The number fields that are ${O}^{*}$-fields

Abstract

Using the theory on infinite primes of fields developed by Harrison in [2], the necessary and sufficient conditions are proved for real number fields to be $O^{*}$-fields, and many examples of $O^{*}$-fields are provided.

1 Introduction

Let R be an associative ring. If a partial order on R is contained in a total order on R, then we say that the partial order is extended to a total order. A ring is called an $O^{*}$-ring if each partial order is extended to a total order. This concept was introduced by Fuchs in [1], and as an open question in [1], he asked establishing ring-theoretical properties of $O^{*}$-rings. In [5], Steinberg provided the ring-theoretical characterization of the $O^{*}$-rings, and from his characterizations, the question becomes determining the $O^{*}$-fields [5, Theorem]. An $O^{*}$-field must be a subfield of $\mathbb {R}$, the field of real numbers, and must be algebraic over $\mathbb {Q}$, the field of rational numbers. Steinberg also pointed out that it is sufficient to determine which subfields of $\mathbb {R}$ that is a finite extension of $\mathbb {Q}$ are $O^{*}$-fields [5, p. 2557]. He proved that each real quadratic extension field of $\mathbb {Q}$ is an $O^{*}$-field, and $\mathbb {Q}[\root 4 \of {2}]$ is not an $O^{*}$-field. In [4], it is shown that a sequence of real quadratic extension of $\mathbb {Q}$ is an $O^{*}$-field and $\mathbb {Q}[\root 3 \of {n}]$, where n is a cubic-free positive integer, is an $O^{*}$-field. No other results on $O^{*}$-fields seem available in the literature.

In the present paper, by using the theory on the infinite primes of the number fields developed by Harrison in [2], the connection between the maximal partial orders on a number field R and the subfields of R is established, then necessary and sufficient conditions are obtained for a real number field being $O^{*}$. As a consequence, we show that, for instance, for a real number field R if $[R : \mathbb {Q}]$ is an odd integer, then R is $O^{*}$, where $[R : \mathbb {Q}]$ denotes the dimension of R over $\mathbb {Q}$.

We review a few definitions from partially ordered rings. Let R be an associative ring and P be a subset of R. Then P is said to be the positive cone of a partial order if $P + P \subseteq P$, $PP \subseteq P$, and $P \cap -P = \{0\}$. We often just call P as a partial order on R. In this paper, we only consider partial orders on fields. By Zorn’s Lemma, each partial order is contained in a maximal partial order. We denote by $\mathbb {Q}^{+}$ and $\mathbb {R}^{+}$ the positive cone of the usual total order on $\mathbb {Q}$ and $\mathbb {R}$, respectively. For a maximal partial order P, $\mathbb {Q}^{+} \subseteq P$ and P is division closed in the sense that for any two elements a, b in the field, if $ab > 0$ and $a > 0$, then $b > 0$.

For more information and undefined terminologies on partially ordered rings, the reader is referred to [1, 3].

2 Infinite primes for fields

In this section, we collect some results from [2] that will be used in the following section. Let R be a field. A nonempty subset S is called a preprime if S is closed under the addition and multiplication in R, and $-1 \not \in S$, that is,

$$\begin{aligned} S + S \subseteq S, \ SS \subseteq S, \text{ and } -1 \not \in S. \end{aligned}$$

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 $R = S - S = \{a - b \mid a, b \in S\}$.

Let R be a number field that is n-dimensional over $\mathbb {Q}$. There exist exactly n isomorphisms (or embeddings) $\sigma _{1}, \sigma _{2}, \dots , \sigma _{n}$ of R into $\mathbb {C}$, the field of complex numbers. 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 $r + 2s = n$. Then

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

are these isomorphisms, and $\sigma _{1}, \dots , \sigma _{r}$ are called the real infinite prime divisors of R, and the sets $\{\sigma _{r + 1}, \rho \circ \sigma _{r + 1}\}, \dots , \{\sigma _{r + s}, \rho \circ \sigma _{r + s}\}$ are called the complex infinite prime divisors of R.

Theorem 2.1

(1):

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

(2):

([2, 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.

In the present paper, we only consider the real number fields which are subfields of $\mathbb {R}$ and finite dimensional over $\mathbb {Q}$. Let R be a real number field. For a subset H of R, define $E_{H} = H - H = \{a - b \mid a, b \in H\}$. The following result shows that in a real number field, infinite primes are precisely maximal partial orders.

Lemma 2.2

Let R be a real number field.

(1):

Let H be subset of R such that $H + H \subseteq H$, $HH \subseteq H$, and $\mathbb {Q}^{+} \subseteq H$. Then $E_{H}$ is a subfield of R.

(2):

If S is an infinite prime on R, then S is a maximal partial order on R.

(3):

If P is a maximal partial order on R, then P is an infinite prime on R.

Proof

(1) It is clear that $E_{H}$ is a subring of R and a subspace of R over $\mathbb {Q}$. Take $0 \ne a \in E_{H}$, define $f_{a}: E_{H} \rightarrow E_{H}$ by $f_{a}(u) = au$ for any $u \in E_{H}$. Then $f_{a}$ is a linear transformation from $E_{H}$ to $E_{H}$ over $\mathbb {Q}$. Since $f_{a}$ is one-to-one and $E_{H}$ is finite-dimensional over $\mathbb {Q}$, $f_{a}$ is onto, so there exists $b \in E_{H}$ such that $ab = 1$. Thus $E_{H}$ is a subfield of R.

(2) Let S be an infinite prime on R. From the proof of [2, Proposition 3.5], we have $S \cap -S = \{0\}$, so S is a partial order on R. Assume $S \subseteq P$, where P is a maximal partial order on R. If $-1 \in P$, then $1 \in P \cap -P = \{0\}$, a contradiction. Thus $-1 \not \in P$ and P is a preprime. Since S is a maximal preprime and P is a preprime, $S \subseteq P$ implies that $S = P$ and hence S is a maximal partial order on R.

(3) If P is a maximal partial order, then P is a preprime. Let $P \subseteq S$ which is an infinite prime. By (2), S is a maximal partial order, so $P = S$ is an infinite prime. $\square$

Since in a real number field, the maximal partial orders and the infinite primes are the same, we freely use both names.

3 $O^{*}$-fields

For a field K and a subfield F of K, [K : F] denotes the dimension of K as an F-vector space. A partial order on a real number field R is called directed if each element is a difference of two positive elements. For an embedding $\sigma$ from R to $\mathbb {C}$, it is clear that $\sigma ^{-1}(\mathbb {R}^{+})$ is a partial order on R.

Theorem 3.1

Let R be a real number field. Then following conditions are equivalent.

(1):

R is an $O^{*}$-field,

(2):

For each complex infinite prime divisor $(\sigma , \rho \circ \sigma )$ of R, $\sigma ^{-1}(\mathbb {R}^{+})$ is extended to a total order on R,

(3):

For each maximal partial order P on R, $E_{P} = R$,

(4):

Each maximal partial order P on R is directed,

(5):

Each maximal partial order P on R contains $[R : \mathbb {Q}]$ linearly independent vectors.

Proof

$(1) \Rightarrow (2)$ is clear.

$(2) \Rightarrow (1)$ Let P be a maximal partial order on R. We show that P must be a total order. Since P is an infinite prime on R by Lemma 2.2, Theorem 2.1 implies that $P = \sigma ^{-1}(\mathbb {R}^{+})$, where $\sigma$ is a real or complex infinite prime divisor of R, then P must be a total order by the assumption.

$(1) \Rightarrow (3)$ Assume that there exists a maximal partial order P on R such that $E_{P} \ne R$. Take $z \in R \setminus E_{P}$. Then $z \not \in P$ and $-z \not \in P$, so P is not a total order and R is not $O^{*}$.

$(3) \Rightarrow (1)$ Let P be a maximal partial order on R. We show that P must be a total order. By Lemma 2.2, P is an infinite prime. Since $E_{P} = P - P$ and $E_{P} = R$, $R = P - P$, that is P is a full infinite prime. By Theorem 2.1, $P = \sigma ^{-1}(\mathbb {R}^{+})$ for some real infinite prime divisor $\sigma$ of R. Thus P is a total order on R.

$(3) \Rightarrow (4)$ Let P be a maximal partial order. Then $R = P - P$, so P is a directed partial order.

$(4) \Rightarrow (5)$ Let P be a maximal partial order on R, since P is directed, $R = P - P$, so P is a generating set of R as a vector space over $\mathbb {Q}$. Thus P contains $[R : \mathbb {Q}]$ linearly independent vectors.

$(5) \Rightarrow (3)$ Let P be a maximal partial order on R. Since P contains $[R : \mathbb {Q}]$ linearly independent vectors, $P - P$ is a subspace of $[R : \mathbb {Q}]$-dimensional, and hence $R = P - P$. $\square$

The following result is an immediate consequences of Theorem 3.1.

Theorem 3.2

Let R be a real number field. If $[R : \mathbb {Q}] = n$ is an odd integer, then R is an $O^{*}$-field.

Proof

Let P be a maximal partial order on R and $[E_{P} : \mathbb {Q}] = k$. Then $k \ | \ n$. Since P is a full infinite prime in $E_{P}$, there exists a real infinite prime divisor $\sigma$ of $E_{P}$ such that $P = \sigma ^{-1}(\mathbb {R}^{+})$. It is well-known that $\sigma$ can be extended to $[R : E_{P}] = n / k$ embeddings from R to $\mathbb {C}$, and since n/k is odd, one of those n/k embeddings must be a real infinite prime divisor of R, denoted by $\delta$. Then $P = \sigma ^{-1}(\mathbb {R}^{+}) \subseteq \delta ^{-1}(\mathbb {R}^{+})$, and hence $P = \delta ^{-1}(\mathbb {R}^{+})$ since P is a maximal partial order on R. Thus R is $O^{*}$. $\square$

There exist real number fields R in which $E_{P} = R$ is not true for all the maximal partial orders P, and hence R is not an $O^{*}$-field. The following example is due to Steinberg.

Example 3.3

Let $R = \mathbb {Q}[a]$ with $a = \root 4 \of {2}$. The irreducible polynomial $f(x) = x^{4} - 2$ has four roots: $a, -a, ia, -ia$, where $i^{2} = -1$. Let $\sigma$ be the embedding from R to $\mathbb {C}$ that sends a to ia. Then $-a^{2} \in \sigma ^{-1}(\mathbb {R}^{+})$. If $\sigma ^{-1}(\mathbb {R}^{+}) = P$ is not a maximal partial order, then $P \subsetneq P_{1}$ for some maximal partial order $P_{1}$. If $P_{1}$ is not a full infinite prime on R, then by Theorem 2.1, there exists a complex infinite prime divisor $\{\gamma , \rho \circ \gamma \}$ such that $P_{1} = \gamma ^{-1}(\mathbb {R}^{+})$. However $\{\sigma , \rho \circ \sigma \}$ is the only complex infinite prime divisor of R, so $P = P_{1}$, a contradiction. Hence we must have $E_{P_{1}} = R$ and $P_{1} = \delta ^{-1}(\mathbb {R}^{+})$ for some real infinite prime divisor $\delta$ of R. Then $\delta$ is either the identity mapping or the embedding that sends a to $-a$. In either case, $a^{2} \in P_{1}$. On the other hand, $-a^{2} \in P \subsetneq P_{1}$, so $a^{2} \in P_{1} \cap -P_{1}$, a contradiction. Thus P is a maximal partial order on R. It follows that R is not $O^{*}$ by Theorem 3.1.

Similar to Steinberg’s example, if n is a positive integer divisible by 4, then $\mathbb {Q}[\root n \of {2}]$ is not an $O^{*}$-field since the partial ordered $P = \mathbb {Q}^{+} + \mathbb {Q}^{+}(-a^{\frac{n}{2}})$, where $a = \root n \of {2}$, cannot be extended to a total order.

Let R be a real number field with $[R : \mathbb {Q}] = 4$. It is possible for R to be an $O^{*}$-field as shown in Example 3.6.

Theorem 3.4

If $R = \mathbb {Q}[\root 2n \of {p}]$, where p is a prime number and n is a odd integer, then R is $O^{*}$.

Proof

Let P be a maximal partial order on R and $a = \root 2n \of {p}$, $b = a^{n} = \sqrt{p}$. Then $b^{2} = p$. We first show that either $b \in P$ or $-b \in P$. Assume $-b \not \in P$. Define $P' = P + Pb$. It is clear that $P' + P' \subseteq P'$ and $P'P' \subseteq P'$. Suppose that $P' \cap -P' \ne \{0\}$. Let $0 \ne w \in P' \cap -P'$. Then $w = \alpha + \beta b$ and $-w = \alpha ' + \beta ' b$, where $\alpha , \alpha ', \beta , \beta ' \in P$. Then $(\alpha + \alpha ') + (\beta + \beta ')b = 0$ and $w \ne 0$ implies $\beta + \beta ' \ne 0$ and hence $-b = (\beta + \beta ')^{-1}(\alpha + \alpha ') \in P$ since P is division closed, a contradiction. Thus $P' \cap -P' = \{0\}$, so $P'$ is a partial order. It follows that $P = P'$ and $b \in P$. Thus either $b \in P$ or $-b \in P$. Without loss of generality, we may assume $b \in P$.

Since $\mathbb {Q}[b]$ is a subfield of R and $\mathbb {Q}[b] \subseteq E_{P}$, $2 \mid [E_{P} : \mathbb {Q}]$, so $[E_{P} : \mathbb {Q}] = 2k$, where k is a positive integer and $k \ | \ n$. It follows that $[R : E_{P}] = n / k$ is an odd integer. By a similar argument of Theorem 3.2, each real infinite prime divisor of $E_{P}$ is extended to a real infinite prime divisor of R, so P must be a total order. $\square$

In the following we collect some conditions that make a real number field being $O^{*}$. Let R be a real number field. By Primitive Element Theorem, R is a simple extension over $\mathbb {Q}$, that is, $R = \mathbb {Q}[a]$ for some $a \in R$.

Theorem 3.5

Let R be a real number field.

(1):

Suppose that $R = \mathbb {Q}[a]$. Let $f(x) \in \mathbb {Q}[x]$ be the irreducible polynomial such that $f(a) = 0$. If each root of f(x) is a real number, then R is an $O^{*}$-field.

(2):

Let R be a real number field. If for each complex infinite prime divisor $\{\sigma , \rho \circ \sigma \}$, $\sigma ^{-1}(\mathbb {R}^{+})$ is not prime, then R is an $O^{*}$-field.

(3):

Let R be a real number field and $R = \mathbb {Q}[a]$ with the minimal polynomial f(x) of a over $\mathbb {Q}$. If f(x) has a pure imaginary root, then R is not an $O^{*}$-field.

Proof

(1) Let P be a maximal partial order on R. If P is not a full infinite prime, then, by Theorem 2.1, $P = \sigma ^{-1}(\mathbb {R}^{+})$ for a complex infinite prime divisor $\{\sigma , \rho \circ \sigma \}$, a contradiction since the minimal polynomial f(x) of a has only real roots. Thus P is full and $E_{P} = R$.

(2) Let P be a maximal partial order on R. Then P is an infinite prime by Lemma 2.2. So, by Theorem 2.1, $P = \sigma ^{-1}(\mathbb {R}^{+})$ for some real infinite prime $\sigma$ by the assumption, so P is a total order on R.

(3) Suppose that $z = ib, b \in \mathbb {R}$ is a root of f(x). Then there exists an embedding $\sigma$ from R to $\mathbb {C}$ that sends a to ib. Let $P = \sigma ^{-1}(\mathbb {R}^{+})$. Then P is a partial order on R. Since $\sigma (-a^{2}) = -(ib)^{2} = b^{2} \in \mathbb {R}^{+}$, $-a^{2} \in P$, so P cannot be extended to a total order and R is not $O^{*}$. $\square$

As an application of Theorem 3.5, we determine all the $O^{*}$-fields R with $[R : \mathbb {Q}] = 4$.

Example 3.6

Let R be a real number field with $[R : \mathbb {Q}] = 4$. Assume that $R = \mathbb {Q}[\alpha ]$, where $\alpha \in \mathbb {R}$ is a root of an irreducible polynomial f(x) of degree 4 over $\mathbb {Q}$. We consider following cases.

(1):

f(x) has 4 real roots. By Theorem 3.5(1), R is $O^{*}$.

(2):

f(x) has 2 real roots and 2 pure imaginary roots. By Theorem 3.5(3), R is not $O^{*}$.

(3):

f(x) has 2 real roots and 2 complex roots with nonzero real part. Let $\{\sigma , \rho \circ \sigma \}$ be the complex infinite prime divisor of R. Let $E = \sigma ^{-1}(\mathbb {R})$. Then E is a subfield of R, so $E = R$, $[E : \mathbb {Q}] = 2$, or $E = \mathbb {Q}$. Let r be a real root of f(x) such that $\sigma (r)$ is a complex root of f(x). Then $r, -r \not \in E$, so $E \ne R$. If $E = \mathbb {Q}$, then $\sigma ^{-1}(\mathbb {R}^{+}) = \mathbb {Q}^{+}$ is not a prime. Assume that $[E : \mathbb {Q}] = 2$. Then $E = \mathbb {Q}[\beta ]$ for some $\beta \in \mathbb {R}$. Then $\sigma ^{-1}(\mathbb {R}^{+}) = E \cap \mathbb {R}^{+}$ or $E \cap -\mathbb {R}^{+}$, and hence $\sigma ^{-1}(\mathbb {R}^{+}) \subsetneq R \cap \mathbb {R}^{+}$ or $R \cap -\mathbb {R}^{+}$, respectively. Thus $\sigma ^{-1}(\mathbb {R}^{+})$ is not a prime. By Theorem 3.5(2), R is $O^{*}$.