Galois extensions and $O^{*}$-fields

Abstract

A field F is $O^{*}$ if each partial order that makes F a partially ordered field can be extended to a total order that makes F a totally ordered field. We use the theory of infinite primes developed by Dubois and Harrison to prove the following. For a subfield F of $\mathbb {C}$ that is finite-dimensional over $\mathbb {Q}$, we prove that when F is Galois over $\mathbb {Q}$, F is an $O^{*}$-field if and only if is a subfield of $\mathbb {R}$. We find other conditions that make F an $O^{*}$-field and provide several examples. As well for an arbitrary field of characteristic 0, we characterize the maximal partial orders that are Archimedean.

1 Introduction

A field F is called $O^{*}$ if each partial order on F making F into a partially ordered field can be extended to a total order on F making F into a totally ordered field, that is, if P is the positive cone of a partial order on F, then there exists a total order on F with the positive cone T such that $P \subseteq T$. The concept of $O^{*}$-rings was introduced by Fuchs in 1963 [5]. Identifying fields that are $O^{*}$-fields is an open question in Steinberg’s book “Lattice-ordered Rings and Modules” [14, Open Problem 22].

A field F is called a number field if F is a subfield of $\mathbb {C}$, the field of complex numbers, and F is finite-dimensional over $\mathbb {Q}$, the field of rational numbers. By using the theory of infinite primes for rings and fields developed by Harrison [6], the necessary and sufficient conditions for real number fields being $O^{*}$-fields have been proved recently [9]. The current paper continues the previous work and considers the number fields in $\mathbb {C}$, not necessarily in $\mathbb {R}$, the field of real numbers. This gives us the ability to use Galois extensions to get some results on $O^{*}$-fields.

Let R be a field. From [6, p. 3], a nonempty subset S of R is called a preprime if S is closed under the addition and multiplication in R, and $-1 \not \in S$, that is, $S + S \subseteq S$, $SS \subseteq S$, and $-1 \not \in S$. A maximal preprime is called a prime. By Zorn’s Lemma, each preprime is contained in a prime. A prime S is called infinite if $1 \in S$, otherwise S is called finite. An infinite prime S of R is called full if $R = S - S = \{a - b \ | \ a, b \in S\}$.

Let F be a number field that is n-dimensional over $\mathbb {Q}$. As observed on [6, p. 37] there exist exactly n embeddings $\sigma _{1}, \sigma _{2}, \ldots , \sigma _{n}$ of F into $\mathbb {C}$. Let $\rho$ be the ordinary complex conjugate on $\mathbb {C}$. Assume $\rho \circ \sigma _{i} = \sigma _{i}$, for $1 \le i \le r$, and $\rho \circ \sigma _{i} = \sigma _{i + s}$ for $r < i \le r + s$ with $r + 2\,s = n$. Then

$$\begin{aligned} \sigma _{1}, \ldots , \sigma _{r}, \ \sigma _{r + 1}, \ldots , \sigma _{r + s}, \rho \circ \sigma _{r + 1}, \ldots , \rho \circ \sigma _{r + s} \end{aligned}$$

are these embeddings, and $\sigma _{1}, \ldots , \sigma _{r}$ are called the real infinite prime divisors of F, and the sets $\{\sigma _{r + 1}, \rho \circ \sigma _{r + 1}\}, \ldots , \{\sigma _{r + s}, \rho \circ \sigma _{r + s}\}$ are called the complex infinite prime divisors of F [6]. An embedding $\sigma$ of F is called imaginary if there exists an element $0 \ne a \in F$ such that $\sigma (a)$ is a pure imaginary complex number.

Theorem 1

(1) ( [6, Proposition 3.5]) Let R be a number field and let $\sigma _{1}, \ldots , \sigma _{r}$ be the real infinite prime divisors of R. Then the sets $\sigma _{1}^{-1}(\mathbb {R}^{+}), \ldots , \sigma _{r}^{-1}(\mathbb {R}^{+})$ are distinct and consist exactly of all the full infinite primes of R.

(2) ([6, Proposition 3.6]) Let R be a number field. Let P be an infinite prime of R which is not full $(i.e., P - P \ne R)$. Then there exists a complex infinite prime divisor $\{\sigma , \rho \circ \sigma \}$ of R with $P = \sigma ^{-1}(\mathbb {R}^{+})$. If R is a normal number field, then this gives a one-one correspondence between all the non-full infinite primes of R and all the complex infinite prime divisors of R.

For a number field F, the infinite primes of F are precisely the maximal partial orders on F. If P is a maximal partial order on F. Then $E_{P} = P - P$ is a subfield of F and P is a total order if and only if $E_{P} = F$. The proofs of these facts are given in [9, Lemma 2.2 & Theorem 3.1].

For more information on partially ordered rings and undefined terminology, the reader is referred to [1, 2, 5, 6, 9, 11, 13, 14]. In the following, $\mathbb {R}^{+}$ and $\mathbb {Q}^{+}$ denote the usual total order on $\mathbb {R}$ and $\mathbb {Q}$, respectively.

2 Galois extension and $O^{*}$-fields

We recall a few definitions and notations from Galois theory. Let L be a number field and K a subfield of L. The Galois group Gal(L/K) is the set of all K-automorphisms of L. Let S be a subset of Galois(L/K). Define

$$\begin{aligned} \mathcal {F}(S) = \{a \in L \ | \ f(a) = a, \forall f \in S\}. \end{aligned}$$

Then $\mathcal {F}(S)$ is a subfield of L, called the fixed field of S, and clearly $K \subseteq \mathcal {F}(S)$. L is called Galois over K if $K = \mathcal {F}(\text{ Gal }(L / K))$. L is Galois over K if and only if $|\text{ Gal }(L / K)| = [L: K]$ [13, Corollary 2.16]. Let $L = K[\alpha ]$ and $[L: K] = n$. Then L is Galois over K if and only if the minimal polynomial of $\alpha$ has n roots in L [13, Corollary 2.17].

Theorem 2

Let F be a number field that is Galois over $\mathbb {Q}$. Then F is $O^{*}$ if and only if F is a subfield of $\mathbb {R}$.

Proof

“$\Rightarrow$” Assume that F is $O^{*}$. Since F is Galois over $\mathbb {Q}$, F is the splitting field of a set of irreducible polynomials $\{f_{j}\}$ over $\mathbb {Q}$. Suppose that one $f_{j}$ has a complex root $\alpha = a + ib$ with $a, b \in \mathbb {R}$ and $b \ne 0$. Then $\overline{\alpha } = a - ib$ is a root of $f_{j}$ as well. So $\alpha - \overline{\alpha } = 2ib \in F$ and $(\alpha - \overline{\alpha })^{2} = -(2b)^{2} \ne 0$.

Let $P = \mathbb {R}^{+} \cap F$. Then P is a partial order on F and $-(\alpha - \overline{\alpha })^{2} \in P$. Since F is $O^{*}$, $P \subseteq T$, where T is a total order on F. Thus $(\alpha - \overline{\alpha })^{2} \in T$, so $0 \ne (\alpha - \overline{\alpha })^{2} \in T \cap -T$, a contradiction. Hence each $f_{j}$ only contains real roots and hence $F \subseteq \mathbb {R}$.

“$\Leftarrow$” Assume $F \subseteq \mathbb {R}$. Then all infinite prime divisors of F are real infinite prime divisors since for any $a \in F$, the roots of the minimal polynomial of a over $\mathbb {Q}$ are in $F \subseteq \mathbb {R}$ [13, Proposition 3.28]. Let P be a maximal partial order on F. Then P is an infinite prime of F by [9, Lemma 2.2], so there exists a real infinite prime divisor $\delta$ of F such that $P = \delta ^{-1}(\mathbb {R}^{+})$ by Theorem 1(1). Thus P is a total order on F and F is $O^{*}$. $\square$

Let F be a number field. The normal closure $F_{nc}$ of F over $\mathbb {Q}$ is the splitting field over $\mathbb {Q}$ of the set of minimal polynomials of elements of F. If $F = \mathbb {Q}[\alpha ]$, then $F_{nc}$ is the splitting field of the minimal polynomial of $\alpha$ over $\mathbb {Q}$ and $F_{nc}$ is Galois over $\mathbb {Q}$ [13, Proposition 5.9]. If the minimal polynomial of $\alpha$ over $\mathbb {Q}$ has only real roots, then $F_{nc} \subseteq \mathbb {R}$. Thus $F_{nc}$ is $O^{*}$ by Theorem 2 and hence F is $O^{*}$ as well. On the other hand that F is $O^{*}$ may not imply that $F_{nc}$ is $O^{*}$. After proving the following result, we will give an example of such a field F.

Lemma 1

Let E be a number field with a total order P and let F be an extension field of E such that [F : E] is odd. Then P can be extended to a total order on F, that is, there exits a total order $P_{1}$ on F such that $P \subseteq P_{1}$.

Proof

Since $E = P \cup -P = P - P$, by Theorem 1(1), there exists a real infinite prime divisor $\delta$ of E such that $P = \delta ^{-1}(\mathbb {R}^{+})$. Let $F = E[\alpha ]$ and $f(x) = a_{0} + a_{1}x + \cdots + x^{n}$ be the minimal polynomial of $\alpha$ over E, where $a_{0}, a_{1}, \ldots$ are in E. Define

$$\begin{aligned} g(x) = \delta (f(x)) = \delta (a_{0}) + \delta (a_{1})x + \cdots + x^{n}. \end{aligned}$$

Then g(x) is a real irreducible polynomial of degree n over $\delta (E)$ since $\delta$ is a real embedding. Since $[F: E] = n$ is odd, g(x) has a real root $\beta$. Define $\sigma : F \rightarrow \mathbb {C}$ as follows. For all

$$\begin{aligned}{} & {} a_{0} + a_{1}\alpha + \cdots + a_{n - 1}\alpha ^{n - 1} \in F, \\{} & {} \sigma (a_{0} + a_{1}\alpha + \cdots + a_{n - 1}\alpha ^{n - 1}) = \delta (a_{0}) + \delta (a_{1})\beta + \cdots + \delta (a_{n - 1})\beta ^{n - 1}, \end{aligned}$$

where $a_{0}, a_{1}, \ldots , a_{n - 1} \in E$. Then $\sigma$ is a real infinite prime divisor of F. Let $P_{1} = \sigma ^{-1}(\mathbb {R}^{+})$. Then, by Theorem 1(1), $P_{1}$ is a total order on F and for all $w \in P$, $\sigma (w) = \delta (w) \in \mathbb {R}^{+}$. So $w \in P_{1}$, that is, $P \subseteq P_{1}$. $\square$

Example 1

Let $\omega = e^{2\pi i/3}$ and $F = \mathbb {Q}[\omega \root 3 \of {2}]$. Let P be a maximal partial order on F and $E_{P} = P - P$. Then $E_{P}$ is a subfield of F [9, Theorem 2.2] and hence because $[E_{P}: \mathbb {Q}]$ divides $[F: \mathbb {Q}] = 3$, $[E_{P}: \mathbb {Q}]$ is either 3 or 1. We claim $[E_{P}: \mathbb {Q}] \ne 1$. If $[E_{P}: \mathbb {Q}] = 1$, then $E_{P} = \mathbb {Q}$. Since the only total order on $\mathbb {Q}$ is $\mathbb {Q}^{+}$, $P = \mathbb {Q}^{+}$. However by Lemma 1, $P = \mathbb {Q}^{+}$ can be extended to a total order on F, a contradiction. Thus we must have $[E_{P}: \mathbb {Q}] = 3$, so $E_{P} = F$ and P is a total order on F. Hence F is $O^{*}$. By [13, Example 1.27], $F_{nc} = \mathbb {Q}[\omega , \root 3 \of {2}]$, so $F_{nc}$ is not $O^{*}$ by Theorem 2.

Let F be a number field. If F is $O^{*}$, then F does not have imaginary embeddings. In fact, if $\delta : F \rightarrow \mathbb {C}$ is an embedding such that $\delta (a) = ib$ for some $0 \ne a \in F$ and $b \in \mathbb {R}$, where $i = \sqrt{-1}$. Then $P = \delta ^{-1}(\mathbb {R}^{+})$ is a partial order on F that cannot be extended to a total order on F since $-a^{2} \in P$.

However it is not clear that F having no imaginary embeddings is a sufficient condition for F being $O^{*}$. We know that if $[F: \mathbb {Q}]$ is odd, then F is always $O^{*}$ from Lemma 1 or [9, Theorem 3.2].

In the following result, we collect some equivalent conditions equivalent to F not having imaginary embeddings.

Theorem 3

Let F be a number field. The following statements are equivalent.

• (1) F does not have imaginary embeddings,

• (2) For any pair (E, P), where E is a subfield of F and P is a total order on E, and any subfield K of F which is a quadratic extension over E, there exists $\alpha \in K$ such that $K = E[\alpha ]$ and $\alpha$ is a root of irreducible polynomial $f(x) = x^{2} - a$ over E for some $a \in P$,

• (3) For any subfields E and K of F such that $E \subseteq K$ and $[K: E] = 2$, each total order on E can be extended to a total order on K.

Proof

(1) $\Rightarrow$ (2) Since P is a total order on E, by Theorem 1(1), there exists a real infinite prime divisor $\delta$ of E such that $P = \delta ^{-1}(\mathbb {R}^{+})$. Assume that $K = E[\alpha ]$ and $\alpha$ satisfies the irreducible polynomial $f(x) = x^{2} - a$ for some $a \in E$. If $a \not \in P$, then $-a \in P$, so $-\delta (a) \in \mathbb {R}^{+}$. It follows that $g(x) = x^{2} - \delta (a)$ has pure imaginary roots $\theta = \pm i\sqrt{-\delta (a)}$. Then $\phi : K \rightarrow \mathbb {C}$, defined by

$$\begin{aligned} \phi (c + d\alpha ) = \delta (c) + \delta (d)\theta , \ \forall c, d \in E, \end{aligned}$$

is an imaginary embedding of K, and it can be extended to an imaginary embedding of F [12, Theorem 50], a contradiction. Therefore we must have $a \in P$.

(2) $\Rightarrow$ (3) Let P be a total order on E, by Theorem 1(1) there exists a real infinite prime divisor $\delta$ of E such that $P = \delta ^{-1}(\mathbb {R}^{+})$. By the assumption, $K = E[\alpha ]$ for some $\alpha$ satisfying the irreducible polynomial $f(x) = x^{2} - a$ with $a \in P$. Let $\theta$ be a root of $g(x) = x^{2} - \delta (a)$. Since $a \in P$, $\delta (a) \in \mathbb {R}^{+}$ and thus $\theta \in \mathbb {R}$.

Then $\phi : K \rightarrow \mathbb {C}$, defined by

$$\begin{aligned} \phi (c + d\alpha ) = \delta (c) + \delta (d)\theta , \ \forall c, d \in E, \end{aligned}$$

is a real embedding from K to $\mathbb {C}$ that extends $\delta$, so $\phi$ is a real infinite prime divisor of K that extends $\delta$. Thus $P_{1} = \phi ^{-1}(\mathbb {R}^{+})$ is a total order on K and $P = \delta ^{-1}(\mathbb {R}^{+}) \subseteq \phi ^{-1}(\mathbb {R}^{+}) = P_{1}$.

(3) $\Rightarrow$ (1) Let $\sigma$ be an imaginary embedding of F. Then there exists $0 \ne a \in F$ such that $\sigma (a) = ib$, where $b \in \mathbb {R}$. Define $P = \sigma ^{-1}(\mathbb {R}^{+})$ and $E_{P} = P - P$. Then P is a total order on the subfield $E_{P}$ of F. Since $\sigma (-a^{2}) = -\sigma (a)^{2} = b^{2} \in \mathbb {R}^{+}$, $-a^{2} \in P$, so $a^{2} \in E_{P}$. Thus $[E_{P}[a]: E_{P}] = 2$. By (3), P can be extended to a total order $P_{1}$ on $E_{P}[a]$, so $a^{2} \in P_{1}$. On the other hand, $-a^{2} \in P \subseteq P_{1}$. Therefore $a^{2} \in P_{1} \cap -P_{1} = \{0\}$, a contradiction. Hence F cannot have imaginary embeddings. $\square$

The following direct consequence of Theorem 3 is useful when showing number fields are $O^{*}$.

Corollary 1

Let F be a number field that does not have imaginary embeddings. For a maximal partial order P on F, $E_{P}$ cannot be contained in a subfield K of F such that $[K: E_{P}] = 2$ or m, where $m > 1$ is an odd positive integer.

Proof

Assume $E_{P} \subseteq K$ and K is a subfield of F with $[K: E_{P}] = 2$ or m, where $m > 1$ is odd. By Theorem 3 and Lemma 1, the total order P on $E_{P}$ can be extended to a total order $P_{1}$ on K. Since P is a maximal partial order on F and $P_{1}$ is also a partial order on F, $P = P_{1}$, so $K = P_{1} \cup (-P_{1}) = P \cup (-P) = P - P = E_{P}$, a contradiction. $\square$

Theorem 4

Let F be a number field that does not have imaginary embeddings. Assume $[F_{nc}: \mathbb {Q}] = 2^{a}n$, where $a \ge 1$ and n is a positive odd integer. If G = Gal$(F_{nc} / \mathbb {Q})$ has a normal subgroup of order n, then F is an $O^{*}$-field.

Proof

Let P be a maximal partial order on F and $E_{P} = P - P$. We assume $E_{P} \ne F$, and get a contradiction. Let

$$\begin{aligned} N = \text{ Gal }(F_{nc} / F), \ H = \text{ Gal }(F_{nc} / E_{P}). \end{aligned}$$

Then $N \subsetneq H \subseteq G$. Since |H| divides |G|, $|H| = 2^{b}m$, where $0 \le b \le a$ is an integer and m is a positive integer that is odd and divides n. Let M be a maximal proper subgroup of H containing N and $K = \mathcal {F}(M)$. Then $|M| = [F_{nc}: K]$ and $E_{P} \subsetneq K \subseteq F$.

If $m = 1$ and $b = 0$, then $|H| = [F_{nc}: E_{P}] = 1$, so $E_{P} = F_{nc}$ and hence $E_{P} = F$, a contradiction. If $m = 1$ and $b \ge 1$, then $|H| = 2^{b}$ and $|M| = 2^{b - 1}$ [8, Corollary 5.26]. Thus

$$\begin{aligned}{}[K: E_{P}] = \frac{[F_{nc}: E_{P}]}{[F_{nc}: K]} = \frac{|H|}{|M|} = 2, \end{aligned}$$

a contradiction by Corollary 1. If $b = 0$ and $m > 1$, then $|H| = m$, and hence

$$\begin{aligned}{}[K: E_{P}] = \frac{[F_{nc}: E_{P}]}{[F_{nc}: K]} = \frac{|H|}{|M|} \end{aligned}$$

is an odd integer $> 1$, a contradiction by Corollary 1.

In the following, assume that $b \ge 1$ and $m > 1$. Let Q be a normal subgroup of G with $|Q| = n$, and $n = p_{1}^{t_{1}} \ldots p_{k}^{t_{k}}$, where $p_{1}, \ldots , p_{k}$ are distinct odd prime numbers, $k \ge 1$, and $t_{i} \ge 1$, $i = 1, \ldots , k$. For $i = 1, \ldots , k$, let $Q_{i}$ be a Sylow $p_{i}$-subgroup of G. Then $QQ_{i}$ is a subgroup and $Q \subseteq QQ_{i} \subseteq G$. Since

$$\begin{aligned} |QQ_{i}| = \frac{|Q||Q_{i}|}{|Q \cap Q_{i}|} \ \text{ and } \ |QQ_{i}| \text{ divides } |G|, \end{aligned}$$

we must have $|QQ_{i}| = n$, so $Q = QQ_{i}$ and $Q_{i} \subseteq Q$ for $i = 1, \ldots , k$. Let $W_{i}$ be a Sylow $p_{i}$-subgroup of H, $i = 1, \ldots , k$. Then $W_{i}$ is contained in a conjugate of $Q_{i}$ in G, and hence because Q is normal, $W_{i} \subseteq Q$ for $i = 1, \ldots , k$. Let $W = Q \cap H$. Then W is a normal subgroup of H and $W_{i} \subseteq W$ for $i = 1, \ldots , k$, so $|W_{i}|$ divides |W| for $i = 1, \ldots , k$, and hence $|W| = m$. Since WM is a subgroup of H, either $WM = H$ or $WM = M$.

If $WM = H$, then

$$\begin{aligned} 2^{b}m = |H| = |WM| = \frac{|W||M|}{|W \cap M|} = \frac{m|M|}{|W \cap M|}, \end{aligned}$$

so $|M| = 2^{b}m_{1}$, where $m_{1}$ is a positive integer and $m_{1} | m$. Hence,

$$\begin{aligned}{}[K: E_{P}] = \frac{[F_{nc}: E_{P}]}{[F_{nc}: K]} = \frac{|H|}{|M|} = \frac{m}{m_{1}} \end{aligned}$$

is an odd integer $> 1$, a contradiction by Corollary 1.

If $WM = M$, then $W \subseteq M$. The quotient group H/W is a 2-group and M/W is a maximal subgroup of H/W. Hence $|M / W| = 2^{b - 1}$ [8, Corollary 5.26], so $|M| = 2^{b - 1}m$ and

$$\begin{aligned}{}[K: E_{P}] = \frac{[F_{nc}: E_{P}]}{[F_{nc}: K]} = \frac{|H|}{|M|} = 2, \end{aligned}$$

a contradiction by Corollary 1.

Therefore if P is a maximal partial order of F, we must have $E_{P} = F$, so P is a total order on F by Theorem 1(1). $\square$

Corollary 2

Let F be a number field that does not have imaginary embeddings. Assume $[F_{nc}: \mathbb {Q}] = 2^{a}p^{b}$, where a, b are positive integers and p is an odd prime number. If $p \ge 2^{a}$, then F is an $O^{*}$-field.

Proof

Since $p \ge 2^{a}$, the Sylow p-subgroup of G is normal, so by Theorem 4, F is $O^{*}$. $\square$

Let F be a number field that does not have imaginary embeddings. By Theorem 4, if $[F_{nc}: \mathbb {Q}] = 4p^{b}$ with $b \ge 1$ and prime number $p > 3$, then F is $O^{*}$. We show that if $[F_{nc}: \mathbb {Q}] = 4(3^{b})$ with $b \ge 1$, then F is $O^{*}$ as well. We first have the following result that will be used in the proof.

Lemma 2

Let G be a group of order $|G| = 4(3^{k})$ with $k \ge 2$. Then G has either the normal Sylow 3-subgroup or a normal subgroup of order $3^{k - 1}$.

Proof

If the number of Sylow 3-subgroups of G is 1, then G has the normal Sylow 3-subgroup. Now suppose that N and M are two different Sylow 3-subgroups of G. Then

$$\begin{aligned} |N \cap M| = \frac{|N||M|}{|NM|} \ge \frac{|N||M|}{|G|} = \frac{(3^{k})(3^{k})}{4(3^{k})} = \frac{9}{4}(3^{k - 2}), \end{aligned}$$

so because $|N \cap M|$ divides $3^{k}$, $|N \cap M| = 3^{k - 1}$.

We claim that $N \cap M$ is normal in G. It is well known that $N \cap M$ is normal in N and M [8, Corollary (5.26)]. Let K = $N_{G}(N \cap M)$ be the normalizer of $N \cap M$ in G. Then $NM \subseteq K$ and hence $3^{k + 1} = (|N||M|) / |N \cap M|= |NM| \le |K|$. So since |K| divides |G|, $|K| = 4(3^{k})$. Therefore $K = G$, so $N \cap M$ is normal in G. $\square$

Example 2

Let F be a real number field that does not have imaginary embeddings. If $[F_{nc}: \mathbb {Q}] = 4(3^{b})$ with $b \ge 1$, then F is $O^{*}$.

Let P be a maximal partial order on F and $E_{P} = P - P$. Define

$$\begin{aligned} G = \text{ Gal }(F_{nc} / \mathbb {Q}), \ H = \text{ Gal }(F_{nc} / E_{P}), \text{ and } N = \text{ Gal }(F_{nc} / F). \end{aligned}$$

Then $N \subseteq H \subseteq G$, so |H| divides $|G| = 4(3^{b})$. In the following, we consider the case that $|H| = 4(3^{c})$ with $1 \le c < b$, and leave the similar verifications of the other cases to the reader.

Assume $E_{P} \ne F$. We derive a contradiction. Since $E_{P} \ne F$, $N \ne H$. Let M be a maximal proper subgroup of H containing N and define $E = \mathcal {F}(M)$. Then

$$\begin{aligned}{}[E: E_{P}] = \frac{[F_{nc}: E_{P}]}{[F_{nc}: E]} = \frac{|H|}{|M|}, \end{aligned}$$

by the Fundamental Theorem of Galois Theory [13, Theorem 5.1]. We will use this fact repeatedly in the argument.

(1) Assume $c = 1$, so $|H| = 12$. Then H has either a normal Sylow 2-subgroup or a normal Sylow 3-subgroup [8, Theorem (5.14)].

First assume that H has a normal Sylow 3-subgroup Q. Since QM is a subgroup and $M \subseteq QM \subseteq H$, $QM = M$ or $QM = H$. If $QM = M$, then $Q \subseteq M$ and M/Q is a maximal subgroup in H/Q. Since $|H / Q| = 4$, $|M / Q| = 2$ [8, Corollary 5.26], so $|M| = 6$, and hence $[E: E_{P}] = |H| / |M| = 2$, a contradiction by Corollary 1. Suppose that $H = QM$. If $|Q \cap M|=3$ then $Q \subseteq M$ and thus $QM = M$, a contradiction. So $|Q \cap M| = 1$, and hence

$$\begin{aligned} |H| = \frac{|Q||M|}{|Q \cap M|} = 3|M| \end{aligned}$$

so that $[E: E_{P}] = |H| / |M| = 3$, again a contradiction by Corollary 1.

Now assume that H has the normal Sylow 2-subgroup W. If $WM = M$, then $W \subseteq M$, so $|M| = 4$. Then $[E: E_{P}] = |H| / |M| = 3$, a contradiction by Corollary 1. If $H = WM$ and $|W \cap M| = 2$, then

$$\begin{aligned} 12 = |H| = \frac{|W||M|}{|W \cap M|} = 2|M| \ \Rightarrow \ |M| = 6, \end{aligned}$$

so $[E: E_{P}] = |H| / |M| = 2$, a contradiction by Corollary 1. If $H = WM$ and $|W \cap M| = 1$, then $|M| = 3$. Let $K = \mathcal {F}(W)$. We have

$$\begin{aligned}{}[K: E_{P}] = \frac{|H|}{|W|} = 3 \ \text{ and } \ [E: E_{P}] = \frac{|H|}{|M|} = 4, \end{aligned}$$

so $[KE: E_{P}] = 12$ [13, Problem 17, p. 14]. We also have

$$\begin{aligned}{}[E_{P}: \mathbb {Q}] = \frac{[F_{nc}: \mathbb {Q}]}{[F_{nc}: E_{P}]} = \frac{4(3^{b})}{4(3)} = 3^{b - 1}. \end{aligned}$$

It follows that $[KE: \mathbb {Q}] = [KE: E_{P}][E_{P}: \mathbb {Q}] = 4(3^{b})$, and hence $KE = F_{nc}$.

Since $[K: E_{P}] = 3$, $K = E_{P}[\alpha ]$ and $\alpha$ is a root of an irreducible polynomial f(x) of degree 3 over $E_{P}$. Since $E_{P} \subseteq F \subseteq \mathbb {R}$, we may take $\alpha$ as a real root of f(x), so $K \subseteq \mathbb {R}$, and hence since $E \subseteq F \subseteq \mathbb {R}$, $F_{nc} = KE \subseteq \mathbb {R}$. So $F_{nc}$ is $O^{*}$ by Theorem 2 and thus F is also $O^{*}$. But then $E_{p}=F$, a contradiction of our initial assumption.

(2) Assume $|H| = 4(3^{c})$ with $c \ge 2$. By Lemma 2, H has a normal subgroup Q of order $3^{c}$ or $3^{c - 1}$.

(2a) Suppose that $|Q| = 3^{c}$. If $QM = M$, then $Q \subseteq M$, so $|M| = 2(3^{c})$. Thus $[E: E_{P}] = |H| / |M| = 2$, a contradiction by Corollary 1. If $QM = H$, then

$$\begin{aligned} 4(3^{c}) = |H| = |QM| = \frac{|Q||M|}{|Q \cap M|} \Rightarrow |M| = 4|Q \cap M|, \end{aligned}$$

so $[E: E_{P}] = 3^{c} / |Q \cap M|$ is an odd integer $> 1$ since $|Q \cap M| \ne 3^{c}$, a contradiction by Corollary 1.

(2b) Suppose that $|Q| = 3^{c - 1}$. If $QM = M$, then $Q \subseteq M$. Since Q is contained in a Sylow 3-subgroup of H and M is maximal, $Q \ne M$. Since $|H / Q| = 12$ and M/Q is maximal in H/Q, $|M / Q| = 3, 6$, or 4, and hence $|M| = 3^{c}, 2(3^{c})$ or $4(3^{c - 1})$. If $|M| = 2(3^{c})$ or $4(3^{c - 1})$, then $[E: E_{P}] = |H| / |M|$ is either 2 or 3, a contradiction by Corollary 1.

Consider the case that $|M| = 3^{c}$. Let W be a Sylow 2-subgroup of H and $K = \mathcal {F}(W)$. Then $[K: E_{P}] = |H| / |W| = 3^{c}$. It follows that $[EK: E_{P}] = [E: E_{P}][K: E_{P}] = 4(3^{c})$ [13, Problem 17, p. 14]. Since

$$\begin{aligned}{}[E_{P}: \mathbb {Q}] = \frac{[F_{nc}: \mathbb {Q}]}{[F_{nc}: E_{P}]} = \frac{4(3^{b})}{4(3^{c})} = 3^{b - c}, \end{aligned}$$

we have $[EK: \mathbb {Q}] = 4(3^{b})$, so $EK = F_{nc}$. Similar to the argument used in (1), it follows that $F_{nc} \subseteq \mathbb {R}$, a contradiction.

If $QM = H$, then

$$\begin{aligned} 4(3^{c}) = |H| = \frac{|Q||M|}{|Q \cap M|} = \frac{3^{c - 1}|M|}{|Q \cap M|} \ \Rightarrow \ |M| = 12|Q \cap M|, \end{aligned}$$

so $[E: E_{P}] = |H| / |M|$ is an odd integer $> 1$, a contradiction again by Corollary 1.

Therefore, we have proved that if $[F_{nc}: \mathbb {Q}] = 4(3^{b})$ with $b \ge 1$, then F is $O^{*}$.

Theorem 5

Let F be a real number field that does not have any imaginary embeddings and $[F_{nc}: \mathbb {Q}] = 4p^{b}$, where p is an odd prime number and $b \ge 1$, then F is $O^{*}$.

Proof

The case for $p=3$ is demonstrated in Example 2, leaving only $p \ge 4$, for which the Sylow p-subgroup of G is normal so by Theorem 4, F is $O^{*}$ $\square$

Let F be a number field and $F_{nc}$ be its normal closure. Define

$$\begin{aligned} G = \text{ Gal }(F_{nc} / \mathbb {Q}) \ \text{ and } \ N = \text{ Gal }(F_{nc} / F). \end{aligned}$$

Suppose that $|N| = 2^{a}n$ where $a \ge 1$ and n is a positive odd integer. Let S be a Sylow 2-subgroup of N and $E = \mathcal {F}(S)$. Then $F \subseteq E \subseteq F_{nc}$ with $[E: F] = n$ and $[F_{nc}: E] = 2^{a}$ [13, Theorem 5.1].

Although it is not certain if F is $O^{*}$ when F does not have imaginary embeddings, F is indeed $O^{*}$ in case that E does not have imaginary embeddings.

Theorem 6

Let E be defined as above. If E does not have imaginary embeddings, then F is an $O^{*}$-field.

Proof

Let P be a maximal partial order on F and $E_{P} = P - P$. Define H = Gal$(F_{nc} / E_{P})$ and assume $|H| = 2^{t}k$ where $t \ge 1$ and k is a positive odd integer. Since $N \subseteq H$, |N| divides |H| implies that $a \le t$ and $n \ | \ k$. Let W be a Sylow 2-subgroup of H such that $S \subseteq W$, where S is a Sylow 2-subgroup of N, and let $K = \mathcal {F}(W)$. So $|W| = 2^{t}$. Then there exists a chain of subgroups of W:

$$\begin{aligned} S \subseteq S_{1} \subseteq \ldots \subseteq S_{t - a - 1} \subseteq W, \end{aligned}$$

such that $|S_{i}| = 2^{a + i}$ for $i = 1, \ldots , t - a - 1$ [8, Corollary 5.23]. Let $E_{i} = \mathcal {F}(S_{i})$ for $i = 1, \ldots , t - a - 1$. We have

$$\begin{aligned} K \subseteq E_{t - a - 1} \subseteq \ldots \subseteq E_{1} \subseteq E \end{aligned}$$

and $[E: E_{1}] = 2, [E_{1}: E_{2}] = 2, \ldots , [E_{t - a - 1}: K] = 2$.

Now that P is a total order on $E_{P}$ and $[K: E_{P}] = |H| / |W| = k$ is odd implies that P can be extended to a total order $P_{K}$ on K by Lemma 1. Since E does not have imaginary embeddings, by Theorem 3, $P_{K}$ can be extended to a total order $P_{t - a - 1}$ on $E_{t - a - 1}$, so P can also be extended to the total order $P_{t - a - 1}$. Continuing this process, we have that P can be extended to a total order $P_{E}$ on E. Thus P can be extended to a total order $P_{E} \cap F$ on F since $F \subseteq E$. Therefore, since P is maximal on F, $P = P_{E} \cap F$ is a total order on F and F is $O^{*}$. $\square$

The following is a direct consequence of Theorem 6.

Corollary 3

If a number field F has no imaginary embeddings and $[F_{nc}: F]=2^k$ where k is a positive integer, then F is $O^{*}$.

3 Archimedean maximal partial orders on fields

Harrison and Dubois obtained many important results not only for the infinite primes on the number fields, but also for the infinite primes on arbitrary fields as well [2,3,4, 6, 7]. In this section, we use the connection between Archimedean maximal partial orders and infinite primes on an arbitrary field of characteristic 0 to study Archimedean maximal partial orders.

Let $(R, \le )$ be a partially ordered field. The partial order $\le$ on R is called Archimedean if for any $a, b \in R$, $\mathbb {Z}a \le b$ implies that $a = 0$, where $\mathbb {Z}$ is the set of all integers and $\mathbb {Z}a \le b$ means that $ma \le b$ for all integers m. The partial order $\le$ on R is called strong Archimedean if $0 \le 1$ and for any $0 \le a \in R$, there is a positive integer n such that $0 \le n - a$. The above definition for Archimedean is widely used in partially ordered groups and rings, see [1, 5, 9, 15] and the above definition for strong Archimedean is called Archimedean in [2, 6].

Lemma 3

Let $(R, \le )$ be a partially ordered field.

• (1) If $\le$ is a total order, then $\le$ is Archimedean if and only if it is strong Archimedean.

• (2) If $\le$ is a lattice order that is strong Archimedean, then it is a total order. However a lattice order that is Archimedean may not be a total order.

Proof

(1) Since $\le$ is a total order, $1 \ge 0$. Assume that $(R, \le )$ is Archimedean and $0 \le a$. Then $\mathbb {Z}1 \not \le a$, so there exists a positive integer n such that $n > a$, so $0 \le n - a$. Thus $(R, \le )$ is strong Archimedean. Now assume that $(R, \le )$ is strong Archimedean and $\mathbb {Z}a \le b$ for some $a, b \in R$. Suppose that $a \ne 0$. If $a > 0$, then for any positive integer n, $na \le b$ implies that $n \le a^{-1}b$ for all integers n, so there is no positive integer m such that $0 \le m - a^{-1}b$, a contradiction. If $a < 0$, then for all positive integers n, $n(-a) \le b$ implies that $n \le (-a)^{-1}b$, a contradiction again. Thus we must have $a = 0$, that is, $(R, \le )$ is Archimedean.

(2) Let $(R, \le )$ be a lattice-ordered field that is strong Archimedean and $a \in R$. Since $a^{+} = a \vee 0 < n$ and $a^{-} = -a \vee 0 < m$ for some positive integers n, m, we have

$$\begin{aligned} a^{+}a^{-} = a^{+}a^{-} \wedge a^{+}a^{-} \le na^{-} \wedge ma^{+} \le nm(a^{+} \wedge a^{-}) = 0. \end{aligned}$$

Hence $a^{+}a^{-} = 0$, so $a^{+} = 0$ or $a^{-} = 0$, that is, $(R, \le )$ is a totally ordered field. $\square$

An example of an Archimedean lattice order that is not a total order is $R = \mathbb {Q}[\sqrt{2}]$ with the coordinate-wise ordering $R^{+} = \{a + b\sqrt{2} \ | \ a, b \in \mathbb {Q}^{+}\}$.

Let P be a partial order on R and suppose that $1 \in P$. Define

$$\begin{aligned} J_{P} = \{a \in R \ | \ 1 + \mathbb {Z}a \subseteq P\}, \text{ where } \mathbb {Z}a = \{na \ | \ n \in \mathbb {Z}\}. \end{aligned}$$

Lemma 4

For any Archimedean partial order P with $1 \in P$, $J_{P} = \{0\}$.

Proof

It is clear that $0 \in J_{P}$. Let $a \in J_{P}$. Then $1 + \mathbb {Z}a \ge _{P} 0$, where $\ge _{P}$ is the partial order with the positive cone P. Then we have $\mathbb {Z}a \le _{P} 1$, so $a = 0$ since P is Archimedean. Therefore $J_{P} = \{0\}$. $\square$

Theorem 7

Let R be a field and let P be a maximal partial order on R.

• (1) Then $1 \in P$, P is an infinite prime of R, and $E_{P} = P - P$ is a subfield of R.

• (2) If P is Archimedean, then P is strong Archimedean.

Proof

(1) We first show that if $0 \ne a \in P$, then $a^{-1} \in P$. It is easy to see that,

$$\begin{aligned} P' = \{x \in R \ | \ wx \in P \text{ for } \text{ some } 0 \ne w \in P\} \end{aligned}$$

is a partial order on R and $P \subseteq P'$. Since P is maximal, $P = P'$. Then $a^{-1} \in P' = P$ because $a^2 a^{-1} = a \in P$. In particular, since P is maximal, there exists $0 \ne a \in P$, and thus $1 = aa^{-1} \in P$.

It is clear that P is a preprime. Assume that $P \subsetneq P_{1}$, where $P_{1}$ is a preprime. We derive a contradiction. Take $t \in P_{1} {\setminus }P$. Define

$$\begin{aligned} P[t] = \{f(t) \ | \ f(t) \text{ is } \text{ a } \text{ polynomial } \text{ in } t \text{ with } \text{ coefficients } \text{ in } P\}. \end{aligned}$$

It is clear that $P \subseteq P[t]$ and P[t] is closed under the addition and multiplication of R. If $P[t] \cap (-P[t]) = \{0\}$, then P[t] is a partial order on R, so $P = P[t]$ since P is a maximal partial order. Then $t \in P$, a contradiction. Thus $P[t] \cap (-P[t]) \ne \{0\}$. Let $w \in P[t] \cap (-P[t])$ and $w \ne 0$. Then $w = f(t)$ and $-w = g(t)$ for some $f(t), g(t) \in P[t]$, so $f(t) + g(t) = 0$. Thus we have $c_{n}t^{n} + \ldots + c_{1}t + c_{0} = 0$ with $c_{i} \in P$ and $c_{0} \ne 0$, so $c_{0}^{-1} \in P$ implies that

$$\begin{aligned} -1 = c_{0}^{-1}c_{n}t^{n} + \ldots + c_{0}^{-1}c_{1}t \in P[t] \subseteq P_{1}, \end{aligned}$$

a contradiction. Therefore P is a maximal preprime with $1 \in P$, that is, P is an infinite prime.

To see that $E_{P}$ is subfield of R, take $0 \ne t \in E_{P}$. If $t \in P$, then $t^{-1} \in P$. Assume that $t \not \in P$. Define P[t] as above. If $P[t] \cap (-P[t]) = \{0\}$ then P[t] is a partial order on R that contains P, so $P = P[t]$ and hence $t \in P$, a contradiction. Therefore, $P[t] \cap (-P[t]) \ne \{0\}$. So the same argument as in the previous paragraph gives

$$\begin{aligned} 1 = -(c_{0}^{-1}c_{n}t^{n - 1} + \cdots + c_{0}^{-1}c_{1})t, \end{aligned}$$

where $c_{0} \ne 0, c_{1}, \ldots , c_{n} \in P$ and $n \ge 1$. So $t^{-1} = -(c_{0}^{-1}c_{n}t^{n - 1} + \cdots + c_{0}^{-1}c_{1}) \in E_{P}$, and hence $E_{P}$ is a subfield of F.

(2) By Lemma 4, $J_{P} = \{0\}$, so P is strong Archimedean [7, Corollary 1.4]. $\square$

For a number field F and a maximal partial order P on F, P is a strong Archimedean infinite prime divisor of F [6, 9]. Thus if P is directed, $P = \delta ^{-1}(\mathbb {R}^{+})$ for a real embedding $\sigma$ of F, and if P is not directed, $P = \sigma ^{-1}(\mathbb {R}^{+})$ for a complex embedding $\delta$ on F by Theorem 1. We generalize this result to Archimedean maximal partial orders on an arbitrary field of characteristic 0.

Corollary 4

Let R be a field of characteristic 0 and suppose that P is a maximal partial order on R that is Archimedean.

• (1) If P is directed, then there exists an embedding $\sigma$ from R to $\mathbb {R}$ such that $P = \sigma ^{-1}(\mathbb {R}^{+})$. In particular, P is a total order on R.

• (2) If P is not directed, then there exists an embedding $\delta$ from R to $\mathbb {C}$ such that $P = \delta ^{-1}(\mathbb {R}^{+})$ and $\delta (R) \not \subseteq \mathbb {R}$.

Proof

(1) By Theorem 6, P is a strong Archimedean infinite prime of R. Since P is directed, $R = P - P$. It follows from [6, Proposition 1.7] or [2, 4.9] that there exists an embedding $\sigma : R \rightarrow \mathbb {R}$ such that $P = \sigma ^{-1}(\mathbb {R}^{+})$, so P is a total order on R.

(2) By [7, Corollary 1.4], there is an embedding $\delta$ from R to $\mathbb {C}$ such that $P = \delta ^{-1}(\mathbb {R}^{+})$. Since P is not directed, $\delta (R) \not \subseteq \mathbb {R}$. $\square$

Example 3

A directed maximal partial order may not be a total order if it is not Archimedean. For instance, let $\mathbb {R}$ be equipped with a non-Archimedean total order $\le$. So there exists $z \in \mathbb {R}$ such that $n \le z$ for all positive integers n. Define the positive cone on $\mathbb {C}$ as follows.

$$\begin{aligned} P = \{a + bi \ | \ 0 \le a \text{ and } n|b| \le a \text{ for } \text{ all } n > 0, a, b \in \mathbb {R}\}, \end{aligned}$$

where $|b| = b$ if $b \ge 0$ in $\mathbb {R}$ and $|b| = -b$ if $b < 0$ in $\mathbb {R}$. Then P is a directed partial order that is not Archimedean [10, Theorem 1]. This maximal partial order is not a total order because $\mathbb {C}$ cannot have a total order with respect to which it is a totally ordered field.