Let $K = \mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\dots,\sqrt{p_n})$ for distinct primes $p_1,p_2,\dots,p_n$. Prove $[K:\mathbb{Q}] = 2^n$.

Question

I'm new to this forum so I hope you can forgive any errors (if any) present in this post.

I already have a solution for the question in my title, but my textbook mentioned that I should Galois Theory. My solution makes no use of Galois Theory, and I was hoping to figure out what the question wanted me to do.

My solution:

Consider the intermediate field $F_{k} = \mathbb{Q}(\sqrt{p_1},\dots, \sqrt{p_k})$. We claim that for any subset $S \subseteq \{p_{k+1},\dots,p_n\}$, $\prod\limits_{x \in S}\sqrt{x} \notin F_{k}$.

Let $S = \{p_{i_1},p_{i_2},\dots, p_{i_{m}}\}$. Suppose $\sqrt{p_{i_1}p_{i_2}\dots p_{i_{m}}} = a + b\sqrt{p_1}$. Then, $$0 = (a^2 + b^2p_1 - p_{i_1}p_{i_2}\dots p_{i_{m}}) + 2ab\sqrt{p_1}.$$ If $a = 0$, $b = \sqrt{\frac{p_{i_1}p_{i_2}\dots p_{i_{m}}}{p_1}}$, which contradicts that $b$ is rational. If $b = 0$, $a = \sqrt{p_{i_1}p_{i_2}\dots p_{i_{m}}}$, which contradicts that $a$ is rational. The desired result follows.\

Suppose the statement holds for $k-1$, and let $S = \{p_{i_1},p_{i_2},\dots, p_{i_{m}}\}$. Then, note $F_k = F_{k-1}(\sqrt{p_{k}})$. Suppose $\sqrt{p_{i_1}p_{i_2}\dots p_{i_{m}}} = a + b\sqrt{p_k}$ with ($a,b \in F_{k-1}$). Then, $$0 = (a^2 + b^2p_k - p_{i_1}p_{i_2}\dots p_{i_{m}}) + 2ab\sqrt{p_k}.$$ If $a = 0$, we have that $bp_k = \sqrt{p_{i_1}p_{i_2}\dots p_{i_{m}}p_k}$, which contradicts the inductive hypothesis, and, if $b = 0$, $a = \sqrt{p_{i_1}p_{i_2}\dots p_{i_{m}}}$, contradicting the inductive hypothesis. It follows that $\sqrt{p_{k+1}} \notin F_{k}$, and, hence, $[K : \mathbb{Q}] = 2^n$.

If the validity of this solution could also be verified, I would be grateful.

In your inductive step: why must you have $ab = 0$? Since $a, b \in F_{k-1}$, it is not necessary that $a^2, b^2 \in \mathbb{Q}$, right? — Aryaman Maithani, May 12 '20 at 08:52
Did you compare with the answers at this duplicate? Some of them use Galois theory. — Dietrich Burde, May 12 '20 at 08:58
@AryamanMaithani Shouldn't it hold that, by the inductive hypothesis, ${1, \sqrt{p_k} }$ are linearly independent over $F_{k-1}$? Then, we have either $a$ or $b$ is $0$ — PolyBeanDip, May 12 '20 at 09:02

score 1 · Accepted Answer · answered May 19 '20 at 17:48

We claim first that $K/\mathbb{Q}$ is a Galois extension. This follows because $K$ is the splitting field of $f(x) = (x^2 - p_1)\ldots(x^2-p_n)$ over $\mathbb{Q}$, so it is a normal extension. Any finite extension of $\mathbb{Q}$ is separable, so $K/\mathbb{Q}$ is normal and separable, hence Galois.

Thus, by standard results of Galois Theory, $[K:\mathbb{Q}] = \lvert \text{Aut}_\mathbb{Q}(K)\rvert$.

Suppose that $\varphi:K \to K$ is a $\mathbb{Q}$-automorphism of $K$. Then for each $i$, $\varphi(\sqrt{p_i})^2 = \varphi(\sqrt{p_i}^2) = \varphi(p_i) = p_i$, so $\varphi(\sqrt{p_i}) = \pm \sqrt{p_i}$. Then since $K$ is generated as a $\mathbb{Q}$-algebra by the $\sqrt{p_i}$, $\varphi$ is determined uniquely by $n$ choices from $\{\sqrt{p_i}, -\sqrt{p_i}\}$. There are at most $2^n$ such choices, hence $\lvert \text{Aut}_\mathbb{Q}(K) \rvert \leq 2^n$.

It is clear that each of these $2^n$ choices actually does give rise of a well-defined $\mathbb{Q}$-automorphism of $K$, so indeed $[K : \mathbb{Q}] = \lvert \text{Aut}_\mathbb{Q}(K)\rvert = 2^n$.

NB: My proof is somewhat disingenuous, because the fact that each of the $2^n$ choices induces a well-defined $\mathbb{Q}$-automorphism relies implicitly on an inductive argument, hence making this solution no better than yours. However, I think the book probably seeks to highlight the connection between the degree of the extension and the Galois group, rather than to give an improved method of proof.

Let $K = \mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\dots,\sqrt{p_n})$ for distinct primes $p_1,p_2,\dots,p_n$. Prove $[K:\mathbb{Q}] = 2^n$.

1 Answers1