Math 541 Archives - Page 2 of 8 - Shawn Zhong

Math 541 – 4/23

$Polynomial Ring • Polynomials over a ring ○ Let R be a commutative ring ○ A polynomial over R is a sum a_n x^n+a_(n−1) x^(n−1)+…+a_1 x+a_0 ○ Where x is a variable, and a_i∈R • Degree ○ If f=a_n x^n+a_(n−1) x^(n−1)+…+a_1 x+a_0 is a polynomial over R ○ The degree of f, denoted deg⁡(g), is sup⁡{n≥0│a_n≠0} ○ Note: deg⁡(0)=−∞ • Leading term and leading coefficient ○ If deg⁡(f)=n≥0 ○ The leading term of f is a_n x^n ○ The leading coefficient of f is a_n • Notation ○ Let R[x]≔{Polynomials over a commutative ring R} ○ Then R[x] is a commutative ring with ○ ordinary addition and multiplication of polynomials • R is a subring of R[x] ○ R is identified with the constant polynomials ○ There is a ring homomorphism i:R→R[x] defined as ○ mapping the ring element r∈R to the constant polynomial r ○ The constant polynomials in R[x] form a subring ○ And i gives an isomorphism between R and the subring • Polynomial ring with multiple variables ○ We define polynomial rings in several variables inductively ○ R[x_1,x_2 ]=(R[x_1 ])[x_2 ] ○ ⋮ ○ R[x_1,…,x_n ]=(R[x_1,…,x_(n−1) ])[x_n ] Proposition 67: Properties of Polynomial Rings • Statement ○ Let R be a domain. Let p,q∈R[x]∖{0}, then 1. deg⁡(pq)=deg⁡(p)+deg⁡(q) 2. (R[x])^×=R^× 3. R[x] is a domain • Proof ○ Write § p=a_n x^n+…+a_1 x+a_0, where deg⁡(p)=n § q=b_m x^m+…+b_1 x+b_0, where deg⁡(q)=m ○ Then a_n≠0 and b_m≠0 ○ Since R is a domain, a_n m_m≠0 ○ So, the leading term of pq is a_n b_m x^(m+n), which verifies (1) ○ Also, a_n b_m x^(m+n)≠0. This proves (3) ○ For (2), suppose pq=1, then § deg⁡(p)+deg⁡(q)=deg⁡(pq)=0 by (1) § Thus, deg⁡(p)=0=deg⁡(q) i.e. p,q∈R § Since pq=1, p,q∈R^× § Thus (R[x])^×⊆R^× § Also, R^×⊆(R[x])^× § Therefore (R[x])^×=R^× Ideal • Definition ○ Let R be a ring, let I be a subset of R, and let r∈R ○ rI≔{rx│x∈I} is a left ideal of R if § I is a additive subgroup of R § rI=I, ∀r∈R ○ Right ideal Ir≔{xr│x∈I} are defined similarly ○ I is an ideal if I is both a left and right ideal • Intuition ○ Normal subgroups are to groups as ideals are to rings • Example ○ If R is a ring, then R and {0} are both ideals Proposition 68 • Statement ○ If I⊆R is an ideal, then I=R⟺1∈I • Proof ○ (⟹) Trivial ○ (⟸) Let r∈R. ○ By definition of ideal, rI=I ○ So r=r⋅1∈I ○ Thus R=I • In particular, if R is a ring, and S is a subring of R ○ S is an ideal in R⟺S=R (subrings contain 1) ○ Similarly, if I⊆R is an ideal, and I is a subring of R⟺I=R Principal Ideal • Definition ○ Let R is a commutative ring, and let r∈R, then ○ (r)≔{ar│a∈R} is called the principal ideal generated by r • Proof: principal ideals are ideals ○ 0=0⋅r∈(r), so (r) is not empty ○ Let ar,br∈(r), then § ar−br=(a−b)r∈(r) § Therefore, (r) is an additive subgroup of R ○ Let a∈R, br∈(r), then § a(br)=abr∈(r) § (br)a=bra=abr∈(r) § So a(r)=(r)a,∀a∈R • Example ○ If n∈Z, (n) is just the cyclic subgroup generated by n$

Math 541 – 4/20

$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)$

Math 541 – 4/18

$Proposition 64 • Statement ○ Let n 0 ○ Every nonzero element in Z\/nZ is either a unit or a zero-divisor • Note ○ We don’t have this property in Z ○ In Z, the units are ±1, there are no zero-divisor ○ 2∈Z is not 0 or unit or zero-divisor • Proof ○ Suppose a ̅∈Z\/nZ is nonzero and not a unit ○ Then d≔(a,n) 1 ○ Write cd=a,md=n ○ Then a ̅m ̅=c ̅d ̅m ̅=c ̅n ̅=0 ̅ ○ Moreover, m ̅≠0 ̅ § Since md=n,1≤m≤n, and d 1 § m cannot be a multiple of n Field • Definition ○ Communitive ring R is called a field if ○ Every nonzero element of R is a unit ○ i.e. Every nonzero element of R have a multiplicative inverse • Examples ○ Q,R ○ ℂ § But not true for R2 with (r_1,r_2 )(r_1^′,r_2^′ )=(r_1 r_1^′,r_2 r_2^′ ) ○ Z\/pZ (p prime) § 1≤a≤p−1,(a,p)=1⇒a ̅∈Z\/pZ § Note: Z\/nZ is a field ⟺ n is prime Product Ring • If R_1,R_2 are rings, R_1×R_2 has the following ring structure • For addition, it s just the product as groups • For multiplication, (r_1,r_2 )(r_1^′,r_2^′ )=(r_1 r_1^′,r_2 r_2^′ ) with identity (1_(R_1 ),1_(R_2 ) ) Integral Domain • Definition ○ A communicative ring R is an integral domain (or just domain) if ○ R contains no zero-divisors • Example ○ Unites are not zero-divisors, so fields are domains ○ Z is a domain ○ Z\/nZ is a domain ⟺ it is a field ○ R_1×R_2 is a domain ⟺ one of them is trivial, and the other is a domain$