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Homework 2

Math 447: Real Variables

Answer Exercise 1 and one other execise in this assignment. Note that you should do DO ALL PARTS of Ex. 1 and of the other exercise you choose.

Exercise 1 A totally ordered semigroup is a semigroup (A, +) with a relation ≤ that satisfies all of the following properties:

t1) ∀a ∈ A a ≤ a;

t2) a ≤ b and b ≤ a =⇒ a = b;

t3) a ≤ b and b ≤ c =⇒ a ≤ c;

t4) either a ≤ b or b ≤ a for all a, b ∈ A;

t5) a ≤ b =⇒ a + c ≤ b + c for all a, b, c ∈ A.

(a) Define the relation ≤ on N by stating that x ≤ y if either x = y or ∃z ∈ N such that x + z = y. Prove that z is necessarily unique (so we can define y − x = z when y > x).

(b) Prove that ≤ satisfies properties t1-5 (so tthat ≤ makes N into a totally ordered semigroup).

(c) Let n ∈ N be any natural number (defined as in homework 1, assuming S is injective). Prove that the set X = { x ∈ N | x ≤ n } is finite (using the definition of “finite” given in homework 1).

Exercise 2 Let S be a commutative semigroup with cancellation and G(S) the Grothendieck group of S.

(a) Prove that the function ι: S → G(S) defined by s 7→ [s + s, s] is injective and satisfies ι(s + t) = ι(s) + ι(t) for all s, t ∈ S.

Note: this justifies the claim that G(S) contains S, since we may identify S with ι(S).

(b) Suppose (G, •) is a group and ϕ: S → G is a function such that ϕ(x + y) = ϕ(x) • ϕ(y). Prove that there is a function ψ: G(S) → G such that ψ ◦ ι = ϕ and ψ(g + h) = ψ(g) • ψ(h) for all g, h ∈ G(S).

Note: this justifies the claim that G(S) is the “smallest” group containing S.

Exercise 3 Assume that (S, +, ≤) is a totally ordered commutative semigroup with cancellation. Define the order relation ≤ on G(S) by

[a, b] ≤ [c, d] ⇐⇒ a + d ≤ c + b.

Prove that

(a) (∀s, t, c ∈ S)   s + c ≤ t + c =⇒ s ≤ t; (this is the converse of property t5 in Ex. 1.)

(b) (∀s, t ∈ S) [s + s, s] ≤ [t + t, t] ⇐⇒ s ≤ t;

(c) (G(S), ≤ ) is a totally ordered group.

(The properties for a totally ordered group are the same as for a semigroup; see Ex. 1.)

Exercise 4 In this exercise you will verify the correspondence between G(N, +) and Z given in Lecture 2. Let

Define addition as follows:

(Recall from Ex. 1 that x − y is the unique element of N such that x = y + (x − y).)

(a) Prove that the function ϕ: G(N, +) → Z defined by

is bijective and satisfies ϕ([a + c, b + d]) = ϕ([a, b]) + ϕ([c, d]).