# primes of the form $p = x^2 + ny^2$: primitive element for the Hilbert class field of $\mathbb{Q}(\sqrt{-n})$ is a real algebraic integer?

The following proposition is found in $$\S 5$$ of Primes of the form $$p = x^2 + ny^2$$: Fermat, Class Field Theory, and Complex Multiplication by David A. Cox:

Proposition 5.29. Let $$K$$ be an imaginary quadratic field and let $$L$$ be a finite extension of $$K$$ which is Galois over $$\mathbb{Q}$$. Then

(i) There is a real algebraic integer $$\alpha$$ such that $$L = K(\alpha)$$.

(ii) Let $$\alpha$$ be as in (i) and $$f(x) \in \mathbb{Z}[x]$$ denote its minimal polynomial. If $$p$$ is a prime not dividing the discriminant of $$f(x)$$, then $$p\text{ splits completely in } L \Leftrightarrow \cases{(d_K/p) = 1 \text{ and } f(x) \equiv 0\bmod{p}\cr \text{has an integer solution.} }$$ My trouble concerns (i). In his proof Cox uses the fact that $$[L \cap \mathbb{R} : \mathbb{Q}] = [L : K] < \infty$$ to conclude there exists $$\alpha \in \mathscr{O}_L \cap \mathbb{R}$$ such that $$L \cap \mathbb{R} = \mathbb{Q}(\alpha)$$.

The Primitive Element Theorem guarantees the existence of $$\alpha \in L \cap \mathbb{R}$$ such that $$L \cap \mathbb{R} = \mathbb{Q}(\alpha)$$, but I do not understand how we are able to select $$\alpha$$ so that $$\alpha \in \mathscr{O}_L$$. Perhaps I am missing something obvious?

• in the first edition, this is Proposition 5.29; the step you dislike refers to Exercise 5.19 a few pages later Jan 14, 2021 at 19:58
• @WillJagy Cox refers to exercise 5.19 in his proof. The relevant part is (ii), which states For $\alpha \in L \cap \mathbb{R}$, $L \cap \mathbb{R} = \mathbb{Q}(\alpha) \Leftrightarrow L = K(\alpha)$. He uses this and the Primitive Element Theorem to establish $L = K(\alpha)$ under the assumption $[L : K] < \infty$. There is nothing about $\alpha \in \mathscr{O}_L$ in exercise 5.19, which can be solved without establishing this stronger claim. Jan 14, 2021 at 20:17
• Multiply $\alpha$ by a suitably chosen rational integer.
– user23365
Jan 15, 2021 at 7:36

Let $$A$$ be an integral domain and $$L$$ be a field containing $$A$$. Then we have
Proposition 2.6. Let $$K$$ be the field of fractions of $$A$$ and let $$L$$ be a field containing $$K$$. If $$\alpha \in L$$ is algebraic over $$K$$, then there exists a $$d \in A$$ such that $$d\alpha$$ is integral over $$A$$.
Taking $$A = \mathbb{Z}$$, $$K = \mathbb{Q}$$, and $$L \cap \mathbb{R} = \mathbb{Q}(\alpha)$$ the field containing $$K$$ allows us to select $$\alpha \in \mathscr{O}_L \cap \mathbb{R}$$.