sqrt3+\sqrt5$ over the following fields.

(a) $\mathbb{Q}$ (b) $\mathbb{Q(\sqrt5)}$ (c) $\mathbb{Q(\sqrt10)}$ (d) $\mathbb{Q(\sqrt15)}$

Attempt

I'm pretty sure that that irreducible polynomial for $α$ over $\mathbb{Q(\sqrt10)}$ is $f(x)=x^4-16x^2+4$, but of course, I have to prove it.

I see that $α^3-α-1=0$, so it suffices to show that $[\mathbb{Q}(\sqrt10,α):\mathbb{Q(\sqrt10)}]=4$. I see that $[\mathbb{Q}(\sqrt10,α):\mathbb{Q(\sqrt10)}][\mathbb{Q}(\sqrt10):\mathbb{Q}]=[\mathbb{Q}(\sqrt10,α),\mathbb{Q}(α)][\mathbb{Q}(α):\mathbb{Q}]$.

To show that $[\mathbb{Q}(\sqrt10,α):\mathbb{Q(\sqrt10)}]=4$, it is enough to show that $[\mathbb{Q}(\sqrt10,α),\mathbb{Q}(α)]=2$.

Now, I see that $[\mathbb{Q}(\sqrt10,α),\mathbb{Q}(α)]$ must be either $2$ or $1$. Suppose it is $1$. Then $\sqrt10\in \mathbb{Q(\alpha)}$, and from this we obtain $\sqrt10\in \mathbb{Q}(\sqrt3,\sqrt5)$, and hence $\sqrt2\in \mathbb{Q}(\sqrt3,\sqrt5)$.

Now, it seems to me that this is a contradition, although I do not see how to rigorously prove it. If I can prove the following Lemma, I should be done.

I want to prove the Lemma: $\sqrt2\notin \mathbb{Q}(\sqrt3,\sqrt5)$

Attempt to prove Lemma

Suppose $\sqrt2\in \mathbb{Q}(\sqrt3,\sqrt5)$. I see that, for some $a,b,c,d\in \mathbb{Q}$, we have $\sqrt2=a+b\sqrt5+c\sqrt3+d\sqrt3 \sqrt5$. I predict that a contradiction can be obtained by playing around with this equation, but I do not know. Help would be appreciated.

Answer 1

Square both sides of the equation in Michael's comment to get

$2a^2 = 3b^2+5c^2+15d^2+e^2+2be\sqrt{3}+2ce\sqrt{5}+2de\sqrt{15}+6bd\sqrt{5}+10cd\sqrt{3}+2bc\sqrt{15}$

We write this as $m = n\sqrt{3}+l\sqrt{5}+k\sqrt{15}$ for some integers $m,n,l,k$. So,

$m-n\sqrt{3} = \sqrt{5}(l+k\sqrt{3}) \implies m^2-2mn\sqrt{3}+3n^2 = 5(l^2+2lk\sqrt{3}+3k^2)$.

Since $\sqrt{3}$ is not rational, this means that $2mn+10lk = 0$ and thus $m^2+3n^2 = 5l^2+15k^2$. Looking at this last equation mod $5$ means that $m \equiv n \equiv 0 \pmod{5}$ (by checking the different possibilities). Writing $m=5m', n=5n'$ and then dividing by $5$ gives that $5m'^2+15n'^2 = l^2+3k^2$. But now $l,k$ must be multiples of $5$ by the same reasoning as before. We get the classical infinite descent contradiction.

Answer 2

Here are some ways to do it:

1. Calculation: If $\sqrt{2}=\alpha+\beta\sqrt{3}+\gamma\sqrt{5}+\delta \sqrt{15}$ is a sum with rational coefficients, let's rearrange/square things so the number of terms in the equation goes down each time. First, square and rearrange to get $$q\ =\ a\sqrt{3\cdot 5}+b\sqrt{5\cdot 15}+c\sqrt{15\cdot 3}$$ (all coefficients are rational). Square again: $$q^2/2 \ =\ 15bc \sqrt{3\cdot 5}+3ac\sqrt{5\cdot 15}+5ab\sqrt{15\cdot 3}$$ and subtract one from a multiple of the other to get $$0\ =\ a'\sqrt{3\cdot 5}+b'\sqrt{5\cdot 15}+c'\sqrt{15\cdot 3}$$ One can show that $a',b',c'$ are not all zero (here we assume not all $a,b,c$ are zero, because the zero case is easy). Say $a'\ne 0$, then $$\sqrt{3\cdot 5}=A\sqrt{5\cdot 15}+B\sqrt{15\cdot 3}$$ Squaring one last time, we arrive at a contradiction $$\sqrt{3\cdot 5}=A'$$
2. Automorphisms: consider the automorphisms $\sigma, \tau$ of $\mathbb{Q}(\sqrt{3},\sqrt{5})$ over $\mathbb{Q}$, defined by $\sigma(\sqrt{3})=-\sqrt{3}$, $\sigma(\sqrt{5})=\sqrt{5}$, and vice-versa for $\tau$. If $\eta$ is any automorphism, $\sqrt{2}^2-2$ implies $\eta(\sqrt{2})^2-2=0$, therefore $\eta(\sqrt{2})=\pm \sqrt{2}$. Write $\pm=(-1)^\eta$. $$\text{}$$ Now apply the four automorphisms to the equation: $$\sqrt{2}=\alpha+\beta\sqrt{3}+\gamma\sqrt{5}+\delta \sqrt{15}$$ $$(-1)^\sigma\sqrt{2}=\alpha-\beta\sqrt{3}+\gamma\sqrt{5}-\delta \sqrt{15}$$ $$(-1)^\tau\sqrt{2}=\alpha+\beta\sqrt{3}-\gamma\sqrt{5}-\delta \sqrt{15}$$ $$(-1)^{\sigma\tau}\sqrt{2}=\alpha-\beta\sqrt{3}-\gamma\sqrt{5}+\delta \sqrt{15}$$ one of these signs will be $+1$, say $(-1)^\sigma=+1$. Then $$2\sqrt{2}=2\alpha+2\gamma\sqrt{5}$$ from which a contradiction easily arises.

3. Galois theory: $\mathbb{Q}(\sqrt{a},\sqrt{b})/\mathbb{Q}$ has Galois group $C_2\times C_2$ (the extension has degree $4$ and you can write down four automorphisms explicitly; alternatively, use that it's the splitting field of $(t^2-a)(t^2-b)$). $\mathbb{Q},\mathbb{Q}(\sqrt{a}), \mathbb{Q}(\sqrt{b})$ and $\mathbb{Q}(\sqrt{ab})$ are four proper subextensions of $\mathbb{Q}(\sqrt{a},\sqrt{b})/\mathbb{Q}$. But $G$ only has four proper subgroups, so there are all the subextensions, none of which are $\mathbb{Q}(\sqrt{2})$, unless $a$ or $b$ is $2$.

4. Check it locally: Pick a prime $p$ so that, modulo $p$, $\sqrt{3}$ and $\sqrt{5}$ are quadratic residues, which will imply that $\sqrt{15}$ is too, and $\sqrt{2}$ is not . Also arrange that $p$ does not divide the denominator of $\alpha, \beta, \gamma, \delta$. Then if $\sqrt{3}=a$ and $\sqrt{5}=b$ in $\mathbb{F}_p$, $$\alpha+\beta a+\gamma b+\delta ab\in \mathbb{F}_p$$ is a square root of $2$ in $\mathbb{F}_p$, which is a contradiction. $$\text{}$$ In this case, $p=61$ works. Why? Our problem is $\left( \frac{2}{p}\right)=-1,\left( \frac{3}{p}\right)=\left(\frac{5}{p}\right)=1$, which one solves using quadratic reciprocity, giving a solution of the form $p=61\text{ mod }120$. $61$ may divide the denominators of $\alpha, \beta, \gamma$ or $\delta$, but use Dirichlet's theorem to find another prime of this form not dividing them.