Jun 15, 2018 - It is about the case of rings with a prime number of elements. ... https://en.wikipedia.org/wiki/Rng_(algebra)#Definition.
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Rings of prime order Document created on June 15, 2018 Most recent update: August 10, 2018 This document is about the simplest case of Classification of finite rings: https://www.researchgate.net/publication/325527064_Classification_of_finite_rings
It is about the case of rings with a prime number of elements. https://en.wikipedia.org/wiki/Ring_(mathematics) https://en.wikipedia.org/wiki/Prime_number
In this part, I first recall the definitions of rng and ring. Then I prove that if ℜ is a rgn whose number of elements is a prime p, then either ℜ is a ring where all products are equal to zero, or ℜ is isomorphic the field Fp. For the case where not all products are zero, the main steps of my proof of are the following: 1. Prove that every product of two non-zero elements of ℜ is different from zero. This is done below in Theorem 15. 2. Prove that ℜ has a left identity element for at least one element of ℜ. This is done in Theorem 18. 3. Prove that this left identity is a left identity for all the elements of ℜ. This is done in Theorem 19. 4. Prove that R has a right identity element for at least one element of ℜ. 5. Prove that this right identity is a right identity for all the elements of ℜ. 6. Prove that ℜ has a multiplicative identity element. 7. Construct an isomorphism from the field Fp onto ℜ. https://en.wikipedia.org/wiki/Field_(mathematics) https://en.wikipedia.org/wiki/Finite_field https://en.wikipedia.org/wiki/Field_(mathematics)#Finite_fields https://en.wikipedia.org/wiki/Cyclic_group http://mathworld.wolfram.com/CyclicGroup.html
1. Definition. A rng ℜ = (S, 0, +, ×) is a set S provided with a special element 0 and two binary operations + and × called addition and multiplication, such that: ● The set S with the operation of addition + and additive identity 0 forms an abelian group G = (S, 0, +). ● The set S with the operation of multiplication × forms a semigroup 𝓢 = (S, ×). This semigroup 𝓢 may be or not be a monoid, that is, may have or not have a multiplicative identity element. https://en.wikipedia.org/wiki/Rng_(algebra) https://en.wikipedia.org/wiki/Rng_(algebra)#Definition https://en.wikipedia.org/wiki/Abelian_group https://en.wikipedia.org/wiki/Semigroup https://en.wikipedia.org/wiki/Monoid https://en.wikipedia.org/wiki/Identity_element
2. Definition. A ring is a rng with a multiplicative identity element usually denoted by 1. https://en.wikipedia.org/wiki/Ring_(mathematics)
3. Theorem. Let G = (S, 0, +) be an abelian group. Then we can extend this group into a rng in the following trivial way: Let × be the multiplication where all products are equal to 0: (∀ a ∈ S)(∀ b ∈ S)(a × b = 0). Then ℜ = (S, 0, +, ×) is a rng. This rng is a ring if and only if S = {0}; in this case, 1 = 0. For the proof, we check that we have associativity of multiplication and left and right distributivity of multiplication with respect to addition. As all the products are zero, these three properties are equivalent to 0 = 0. I call such rng a MZ-rng, that is, a Multiplication-Zero-rng. 4. Notation. Let G1 and G2 be groups. Then we denote by Hom(G1, G2) the set of homomorphisms from G1 into G2. https://en.wikipedia.org/wiki/Group_homomorphism
5. Definition. Let G1 and G2 be groups. Let h ∈ Hom(G1, G2). Then the kernel of h is defined by Ker(h) = {x ∈ G1| h(x) = 0}. https://en.wikipedia.org/wiki/Kernel_(algebra)#Group_homomorphisms
6. Theorem. Let G1 and G2 be groups. Let h ∈ Hom(G1, G2). Then Ker(h) is a subgroup of G1. https://en.wikipedia.org/wiki/Kernel_(algebra)#Group_homomorphisms
7. Notation. Let G be a group. Then we denote by End(G) the set of endomorphisms from G into G, and we denote by Aut(G) the set of automorphisms from G into G, https://en.wikipedia.org/wiki/Group_homomorphism#Types_of_group_homomorphism
8. Theorem. Let G be group. Let h ∈ End(G) be such that Ker(h) = {0}. Then h ∈ Aut(G). https://en.wikipedia.org/wiki/Kernel_(algebra)#Group_homomorphisms
9. Notation. Let S be a finite set. Then |S| denotes the number of elements of S. https://en.wikipedia.org/wiki/Cardinal_number#Motivation
10. Lagrange’s Theorem. Let G be a finite group and let H be a subgroup of G. Then |H| divides |G|. https://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory)
11. Theorem. Let p be a prime. Let G = (S, 0, +) be a group such that |S| = p. Let h ∈ End(G) be such that h ≠ 0. Then h ∈ Aut(G). Proof. By Theorem 6, Ker(h) is subgroup of G. By Lagrange’s Theorem [Theorem 10], |Ker(h)| divides |G|. From |Ker(h)| divides |G| and |G| = p, we deduce that |Ker(h)| divides p. From |Ker(h)| divides p and p is prime, we deduce that |Ker(h)| = 1 or |Ker(h)| = p. From h ≠ 0, we deduce by Definition 5 that Ker(h) ≠ G. From Ker(h) ≠ G, and Ker(h) is a subgroup of the finite group G = (S, 0, +), we deduce that |Ker(h)| < |S|. From |Ker(h)| < |S| and |S| = p, we deduce that |Ker(h)| < p From [|Ker(h)| = 1 or |Ker(h)| = p] and Ker(h) < p, we deduce that |Ker(h)| = 1. From Definition 5, we deduce that 0 ∈ Ker(h). From |Ker(h)| = 1 and 0 ∈ Ker(h), we deduce that Ker(h) = {0}. From h ∈ End(G) and Ker(h) = {0}, we deduce by Theorem 8 that h ∈ Aut(G)∎ 12. Notation. Let ℜ = (S, 0, +, ×) be a rng. Let s ∈ S. Then Ls denotes the left multiplication by s, that is, (∀ x ∈
S)(Ls(x) = s × x). Then Rs denotes the right multiplication by s, that is, (∀ x ∈ S)(Rs(x) = x × s). 13. Theorem. Let p be a prime. Let S be a set such that |S| = p. Let ℜ = (S, 0, +, ×) be a rng that is not a MZ-rgn. Let {a, b} ⊆ S be such that a × b ≠ 0. Let G = (S, 0, +) denote the abelian additive group of ℜ = (S, 0, +, ×). Then La ∈ Aut(G), where La is defined in Notation 12. Proof. First let us prove that La ∈ End(G). Let {x, y} ⊆ S. Then by left distributivity, La(x + y) = a × (x + y) = a × x + a × y = La(x) + La(y). Thus (∀ {x, y} ⊆ S)[La(x + y) = La(x) + La(y)], that is, La ∈ End(G). From La(b) = a × b and a × b ≠ 0, we deduce that La(b) ≠ 0. This implies that La ≠ 0. From |S| = p, La ∈ End(G), and La ≠ 0, we deduce by Theorem 11 that La ∈ Aut(G)∎
14. theorem. Let p be a prime. Let S be a set such that |S| = p. Let ℜ = (S, 0, +, ×) be a rng that is not a MZ-rgn. Let {a, b} ⊆ S be such that a × b ≠ 0. Let G = (S, 0, +) denote the abelian additive group of ℜ = (S, 0, +, ×). Then Rb ∈ Aut(G) , where Rb is defined in Notation 12. Proof. First let us prove that Rb ∈ End(G). Let {x, y} ⊆ S. Then by right distributivity, Ra(x + y) = (x + y) × b = x × b + y × b = Rb(x) + Rb(y). Thus (∀ {x, y} ⊆ S)[Rb(x + y) = Rb(x) + Rb(y)], that is, Rb ∈ End(G). From Rb(a) = a × b and a × b ≠ 0, we deduce that Rb(a) ≠ 0. This implies that Rb ≠ 0. From |S| = p, Rb ∈ End(G), and Rb ≠ 0, we deduce by Theorem 11 that Rb ∈ Aut(G)∎ 15. Theorem. Let p be a prime. Let S be a set such that |S| = p. Let S* = S ⟍ {0}. Let ℜ = (S, 0, +, ×) be a rng that is not a MZ-rgn. Then (∀ {x, y} ⊆ S*) (x × y ≠ 0). Proof. Let {x, y} ⊆ S*. Let us prove that x × y ≠ 0. Since ℜ = (S, 0, +, ×) is not a MZ-rgn, there exists {a, b} ⊆ S such that a × b ≠ 0. Let G = (S, 0, +) be the abelian additive group of ℜ = (S, 0, +, ×). Since |S| is prime and a × b ≠ 0, we have by Theorem 13 that Rb ∈ Aut(G). Since Rb is an automorphism, it is injective, and consequently, we have (∀ s ∈ S*)(Rb(s) ≠ 0). From x ∈ S* and (∀ s ∈ S*)(Rb(s) ≠ 0), we deduce that Rb(x) ≠ 0, that is, x × b ≠ 0. Since |S| is prime and a × b ≠ 0, we have by Theorem 13 that Lx ∈ Aut(G). Since Lx is an automorphism, it is injective, and consequently, we have (∀ s ∈ S*)(Lx(s) ≠ 0). From y ∈ S* and (∀ s ∈ S*)(Lx(s) ≠ 0), we deduce that Lx(y) ≠ 0, that is, x × y ≠ 0. Thus, we have proved that (∀ {x, y} ⊆ S) (x × y ≠ 0)∎ 16. Lemma. Let ℜ = (S, 0, +, ×) be a rng that is not a MZ-rgn. Then there exists s ∈ S such that s ≠ 0. Proof. Since ℜ = (S, 0, +, ×) is not a MZ-rgn, we have that S ≠ {0}. From 0 ∈ S, we deduce that {0} ⊆ S. From {0} ⊆ S and S ≠ {0}, we deduce that S ⊈ {0}, otherwise, S ⊆ {0} and {0} ⊆ S would imply that S = {0}, contradicting S ≠ {0}. From S ⊈ {0}, we deduce that there exists s ∈ S such that s ∉ {0}. From s ∉ {0}, we deduce that s ≠ 0∎
17. Theorem. Let p be a prime. Let S be a set such that |S| = p. Let ℜ = (S, 0, +, ×) be a rng that is not a MZ-rgn. Let G = (S, 0, +) denote the abelian additive group of ℜ = (S, 0, +, ×). Let s ∈ S be such that s ≠ 0. Then Ls ∈ Aut(G) and Rs ∈ Aut(G). Proof. From s ≠ 0, we deduce by Theorem 15 that s × s ≠ 0. From |S| = p and s × s ≠ 0, we deduce by Theorem 13 that Ls ∈ Aut(G), and we deduce by Theorem 14 that Rs ∈ Aut(G)∎ 18. Theorem. Let p be a prime. Let S be a set such that |S| = p. Let ℜ = (S, 0, +, ×) be a rng that is not a MZ-rgn. Then there exists {e, s} ⊆ S such that s ≠ 0 and e × s =s. Proof. By Lemma 16, there exists s ∈ S such that s ≠ 0. Let G = (S, 0, +) denote the abelian additive group of ℜ = (S, 0, +, ×). By Theorem 14, we have that Rs ∈ Aut(G). This implies that Rs is a surjective function from S onto S, which in turn implies that there exists e ∈ S such that Rs(e) = s, that is, by Notation 12, e × s = s∎ 19. Theorem. Let p be a prime. Let S be a set such that |S| = p. Let ℜ = (S, 0, +, ×) be a rng that is not a MZ-rgn. Then there exists e ∈ S such that (∀ x ∈ S)(e× x = x). Proof. By Theorem 18, there exists {e, s} ⊆ S such that s ≠ 0 and e × s =s. Let x ∈ S. Let us prove that e× x = x. Let G = (S, 0, +) denote the abelian additive group of ℜ = (S, 0, +, ×). From |S| = p and s ≠ 0, we deduce by Theorem 17 that Ls ∈ Aut(G). From Ls ∈ Aut(G), we deduce that Ls is a surjective function from S onto S, which in turn implies that there exists a ∈ S such that Ls(a) = x, that is, by Notation 12, s × a = x. Then e × x = e × (s × a) =(e × s) × a = s × a = x. Thus, we have proved that (∀ x ∈ S)(e× x = x)∎
Under construction
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