# Ordered ring

partially ordered ring

A ring $R$( not necessarily associative) which is a partially ordered group under addition and in which for any elements $a , b , c \in R$ the inequalities $a \leq b$ and $c \geq 0$ imply $a c \leq b c$ and $c a \leq c b$. Every ring is an ordered ring for the trivial order. As examples of ordered rings one may take an ordered field; the ring of real functions on a set $X$, where $f \leq g$ means that $f ( x) \leq g ( x)$ for all $x \in X$; or a matrix ring over an ordered ring $R$, where, by definition, $\| a _ {ij} \| \leq \| b _ {ij} \|$ if $a _ {ij} \leq b _ {ij}$ for all $i , j$. If $R$ is an ordered ring, then the set

$$P = \{ {x } : {x \in R , x \geq 0 } \}$$

is called its positive cone. The positive cone of an ordered ring completely defines the order: $x \leq y$ if and only if $y - x \in P$. A subset $P$ of a ring $R$ can serve as the positive cone for some order if and only if

$$P \cap ( - P ) = \{ 0 \} ,\ \ P + P \subseteq P \ \textrm{ and } \ \ P P \subseteq P .$$

The equation $P \cup ( - P ) = R$ is equivalent to the totality of the order (cf. Totally ordered set).

An ordered ring that is totally ordered or lattice-ordered is accordingly called a totally ordered or lattice-ordered ring (cf. also Archimedean ring). Lattice-ordered rings turn out to be distributive lattices, and their additive groups are torsion-free (cf. Lattice-ordered group). Certain questions in the theory of associative rings and, in particular, in the theory of radicals have analogues in associative lattice-ordered rings. The class of rings which allow a lattice-ordered ring structure is not axiomatizable. If $a , b , c$ are elements of a lattice-ordered ring and $c \geq 0$, then the following relations hold:

$$( a \lor b ) c \geq a c \lor b c ,\ \ c ( a \lor b ) \geq c a \lor c b ,$$

$$( a \wedge b ) c \leq a c \wedge b c ,\ c ( a \wedge b ) \leq c a \wedge c b .$$

Ideals in lattice-ordered rings which are convex subgroups (cf. Convex subgroup) of the additive group are called $l$- ideals. The quotient ring by an $l$- ideal can be made into a lattice-ordered ring in a natural way. The homomorphism theorem holds.

A lattice-ordered ring $R$ is called a functional ring or an $f$- ring if it satisfies any of the following equivalent conditions: 1) $R$ is isomorphic to a lattice-ordered subring of a direct product of totally ordered rings; 2) for any $a , b , x \in R$ one has the implication

$$( a \wedge b = 0 \textrm{ and } x \geq 0 ) \Rightarrow ( a \wedge b x = a \wedge x b = 0 ) ;$$

3) for any subset $X$ of $R$ the set

$$\{ {y } : {y \in R , \forall x \in X x \wedge y = 0 } \}$$

is an $l$- ideal; and 4) for any $a , b \in R$,

$$( a \lor 0 ) ( b \lor 0 ) \wedge ( - a \lor 0 ) = ( b \lor 0 ) ( a \lor 0 ) \wedge ( - a \lor 0 ) = 0 .$$

Condition 4) shows that $f$- rings form a variety of signature $\{ + , - , 0 , \cdot , \lor , \wedge \}$. Neither of the equations in this condition is a consequence of the other. Not every $f$- ring can be imbedded in an $f$- ring with a unit element. If $a , b , c$ are elements of an $f$- ring and $c > 0$, then one has

$$( a \lor b ) c = a c \lor b c ,\ c ( a \lor b ) = c a \lor c b ,$$

$$( a \wedge b ) c = a c \wedge b c ,\ c ( a \wedge b ) = c a \wedge c b ,$$

$$( a \lor ( - a ) ) ( b \lor ( - b ) ) = a b \lor ( - a b ) ,\ a ^ {2} \geq 0 ,$$

as well as the implication $( a \wedge b = 0 )$ $\Rightarrow$ $( a b = 0 )$.

An order of an ordered ring $R$ with a positive cone $P$ can be extended to a total order such that $R$ becomes a totally ordered ring if and only if for any finite set $a _ {1} \dots a _ {n}$ in $R$ one can choose $\epsilon _ {i} = 1$ or $- 1$ such that in the semi-ring generated by $P$ and the elements $\epsilon _ {1} a _ {1} \dots \epsilon _ {n} a _ {n}$ the sum of any two non-zero elements is non-zero. With $P = \{ 0 \}$ one obtains a criterion for the possibility of having a total order on the ring.

#### References

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