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?

  • $\begingroup$ in the first edition, this is Proposition 5.29; the step you dislike refers to Exercise 5.19 a few pages later $\endgroup$
    – Will Jagy
    Jan 14, 2021 at 19:58
  • $\begingroup$ @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. $\endgroup$
    – nohomology
    Jan 14, 2021 at 20:17
  • $\begingroup$ Multiply $\alpha$ by a suitably chosen rational integer. $\endgroup$
    – user23365
    Jan 15, 2021 at 7:36

1 Answer 1


I was missing something obvious. Following the suggestion of @franzlemmermeyer, I found the following proposition in Chapter 2 of J. S. Milne's course notes for Algebraic Number Theory:

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 $\alpha \in \mathscr{O}_L \cap \mathbb{R}$.


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