Try multiplying the successor of 1 with a number c where and see what happens. Also try squaring the successor of 0.
Fact: The ring of integers Z is totally ordered: for any distinct elements a and b in Z, either a>b or a<b.
Fact: The ring of integers is discrete, in the sense that for any element a in Z, there exists an element b in Z such that there is no element c in Z with a<c<b, and the same argument holds with the greater than signs flipped. In other words, successors and predecessors exist in Z (but not in R, for example).
Observation: In Z, the multiplicative identity 1 is the successor of the additive identity 0.
Questions: Is this fact a coincidence? Does this have any significance? Must the multiplicative identity always succeed the additive identity in rings that have the same properties as Z, assuming there are any other?
let's agree on a few things:
we will call a ring with unity R ordered if we have a subset P such that:
1) P is closed under multiplication and addition.
2) for all x in R*, x^{2} is in P
3) 0,-1 are not in P
4) R = P U {0} U -P
we will say x > y if x - y is in P.
note rule (4) assures us that at least one of the following is true:
a) x > y
b) x = y
c) y > x
clearly rule (3) shows if (b) is true, then neither (a) nor (c) can be true; and if (a) or (c) is true, (b) is impossible. now if (a) & (c) are true, we would have (x-y) + (y-x) = 0 in P by rule (1), a contradiction. so we see exactly 1 of (a),(b), or (c) is true. this shows that ">" is irreflexive and anti-symmetric. to see that it is transitive:
suppose x > y, and y > z. then x-y is in P, and y-z is in P, so (x-y) + (y-z) = x-z is in P (by rule (1)), so x > z. thus we have all the usual properties of a total order (trichotomy, irreflexiveness, and anti-symmetry), and this order is consistent with the ring operations in that:
x > 0 and y > 0 means x+y > 0
x > 0 and y > 0 means xy > 0
in such a ring, suppose we have 0 < x < y. is it true that for 0 < z, that 0 < xz < yz?
well clearly 0 < xz, by rule (1). if xz = yz, then (y - x)z = 0, which also contradicts rule (1) (because of rule (3)), since both y - x and z are in P. and if xz > yz, we get xz - yz > 0, so (x - y)z is in P. but y - x and z are in P, so (y - x)z is in P (by rule (1)), so (by the discussion above) (x - y)z = -(y - x)z cannot be in P. so xz < yz must be true.
now, suppose we have such a ordered ring with unity, and that this ring is also discrete; and that the successor of 0 is e ≠ 1. that is, we have:
0 < e < 1.
multiplying through by e, we have:
0 < e^{2} < e, contradicting that e is the successor of 0.
thus, in a discrete ordered ring, the successor of 0 must be 1.