Product Ring and Domain • Statement ○ If R_1 and R_2 are rings, then R_1×R_2 is a domain iff ○ one of the R_1 or R_2 is a domain, and the other is trivial • Proof (⟸) ○ Without loss of generality, assume R_1 is a domain and R_2 is trivial ○ Let (r_1,r_2 ),(r_1′,r_2′ )∈R_1×R_2∖{(0,0)} ○ Then r_1≠0 and r_1^′≠0 ○ Since R_1 is a domain, r_1 r_1^′≠0 ○ Thus, (r_1,r_2 )(r_1′,r_2′ )=(r_1 r_1^′,r_2 r_2^′ )≠0 • Proof (⟹) ○ (1_(R_1 ),0)(0,1_(R_2 ) )=(1_(R_1 )⋅0,0⋅1_(R_2 ) )=(0,0) ○ Since R_1×R_2 is a domain, either (1_(R_1 ),0) or (〖0,1〗_(R_2 ) ) is (0,0) ○ This means either 1_(R_1 ) or 1_(R_2 ) is 0, and thus R_1 or R_2 is trivial ○ Without loss of generality, suppose R_2 is trivial ○ We want to show that R_1 is a domain ○ Let r_1,r_1^′∈R_1∖{0} ○ Then (r_1,0),(r_1^′,0)∈R_1×R_2∖{(0,0)} ○ So (r_1 r_1^′,0)≠0 i.e. r_1 r_1^′≠0 Proposition 66 • Statement ○ A finite domain R is a field • Proof ○ Let a∈R∖{0} ○ We want to show that a has a multiplicative inverse ○ Define a function F:R→R given by r↦ar ○ F is injective § If ar_1=ar_2 § Then a(r_1−r_2 )=0 § Since R is a domain, r_1−r_2=0 § So r_1=r_2 ○ F is surjective since R is finite ○ Choose b∈R s.t. F(b)=1, then ab=1 ○ So b is the inverse of a Subring • Definition ○ A subring of a ring R is a additive subgroup S of R s.t. ○ S is closed under multiplication ○ S contains 1 • Note: A subring of a ring is also a ring • Example 1 ○ A ring is always a subring of itself • Example 2 ○ Mat_n (R has a subring given by diagonal matrices ○ Scalar matices also form a subring • Example 3 ○ Z⊆Q⊆R⊆ℂ is a chain of subring of ℂ • Example 4 ○ R={continuous functions from Rn to Rfor some n≥1} ○ Addition: (f+g)(v)=f(v)+g(v) ○ Multiplication: (fg)(v)=f(v)g(v) with identity of constant function 1 ○ Polynomials in n variables form a subring • Example 5 ○ If f:R→S is a ring homomorphism i.e. § f is a homomorphism of abelian groups under addition § f(r_1 r_2 )=f(r_1 )f(r_2 ),∀r_1,r_2∈R § f(1_R )=1_S ○ Then im(f) is a subring of S ○ Proof § By group theory, im(f) is an additive subgroup of S § 1∈im(f) by assumption § If f(r_1 ),f(r_2 )∈im(f), then f(r_1 )f(r_2 )=f(r_1 r_2 )∈im(f) • Example 6 ○ By HW, ∃! Ring homomorphism f:Z→R for any ring R ○ im(f) is the smallest subring of R ○ Also, im(f)≅Z\nZ, where n=char(R) ○ Note: A ring isomorphism is a ring homomorphism that is bijective • Example 7 ○ {(r_1,0)│r_1∈R_1 }⊆R_1×R_2 is not a subring ○ Since it doesn t contain the identity (1,1)
