Let $X = \{ \sqrt{p} : p \text{ is prime} \}$, $Y \subseteq X$ and $\sqrt{p} \not\in Y$. Show that $[\mathbb{Q}(Y)(\sqrt{p}) : \mathbb{Q}(Y)] = 2$.

Question

I am trying to solve Problem 22 from Chapter 5 of Patrick Morandi's Field and Galois Theory:

22. Let $K = \mathbb{Q}(X)$, where $X = \{ \sqrt{p} : p \text{ is prime} \}$. Show that $K$ is Galois over $\mathbb{Q}$. If $\sigma \in \operatorname{Gal}(K/\mathbb{Q})$, let $Y_\sigma = \{ \sqrt{p} : \sigma(\sqrt{p}) = - \sqrt{p} \}$. Prove the following statements.

    (a) If $Y_\sigma = Y_\tau$, then $\sigma = \tau$.

    (b) If $Y \subseteq X$, then there is a $\sigma \in \operatorname{Gal}(K/\mathbb{Q})$ with $Y_\sigma = Y$.

    (c) If $\mathcal{P}(X)$ is the power set of $X$, show that $\lvert \operatorname{Gal}(K/\mathbb{Q})\rvert = \lvert \mathcal{P}(X) \rvert$ and that $\lvert X \rvert = [K : \mathbb{Q}]$, and conclude that $\lvert \operatorname{Gal}(K/\mathbb{Q}) \rvert > [K : \mathbb{Q}]$.

    (Hint: A Zorn's lemma argument may help in (b). You may want to verify that if $Y \subseteq X$ and $\sqrt{p} \not\in Y$, then $[\mathbb{Q}(Y)(\sqrt{p}):\mathbb{Q}(Y)]=2$. The inequality $\lvert \mathcal{P}(X) \rvert > \lvert X \rvert$ is proved in Example 2.2 of Appendix B.)

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To complete part (c), I need to show that $[K : \mathbb{Q}]$ is not finite. If I can do this, then I will have shown that $[K : \mathbb{Q}]$ is countably infinite, since $K/\mathbb{Q}$ is an algebraic extension. Since $X$ is also countably infinite, this will show that $\lvert X \rvert = [K : \mathbb{Q}]$.

The hint asks me to verify that if $Y \subseteq X$ and $\sqrt{p} \not\in Y$, then $[\mathbb{Q}(Y)(\sqrt{p}):\mathbb{Q}(Y)]=2$. This I am not able to do. I understand that if I show this, then it will imply that $[K : \mathbb{Q}]$ is not finite, because (by induction) for every $n \in \mathbb{N}$, there is an intermediate field $L$ with $[L:\mathbb{Q}] = 2^n$, namely $L = \mathbb{Q}(X_n)$ where $X_n$ is any subset of $X$ of cardinality $n$.

The problem essentially boils down to the question, if $\sqrt{p} \not\in Y$, is it still possible that $\sqrt{p} \in \mathbb{Q}(Y)$? (And we seek to show that the answer is "No".) So, one idea I had was to assume that $\sqrt{p} \in \mathbb{Q}(Y)$ and somehow derive a contradiction, but I had no luck in doing that.

Another idea was to try and show the existence of a non-trivial automorphism of $\mathbb{Q}(Y)(\sqrt{p})$ over $\mathbb{Q}(Y)$. Since $[\mathbb{Q}(\sqrt{p}):\mathbb{Q}]=2$, we have that $[\mathbb{Q}(Y)(\sqrt{p}):\mathbb{Q}(Y)]\leq 2$, so showing that a non-trivial automorphism exists is enough. In fact, we know exactly what this automorphism should look like: it must act as the identity on $Y$ (and $\mathbb{Q}$, trivially) and it must map $\sqrt{p}$ to $-\sqrt{p}$. But I am not able to argue for why such an automorphism must exist. I know that the non-trivial embedding of $\mathbb{Q}(\sqrt{p})$ into $\mathbb{C}$ can be lifted to an embedding of $\mathbb{Q}(Y)(\sqrt{p})$ into $\mathbb{C}$, but there is no reason for this lift to automatically act as the identity on $Y$, right?

I am aware of an earlier question asking to show that for distinct primes $p_1,\dotsc,p_n \in \mathbb{N}$, $\sqrt{p_1},\dotsc,\sqrt{p_n}$ are linearly independent over $\mathbb{Q}$, but in my case I need to show the stronger result that they are algebraically independent over $\mathbb{Q}$, if I'm not mistaken. Any help is appreciated.

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References

• Morandi, Patrick, Field and Galois theory, Graduate Texts in Mathematics. 167. New York, NY: Springer. xvi, 281 p. (1996). ZBL0865.12001.

Answer 1

In the answer below I follow your hint exactly, and show that $[\mathbb{Q}(Y)(\sqrt{p}):\mathbb{Q}(Y)]=2$. As you already noted, it suffices to show that $\sqrt{p}\not\in \mathbb{Q}(Y)$.

When $Y$ is finite, this follows from Bill Dubuque's proof in the accepted answer to the MSE question you linked to.

Suppose now that $Y$ is infinite. Let $\lbrace y_k \rbrace_{k\geq 1}$ be an enumeration of $Y$. Suppose by contradiction that $\sqrt{p}\in \mathbb{Q}(Y)$. By definition of $\mathbb{Q}(Y)$ for an infinite $Y$, this means that $\sqrt{p}\in \mathbb{Q}(y_1,y_2,\ldots,y_N)$ for some finite $N$, and then we can apply Bill Dubuque's result again. This finishes the proof.