All rings in the paper are associative, with 1 ≠ 0 , and of characteristic 0. A partially ordered ring ( R , ≥ ) is a ring R equipped with a partial order ≥ that satisfies

The positive cone of the partially ordered ring ( R , ≥ ) is defined as R + = { a ∈ R | a ≥ 0 } . Clearly R + is closed under the addition and multiplication of R , and R + ∩ − R + = { 0 } . The elements in R + are called positive. On the other hand, let R be a ring and P be a subset of R that satisfies P + P ⊆ P , P P ⊆ P , and P ∩ − P = { 0 } . Then the partial order defined by for all x , y ∈ R , y ≥ x if y − x ∈ P makes R into a partially ordered ring ( R , ≥ ) with the positive cone P . Because of this connection, we will denote a partially ordered ring either by ( R , ≥ ) , where ≥ is a partial order, or by ( R , P ) , where P is the positive cone of a partial order, and we will say that ≥ or P is a partial order on the ring R . A partial order P on a ring R is called directed if for any a ∈ R , there exist b , c ∈ P such that a = b − c . If the partial order P on R is a lattice order, then ( R , P ) is called a lattice-ordered ring ( ℓ -ring). A partial order ≥ is called division closed if a b > 0 (or b a > 0 ) and a > 0 implies b > 0 . In the following, ℤ + , ℚ + , and ℝ + denote the usual total order on the ring ℤ of integers, the field ℚ of rational numbers, and the field ℝ of real numbers, respectively. For more information on partially ordered rings and ℓ -rings, the reader is referred to [1]-[4].

A nonempty subset S of a ring R is called a preprime if S + S ⊆ S , S S ⊆ S , and − 1 ∉ 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 ∈ S , otherwise S is called finite. An infinite prime S of R is called full if R = S − S = { a − b | a , b ∈ S } . An infinite prime S (or a partial order P ) is called strong Archimedean if for any a ∈ S ( a ∈ P ), there exists a positive integer n such that n − a ∈ S ( n − a ∈ P ). A partially ordered ring ( R , ≥ ) is called Archimedean if for any a , b ∈ R , a ≥ n b for all integers n implies that b = 0 . For more information on primes for rings, the reader is referred to [5][6].

In ([7], Corollary 4), it is shown that an Archimedean directed maximal partial order on a field must be a total order. In this section, the result is generalized to integral domains. Let D be an integral domain and P be a partial order on D with 1 ∈ P . Since P \ { 0 } is a multiplicative subset of D , we may form the quotient ring of D by P \ { 0 } , denoted by P − 1 D . Thus

which is an integral domain with the same identity element as D , and D may be considered as a subring of P − 1 D .

Define

Lemma 1. Let D be an integral domain and 1 ∈ P be a partial order on D . Then P q is a partial order on P − 1 D and P ⊆ P q . Moreover, if P is directed on D , then P q is directed on P − 1 D .

Proof. It is clear that P q is closed under the addition and multiplication of P − 1 D . Let x ∈ P q ∩ − P q . Then x = a / b , − x = c / d for some a , b , c , d ∈ P , b ≠ 0 and d ≠ 0 . Then 0 = x + ( − x ) = a / b + c / d = a d + b c / b d implies that a d + b c = 0 , so a = c = 0 . Therefore x = 0 , so P q ∩ − P q = { 0 } and hence P q is a partial order on P − 1 D . For any a ∈ P , in P − 1 D , a = a / 1 ∈ P q , so P ⊆ P q .

Let P be a directed partial order on D . For x ∈ P − 1 D , x = a / b for some a ∈ D and b ∈ P , b ≠ 0 . Since P is directed on D , a = a 1 − a 2 for some a 1 , a 2 ∈ P . Then x = a 1 / b − a 2 / b with a 1 / b , a 2 / b ∈ P q . Therefore, P q is directed on P − 1 D . □

Lemma 2. For an integral domain D and a maximal partial order P on D , P q is a maximal partial order on P − 1 D such that D ∩ P q = P . Moreover, If P is Archimedean on D , then P q is Archimedean on P − 1 D .

Proof. By Lemma 1, P q is a partial order on P − 1 D and by Zorn’s Lemma, P q ⊆ P 1 and P 1 is a maximal partial order on P − 1 D . For x ∈ P 1 , x = a / b for some a ∈ D and b ∈ P , b ≠ 0 . Since b ∈ P ⊆ P q ⊆ P 1 , a = x b ∈ P 1 and hence a ∈ D ∩ P 1 that is a partial order on D . On the other hand, since P ⊆ D ∩ P 1 and P is a maximal partial order on D , we have P = D ∩ P 1 and hence a ∈ P . Hence x ∈ P q . It follows that P q = P 1 is a maximal partial order on P − 1 D and D ∩ P q = P .

Let P be Archimedean on D . We show that P q is Archimedean on P − 1 D . Suppose that x , y ∈ P − 1 D such that n x ≤ P q y for all integer n . Let x = a / b and y = c / d for some a , c ∈ D and b , d ∈ P with b ≠ 0 , d ≠ 0 . Since P ⊆ P q and b d ∈ P , n ( a d ) = n x ( b d ) ≤ P q y ( b d ) = c b for all integers n . From c b − n ( a d ) ∈ D ∩ P q = P for all integers n , we have n ( a d ) ≤ P c b for all integers n . Thus, a d = 0 since P is Archimedean on D , so a = 0 and x = 0 . Therefore, P q is Archimedean on P − 1 D . □

Lemma 3. Let D be an integral domain and P be a maximal partial order on D such that any nonzero element in P is a unit of D . Then any subring of D containing P is a field. In particular, D is a field.

Proof. Let R be a subring of D and P ⊆ R . Take a ∈ R and a ≠ 0 . If a ∈ P , then a is a unit by the assumption. Assume that a ∉ P . Denote by P [ a ] the set of polynomials in a with coefficients in P . It is clear that P [ a ] is closed under the polynomial addition and multiplication. If P [ a ] ∩ − P [ a ] = { 0 } , then P [ a ] is a partial order on D , so P ⊆ P [ a ] and P is maximal implies P = P [ a ] . Hence a ∈ P , a contradiction.

Since P [ a ] ∩ − P [ a ] ≠ { 0 } , there exists 0 ≠ w ∈ P [ a ] ∩ − P [ a ] . Then w = f ( a ) and − w = g ( a ) for some nonzero polynomials f ( a ) , g ( a ) with coefficients in P . Therefore, f ( a ) + g ( a ) = w − w = 0 implies

Since 0 ≠ a 0 ∈ P , a 0 − 1 exists and in P since P is division closed ([8], Lemma 1), so

Thus a − 1 = − a 0 − 1 ( a n a n − 1 + ⋯ + a 1 ) ∈ R and hence R is a field. □

Theorem 1. Let D be an integral domain and P be an Archimedean maximal partial order on D .

1) If P is directed, then there exists an embedding φ : D → ℝ such that P = φ − 1 ( ℝ + ) . In particular, P is a total order on D .

2) If P is not directed, then there exists an embedding ϕ : D → ℂ such that P = ϕ − 1 ( ℝ + ) and ϕ ( D ) ⊆ ℝ .

Proof. By Lemma 2, P q is an Archimedean maximal partial order on P − 1 D . Since each nonzero element in P q is a unit in P − 1 D and P − 1 D is an integral domain, by Lemma 3, P − 1 D is a field.

(1) By Lemma 1, P q is directed on P − 1 D . From ([7], Corollary 4), there exists an embedding σ : P − 1 D → ℝ such that P q = σ − 1 ( ℝ + ) . Define φ = σ | D , the restriction of σ on D . Then φ is an embedding from D to ℝ . For a ∈ D ,

so P = φ − 1 ( ℝ + ) . For a ∈ D , φ ( a ) is either in ℝ + or in − ℝ + , so a ∈ P or − a ∈ P , that is, P is a total order.

(2) If P is not directed, then P q is not directed. Otherwise, by ([7], Corollary 4), P q is a total order on P − 1 D . Since P = D ∩ P q , P is a total order on D as well, so P is directed, a contradiction.

By ([7], Corollary 4), there exists an embedding δ : P − 1 D → ℂ such that P q = δ − 1 ( ℝ + ) and δ ( P − 1 D ) ⊆ ℝ . Define ϕ = δ | D , the restriction of δ on D . Then ϕ is an embedding from D to ℂ . For a ∈ D ,

so P = ϕ − 1 ( ℝ + ) . If ϕ ( D ) ⊆ ℝ , then P = ϕ − 1 ( ℝ + ) is a total order, a contradiction. Therefore, ϕ ( D ) ⊆ ℝ . □

In 2011, Schwartz and Yang proved that for a field if it has transcendence degree d ≥ 1 , then it has a directed partial order ([9], Corollary 4.2). The result was proved by Dubois in 1970 using the language of infinite primes for fields ([10], 4.12). In this section, based on Dubois’ work, we present a direct proof of the result using maximal partial orders. All credit goes to Dubois.

Let F be a field and P be a maximal partial order. Define

Then B P is a subring of F and J P is an ideal of B P ([10], 6). Clearly B P ⊆ E P , E P is a subring of F , and P is directed on F if and only if E P = F .

Lemma 4. Let F be a field and P be a maximal partial order on F .

1) Any subring of F containing P is a subfield of F . In particular, E P is a subfield of F .

2) P is an infinite prime of F .

3) F is algebraic over E P .

4) E P is the quotient field of B P in F .

Proof. (1) Let a ∈ P , a ≠ 0 . Then a a − 1 = 1 and P is division closed implies a − 1 ∈ P . By Lemma 3, (1) is true.

(2) Since P is a preprime and 1 ∈ P , P ⊆ S for some infinite prime S of F . By (1), the subring S − S is a subfield of F , so S ∩ − S = { 0 } since S ∩ − S is an ideal of S − S ([6], Proposition 2.5). Therefore, S is a partial order and hence P = S and P is an infinite prime.

(3) Suppose that a ∈ F \ E P . Denote by P [ a ] the set of all polynomials in a with coefficients in P . Clearly P [ a ] is closed under the addition and multiplication of F . Since P ⊆ P [ a ] and a ∉ P , P [ a ] is not a preprime by (2). So, − 1 ∈ P [ a ] and hence a satisfies a nonzero polynomial over E P . Therefore, F is algebraic over E P .

(4) Let a ∈ B P . Then for some positive integer n , n − a = b ∈ P , so a = n − b ∈ E P . Thus, B P ⊆ E P and hence the quotient field of B P is contained in E P . For 0 ≠ b ∈ P , 1 + b ∈ P , so ( 1 + b ) − 1 ∈ P and hence 1 + ( 1 + b ) − 1 ∈ P . Also ( 1 − ( 1 + b ) − 1 ) ( 1 + b ) = b ∈ P implies that 1 − ( 1 + b ) − 1 ∈ P since P is division closed. Thus, ( 1 + b ) − 1 ∈ B P . Similarly,

imply that 1 − b ( 1 + b ) − 1 ∈ P . Thus b ( 1 + b ) − 1 ∈ B P . Therefore, b = b ( 1 + b ) − 1 ( 1 + b ) − 1 is in the quotient field of B P , so E P is contained in the quotient field of B P . Hence E P is the quotient field of B P . □

Theorem 2. Let F be a field.

1) For a maximal partial order P on F , if J P ≠ { 0 } , then P is directed.

2) (Dubois, Schwartz, Yang) If the transcendence degree of F over ℚ is d ≥ 1 . Then F contains directed partial orders.

Proof. (1) Suppose that E P ≠ F . We derive a contradiction. Since F is algebraic over E P and E P is the quotient field of B P by Lemma 4, we are able to choose an element θ ∈ F \ E P such that θ satisfies an irreducible monic polynomial f ( x ) = x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0 over B P with n ≥ 2 . Define

We show that P 1 is a preprime. Let g ( θ ) = p + p z 1 θ + ⋯ + p z n − 1 θ n − 1 ∈ P 1 \ { 0 } and h ( θ ) = p ′ + p ′ z ′ 1 θ + ⋯ + p ′ z ′ n − 1 θ n − 1 ∈ P 1 \ { 0 } . Then

Since p + p ′ ∈ P , 1 p + p ′ ∈ P and since 1 − p p + p ′ = p ′ p + p ′ ∈ P , p p + p ′ ∈ B P . Similarly, p ′ p + p ′ ∈ B P . Thus

since J P is an ideal of B P . Hence

is in P 1 , so P 1 is closed under the addition of F .

To show that g ( θ ) h ( θ ) ∈ P 1 , we first notice that since θ n = − ( a n − 1 θ n − 1 + ⋯ + a 1 θ + a 0 ) with a i ∈ B P and J P is an ideal of B P , we have

where b 1 , ⋯ , b n − 1 ∈ J P , and b 0 = 1 + b for some b ∈ J P . Since − 1 n ≤ P b ≤ P 1 n for all positive integers n , 1 + b > P 0 and since 2 − 1 1 + b = 1 + 2 b 1 + b ≥ P 0 , 1 1 + b ∈ B P . Thus 1 1 + b b 1 , ⋯ , 1 1 + b b n − 1 are all in J P , so

is in P 1 . Therefore P 1 is closed under the multiplication of F . It is clear that − 1 ∉ P 1 , so P 1 is a preprime. Since P is an infinite prime of F and P ⊆ P 1 , we have P = P 1 .

Let z ∈ J P and z ≠ 0 . Then 1 + z θ ∈ P 1 = P , so z θ ∈ E P and hence θ = z θ z ∈ E P by Lemma 4(4), a contradiction. Hence we must have E P = F , that is, P is a directed partial order on F .

(2) Let X be a transcendental basis of F over ℚ . Then | X | = d ≥ 1 . Take t ∈ X and consider the polynomial ring ℚ [ t ] . Totally order ℚ [ t ] by defining a nonzero polynomial positive if the coefficient of the term of lowest power is a positive rational number. Denote by T the positive cone of the total order on ℚ [ t ] . We note that 1 + n t ∈ T for all integers n . Since T is a partial order on F , T is contained in a maximal partial order P on F . Then t ∈ J P , so J P ≠ { 0 } and by (1), P is directed. □

Remark. Dubois actually proved that there exist at least 2 d directed partial orders on the field F ([10], 4.12).

A partially ordered field ( F , ≥ ) is said satisfying the property (AC): if for any x ∈ F , x + 1 n ≥ 0 for all positive integers n , then x ≥ 0 . The property (AC) was introduced by Dubois [5].

Theorem 3. Let F be a field and P be a directed maximal partial order on F . If P satisfies the property (AC), then P is an Archimedean total order on F .

Proof. Since P satisfies the property (AC),

By ([11], Corollary 1.4), there exists an embedding ϕ : F → ℂ such that P = ϕ − 1 ( ℝ + ) . Since P is directed, P must be a total order. □

A field is called directed O * if each directed partial order can be extended to a total order [12]. A field is directed O * if and only if it is algebraic over ℚ as shown in the following result.

Corollary 1. For a field F , the following are equivalent.

1) F is algebraic over ℚ ;

2) Each directed maximal partial order on F satisfies the property (AC);

3) F is directed O * .

Proof. (1) ⇒ (2) By ([8], Theorem 2), if F is algebraic over ℚ , then each directed maximal partial order is an Archimedean total order. Suppose that x + 1 n ≥ 0 for all positive integers n and x < 0 . Then m ( − x ) ≤ 1 for all integers m , so − x = 0 , a contradiction. Therefore, x ≥ 0 .

(2) ⇒ (3) By Theorem 3.

(3) ⇒ (1) Assume that F is not algebraic over ℚ . Take x ∈ F that is not algebraic over ℚ and form the polynomial ring A = ℤ [ x ] . Define the partial order on A as follows.

Clearly P A is a partial order on A , so a partial order on F . Then P A is extended to a maximal partial order P m on F . Let f ( x ) = 1 + x 2 ∈ ℤ [ x ] . Then for any integer n , 1 + n f ( x ) ∈ P A since 1 + n f ( i ) = 1 ∈ ℤ + . By Theorem 2, P m is a directed maximal partial order since f ( x ) ∈ J P m . Let g ( x ) = − x 2 . Since g ( i ) = 1 ∈ ℤ + , g ( x ) ∈ P A ⊆ P m . Therefore, x 2 < P m 0 , so P m cannot be a total order, a contradiction. □

For an integral domain that is algebraic over ℤ , maximal partial orders and infinite primes are identical ([13], Lemma 2). In fact, that R is algebraic over ℤ is also a necessary condition for maximal partial orders and infinite primes to be identical.

Lemma 5. Let D be an integral domain. If each infinite prime of D is a partial order on D , then D is algebraic over ℤ .

Proof. Suppose that D is not algebraic over ℤ . Take a ∈ D such that a is not algebraic over ℤ . Consider the polynomial ring ℤ [ a ] ⊆ D . Define

Clearly, S is a preprime and 1 ∈ S , so S ⊆ S 1 for an infinite prime S 1 of D . Now a , − a ∈ S ⊆ S 1 implies that a ∈ S 1 ∩ − S 1 , so S 1 ∩ − S 1 ≠ { 0 } and S 1 is not a partial order, a contradiction. □

The following result is a direct consequence of Lemma 5 and ([13], Lemma 2).

Corollary 2. Let D be an integral domain. The maximal partial orders on D and infinite primes of D are identical if and only if D is algebraic over ℤ .

If a commutative ring contains nonzero zero divisors, maximal partial orders and infinite primes are different. In this section, we study the connection between symmetric partial orders, maximal partial orders and full infinite primes for commutative rings that are algebraic over ℤ .

A partial order P on a ring (may not be commutative) is called symmetric if it is directed, division closed, and for any two elements a , b , 0 ≠ a + b ∈ P implies either a ∈ P or b ∈ P . This concept was introduced in [14] and called symmetric cone there. For commutative rings that are algebraic over ℤ , full infinite primes and aymmetric partial orders are associated together as shown in Theorem 5 and a commutative ring with a symmetric partial order has a nice decomposition as shown in Theorem 4.

Let R be a commutative ring and P be a symmetric partial order on R . Define

It is straightforward to check that an element a is in I P if and only if for any 0 ≠ b ∈ P , − b ≤ P n a ≤ P b for all positive integers n , where ≤ P denotes the partial order with the positive cone P . Recall the nil-radical of a commutative ring R is N : = { a ∈ R | a   is   nilpotent } .

Theorem 4. Let R be a commutative ring and P be a symmetric partial order on R .

1) R = P ∪ − P ∪ I P and P ∩ I P = − P ∩ I P = { 0 } .

2) I P is a convex ideal of R and R / I P is a totally ordered ring.

3) If N = { 0 } , then I P is a prime ideal of R .

Proof. (1) Suppose that a ∈ R and a ∉ P ∪ − P . We show that a ∈ I P . Take 0 ≠ b ∈ P . Since a ∉ P and P is division closed, for any positive integers k , if k a ≠ 0 , then k a ∉ P since k 1 ∈ P . Thus b = ( b − k a ) + k a ∈ P implies b − k a ∈ P . Thus b − k a ∈ P for all positive integer k . Similarly, since − a ∉ P , b + k a ∈ P for all positive integers k . Hence b + n a ∈ P for all integers n . Since this is true for any b ∈ P , b ≠ 0 . We have a ∈ I P . It is clear that P ∩ I P = − P ∩ I P = { 0 } .

(2) Take a , b ∈ I P and assume a ≠ 0 , b ≠ 0 , a + b ≠ 0 since a + b ∈ I P under those cases. If a + b ∈ P , then that P is symmetric implies that a ∈ P or b ∈ P , a contradiction by (1). If a + b ∈ − P , then 0 ≠ − ( a + b ) = ( − a ) + ( − b ) ∈ P , so − a ∈ P or − b ∈ P and hence a ∈ − P or b ∈ − P , a contradiction again by (1). Thus a + b ∈ I P . It is clear that a ∈ I P implies that − a ∈ I P . Hence I P is a subgroup of ( R , + ) .

Take a ∈ I P , r ∈ R and a ≠ 0 , r ≠ 0 , r a ≠ 0 . By (1), we consider r ∈ P , or r ∈ − P , or r ∈ I p .

(i) r ∈ P . Then P is division closed implies that r a ∉ P , otherwise a ∈ P , a contradiction by (1). Similarly, r a ∉ − P . Thus r a ∈ I P by (1).

(ii) r ∈ − P . Then − r ∈ P , so by the argument given in (i), ( − r ) a ∈ I P . Thus r a ∈ I P .

(iii) r ∈ I P . Suppose r a ∈ P . We have − r a ≤ P 2 r ≤ P r a and − 1 ≤ P a ≤ P 1 , then

By adding above two inequalities together, we have 2 ( − r a ) ≥ P 0 . Since − r a ≤ P 0 , we must have − r a = 0 , so r a = 0 , a contradiction with r a ≠ 0 . Thus r a ∉ P . Similarly, r a ∉ − P . Hence r a ∈ I P . Therefore, for any r ∈ R and a ∈ I P , r a ∈ I P , so I P is an ideal of R . Let 0 ≤ P a ≤ P b ∈ I P . Clearly, if a ≠ 0 , then a ∉ P ∪ − P , so a ∈ I P . Thus, I P is a convex ideal of ( R , P ) . Since R = P ∪ − P ∪ I P , it is clear that R / I P is a totally ordered ring with positive cone P ¯ = { a + I P | a ∈ P } [13].

(3) If I P = { 0 } , then P is a total order on R . It is well-known that a totally order ring without nonzero nilpotent elements is a domain, so I P is a prime ideal. Now we assume that I P ≠ { 0 } . Let a , b ∈ R such that a b ∈ I P . Suppose that a ∉ I P , b ∉ I P . We derive a contradiction. Since R = P ∪ − P ∪ I P , a , b ∈ P or − P . Without loss of generality, we may assume that a , b ∈ P , so a b ∈ P . Then a b = 0 , since P ∩ I P = { 0 } . It follows from ∀ x ∈ I P , − a ≤ P x ≤ P a and − b ≤ P x ≤ P b that x a = x b = 0 , and hence

Thus x 2 = 0 . Since R contains no nonzero nilpotent elements, x = 0 , so I P = { 0 } , a contradiction. Therefore a b ∈ I P implies a ∈ I P or b ∈ I P , that is, I P is a prime ideal of R . □

We will need following result from [14] that provides a way to construct symmetric partial orders.

Lemma 6. ([14], Theorem 2.3)Let R be a commutative ring, I be a prime ideal of R ,and T be a total order on the quotient ring R / I . Then

is a symmetric partial order on R .

For an infinite prime S of a commutative ring, define P s = [ S \ ( S ∩ − S ) ] ∪ { 0 } .

Theorem 5. Let R be a commutative ring that is algebraic over ℤ . For each full infinite prime S of R , P s is a symmetric partial order on R and S = P s ∪ I P s .

Proof. Let I = S ∩ − S . From ([6], Proposition 2.5), I is a prime ideal of R . Consider the integral domain R / I . Define S ¯ : = { a + I | a ∈ S } in R / I . By ([13], Lemma 5), S ¯ is a full infinite prime of R / I . Since R / I is an integral domain that is algebraic over ℤ , by ([13], Lemma 2 and Theorem 1), S ¯ is a total order on R / I . It is straightforward to verify that

So, by Lemma 6, P s is a symmetric partial order on R . Let 0 ≠ a ∈ I . Then a ∉ P s ∪ − P s , so a ∈ I P s by Theorem 4. Thus I ⊆ I P s . Since S ⊆ P s ∪ I P s and P s ∪ I P s is a preprime, S = P s ∪ I P s . □

Corollary 3. Let R be a commutative ring that is algebraic over ℤ .

1) Each full infinite prime of R is strong Archimedean.

2) Let S be an infinite prime of R . Then there exists a ring homomorphism φ : S − S → ℝ such that Ker ( φ ) = S ∩ − S and S = φ − 1 ( ℝ + ) .

Proof. (1) Let S be a full infinite prime of R . By Theorem 5, S = P s ∪ I P s . Then I P s = S ∩ − S , so I P s is a prime ideal of R . We first show that P s is strong Archimedean. Since R / I P s is an integral domain that is algebraic over ℤ , S ¯ = { a + J s | a ∈ S } is an Archimedean total order ([13], Theorem 1), so S ¯ is strong Archimedean. Let 0 ≠ a ∈ P s . Then a + I P s ∈ S ¯ , so there is a positive integer n such that 0 ≠ n ( 1 + I P s ) − ( a + I P s ) ∈ S ¯ . Thus there exists b ∈ S such that ( n − a ) + I P s = b + I P s . Since b + I P s ≠ 0 , b ∉ I P s and hence b ∈ P s . Then there exists i ∈ I P s such that n − a = b + i . If i = 0 , then n − a = b ∈ P s . If i ≠ 0 , then 0 ≠ − i ∈ I P s , P s ∩ I P s = { 0 } , and b = ( b + i ) + ( − i ) ∈ P s implies that n − a = b + i ∈ P s , since P s is symmetric. Thus, in any case, n − a ∈ P s , so P s is strong Archimedean. Now Take 0 ≠ a ∈ S . If a ∈ P s , then there is a positive integer n such that n − a ∈ P s ⊆ S . If a ∈ I P s , then 1 = ( 1 − a ) + a and a ∉ P s implies that 1 − a ∈ P s ⊆ S , since again P s is symmetric. Therefore S is strong Archimedean.

(2) Since S − S is a commutative ring that is algebraic over ℤ , and S is a full infinite prime of S − S , by (1), S is strong Archimedean. Then, by ([6], Proposition 1.7), there exists a ring homomorphism φ : S − S → ℝ such that Ker ( φ ) = S ∩ − S and S = φ − 1 ( ℝ + ) . □

Lemma 7. Let R be a commutative ring that is algebraic over ℤ and N = { 0 } . Then each symmetric partial order is a maximal partial order.

Proof. Let P be a symmetric partial order on R . Then P ⊆ P m for some maximal partial order on R . Suppose P ≠ P m . We derive a contradiction. Take a ∈ P m \ P . If a ∈ − P , then − a ∈ P ⊆ P m as well, a contradiction. Thus by Theorem 4, a ∈ I P . Let f ( x ) = a n x + ⋯ + a 1 x + a 0 be a nonzero integer polynomial such that f ( a ) = 0 . If a 0 ≠ 0 , then a 0 = − a ( a n a n − 1 + ⋯ + a 1 a ) ∈ I P , so − 1 ≤ P n a 0 ≤ P 1 for all positive integers n , a contradiction. Thus a n a n + ⋯ + a 1 a = 0 for some n ≥ 2 . Suppose that a 1 = 0 . Then a n a n + ⋯ + a k a k = 0 with n > k ≥ 2 and a k ≠ 0 . It follows that

Then R does not contain nonzero nilpotent elements implies that a n a n − 1 + ⋯ + a k a k − 1 = 0 . If k − 1 ≥ 2 , we may repeat the above argument, so without loss of generality, we may assume at the beginning that a n a n + ⋯ + a 1 a = 0 and a 1 ≠ 0 .

Let e = a n a n − 1 + ⋯ + a 1 . Then e 2 = a 1 e since a e = a ( a n a n − 1 + ⋯ + a 1 ) = 0 . If e ∈ P , then − e ≤ P a ≤ P e , so − e ≤ P m a ≤ P m e since P ⊆ P m . If follows from a ∈ P m that − a e = 0 ≤ P m a 2 ≤ P m a e = 0 . Thus a 2 = 0 and a = 0 , a contradiction. Similarly, e ∉ − P . Hence e ∈ I P by Theorem 4. From a , e ∈ I P , we have a 1 ∈ I P , so − 1 ≤ P n a 1 ≤ 1 for all positive integers n , a contradiction. Therefore, we must have P = P m and hence P is a maximal partial order on R . □