By Michael D. Atkinson

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Multiplicatively. Accordingly, one writes 0 = 0ζ and 1 = 0ν . Note that no commutativity assumptions are made here. Using distributivity and the multiplicative unit, it is easy to show that α is commutative. If μ is commutative as well, we say that the ring is commutative. A field is an algebra (F, α, σ, ζ, μ, ι, ν), where (F, α, σ, ζ, μ, ν) is a ring such that 0ζ = 0ν , and the restrictions of μ, ι and ν induce a group on F {0ζ }. 28 B Topological Groups Let (R, α, σ, ζ, μ, ν) be a ring. Then a (right) R-module (or a module over R) is a group (M, α, σ, ¯ ζ¯ ) with an operation μ¯ : M × R → M called multiplication by scalars (from the right) satisfying the following equations for all m, n ∈ M and all r, s ∈ R: ((m, n)α , r)μ¯ = ((m, r)μ¯ , (n, r)μ¯ )α , (m, (r, s)α )μ¯ = ((m, r)μ¯ , (m, s)μ¯ )α , (m, (r, s)μ )μ¯ = ((m, r)μ¯ , s)μ¯ , (m, 0ν )μ¯ = m.

A topological space is called totally disconnected if every connected component consists of a single point. 5 Lemma. Let X and Y be topological spaces, and assume that ϕ : X → Y is continuous and surjective. Then connectedness of X implies that Y is connected. ← Proof. Let A be a nonempty closed open subset of Y . Then Aϕ shares these ← properties, since ϕ is a continuous map. Now connectedness of X implies Aϕ = X ϕ and A = X = Y . 6 Lemma. Let ((Xα , Xα ))α∈A be a family of nonempty topological spaces.

C) Let (R, α, σ, ζ, μ, ν) and (R , α , σ , ζ , μ , ν ) be rings. A (ring) homomorphism from (R, α, σ, ζ, μ, ν) to (R , α , σ , ζ , μ , ν ) is a map ϕ : R → R such that (r, s)αϕ = (r ϕ , s ϕ )α and (r, s)μϕ = (r ϕ , s ϕ )μ hold for all r, s ∈ R, and 0νϕ = 0ν . Note that this implies that ϕ is a group homomorphism from (R, α, σ, ζ ) to (R , α , σ , ζ ). The kernel of ϕ is just the kernel of this group ← homomorphism; that is, ker ϕ = {0ζ }ϕ . A (ring) anti-homomorphism from (R, α, σ, ζ, μ, ν) to (R , α , σ , ζ , μ , ν ) is a map ψ : R → R such that (r, s)αψ = (r ψ , s ψ )α and (r, s)μψ = (s ψ , r ψ )μ hold for all r, s ∈ R, and 0νψ = 0ν .

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