Course Website
Textbook
Abstract Algebra, by Dummit and Foote, Third Edition, 2004
Lecture Notes
| Week 1 | |
|---|---|
| 1/24 | Divides, Equivalence Relations |
| 1/26 | Well-ordering of Z |
| Week 2 | |
| 1/29 | Division Algorithm, gcd |
| 1/31 | Euclidean Algorithm |
| 2/2 | Equivalence Class, Z/nZ, Group |
| Week 3 | |
| 2/5 | Group, Well-definedness, Z/nZ |
| 2/7 | (Z/nZ)*, Properties of Group |
| 2/9 | Order, Symmetric Group |
| Week 4 | |
| 2/12 | Symmetric Group, Cycle |
| 2/14 | Homomorphism, Isomorphism |
| 2/16 | Order, Homomorphism, Subgroup |
| Week 5 | |
| 2/19 | Dihedral Groups, Subgroup |
| 2/21 | Cyclic Group, lcm, Order of g^a |
| 2/23 | Cyclic Subgroup, Generating Set of a Group |
| Week 6 | |
| 2/26 | Finitely Generated Group |
| 2/28 | Coset, Normal Subgroup |
| 3/2 | Exam 1 |
| Week 7 | |
| 3/5 | Quotient Group, Index, Lagrange's Theorem |
| 3/7 | Corollaries of Lagrange's Theorem |
| 3/9 | The First & Second Isomorphism Theorems |
| Week 8 | |
| 3/12 | The Third Isomorphism Theorem |
| 3/14 | Transposition, Sign of Permutation |
| 3/16 | Homework 6, The Correspondence Theorem |
| Week 9 | |
| 3/19 | Sign of Permutation, Alternating Group |
| 3/21 | Subgroups of A_4, Group Action, Orbit, Stabilizer |
| 3/23 | Orbit, Stabilizer, Cayley's Theorem |
| Week 10 | |
| 4/2 | Conjugacy Class, The Class Equation |
| 4/4 | Cauchy's Theorem, Recognizing Direct Products |
| 4/6 | Homework 8, Properties of Finite Abelian Group |
| Week 11 | |
| 4/9 | Fundamental Theorem of Finite Abelian Groups |
| 4/11 | Definition of Ring |
| 4/13 | Exam 2 |
| Week 12 | |
| 4/16 | Properties of Ring, Zero-Divisor, Unit |
| 4/18 | Field, Product Ring, Integral Domain |
| 4/20 | Product Ring, Finite Domain and Field, Subring |
| Week 13 | |
| 4/23 | Polynomial Ring, Ideal, Principal Ideal |
| 4/25 | Examples of Ideals, Quotient Ring |
| 4/27 | Isomorphism Theorems for Rings |
| Week 14 | |
| 4/30 | Generating Ideal, Maximal Ideal, Prime Ideal |
| 5/2 | Prime Ideal, Euclidean Domain |
| 5/4 | Review, Galois Theory |
![Prime Ideal • Definition ○ Suppose R is commutative ○ An ideal P⊊R is prime if ○ a,b∈R,ab∈P⇒a∈P or b∈P • Example ○ Take R=Z ○ The prime ideals are ideals of the form (n), where n is prime or n=0 ○ Proof (⟹) § Let (n)⊆Z be a prime ideal, and n≠0 § We want to show that n is prime § Choose a,b∈Z s.t. n=ab § Then ab∈(n), so either a∈(n) or b∈(n) § Without loss of generality, suppose a∈(n) § Then ├ n┤ |a┤ § Choose q∈Z s.t. nq=a § n=ab⇒n=nqb⇒1=qb⇒b∈{±1} § So n is a prime ○ Proof (⟸) § (0) is prime □ Let a,b∈Z, and ab∈(0) □ Then ab=0 □ ⇒a=0 or b=0 □ ⇒a∈(0) or b∈(0) □ Therefore (0) is prime § (p) is prime for p∈Z prime □ Let a,b∈Z, and say ab∈(p) □ Then p|ab □ Since p is prime, this means p|a or p|b □ ⇒a∈(p) or b∈(p) Proposition 73: Criterion for Prime Ideal • Statement ○ Let R be a commutative ring, P⊆R an ideal, then ○ P is prime ⇔ R\/P is a domain ○ In particular, R is a domain ⇔ 0 ideal is prime • Proof (⟹) ○ Let a+P,b+P∈(R/P)∖\{P} ○ Then (a+P)(b+P)=ab+P=0 ○ So, ab∈P ○ Since P is prime, a∈P or b∈P ○ Therefore a+P=0 or b+P=0 • Proof (⟸) ○ Let a,b∈R, and suppose ab∈P, then ○ 0=ab+P=(a+P)(b+P) ○ Since R\/P is a domain, a+P=0 or b+P=0 ○ So a∈P or b∈P • Example ○ (x^2−1)⊆R[x] is not prime ○ R[x]\/(x^2−1)≅R×R, which is not a domain ○ Also, x^2−1∈(x^2−1), but x−1,x+1∉(x^2−1) Corollary 74 • Statement ○ If R is a commutative ring, and M⊆R is maximal, then M is prime • Proof ○ M is maximal ⇒R\/M is a field ⇒R\/M is a domain ⇒ M is prime Euclidean Domain • Definition ○ Let R be a domain ○ A norm on R is a function N:R→Z(≥0) s.t. N(0)=0 ○ R is called a Euclidean domain if R is equipped with a norm N s.t. ○ ∀a,b∈R with b≠0, ∃q,r∈R s.t. § a=qb+r § Either r=0 or N(r) N(b) • Example 1 ○ Z is a Euclidean domain, N(a)=|a| • Example 2 ○ If F is a field, then F is trivially a Euclidean domain ○ Take N:F→Z(≥0) to be any function s.t. N(0)=0 ○ Then, if a,b∈F, where b≠0, take q=a/b,r=0 • Example 3 ○ If F is a field, then F[x] is a Euclidean domain, with N(p)=degp ○ The division algorithm have is just polynomial division ○ Note § deg0=−∞∉Z(≥0), so this definition isn t quite right § To handle this problem, define a norm to take vales not in Z(≥0) § But any total ordered set in order-preserving bijection with Z(≥0) § (For instance, Z(≥0)∪{−∞}) Proposition 75 • Statement ○ Every ideal in a Euclidean domain R is principal ○ More precisely, if I⊆R is an ideal, then I=(d), where ○ d is an element of I with minimum norm • Proof ○ Let I⊆R be an ideal ○ If I=(0), then I is principal, so assume I≠(0) ○ {N(a)│a∈I∖{0} } has a minimal element, by well-ordering principal ○ Choose d∈I∖{0} s.t. N(d) is minimal ○ Certainly, (d)⊆I ○ Let a∈I, write a=qd+r, where q,r∈R and either r=0 or N(r) N(d) ○ Since r=a−qd∈I, N(r) can t be smaller than N(d) ○ So r=0⇒a=qd⇒a∈(d) • Example 1 ○ We haven t yet proven that F[x] is a Euclidean domain, when F is a field ○ Once we show this, we ll know F[x] has the property that ○ All of its ideals are principal • Definition ○ A domain in which every ideal is principal is called a principal ideal domain • Example 2 ○ Z[x] cannot be a Euclidean domain, since (2,x)⊆Z[x] is not principal Theorem 76 • Statement ○ Let F be a field ○ Then F[x] is a Euclidean domain ○ More specifically, if a,b∈F[x] where b≠0, then ○ ∃!q,r∈F[x] s.t. a=bq+r and degr degb • Proof (Existence) ○ We argue by induction on dega ○ If a=0, take q,r=0, so assume a≠0 ○ Set n≔dega, m≔degb ○ If n m, then take q=0,r=a ○ Assume n≥m ○ Write a=a_n x^n+…+a_1 x+a_0,b=b_m x^m+…+b_1 x+b_0 ○ Set a^′=a−a_n/b_m x^(n−m) b ○ Since a and a_n/b_m x^(n−m) b have the same leading coefficient, dega′ dega ○ By induction, ∃q^′,r∈F[x] with a^′=q^′ b+r with degr degb ○ Set q=q^′+a_n/b_m x^(n−m) b ○ Then a=a^′+a_n/b_m x^(n−m) b=q^′ b+r+a_n/b_m x^(n−m) b=(q^′+a_n/a_m x^(n−m) )b+r=qb+r • Proof (Uniqueness) ○ Suppose bq^′+r^′=a=bq+r where degr degb, and degr′ degb ○ Then deg(a−bq) degb and deg〖(a−bq′) degb 〗 ○ ⇒deg((a−bq)−(a−bq^′ ))=deg(bq^′−bq)=degb+deg(q^′−q) degb ○ ⇒deg(q^′−q) 0⇒q^′=q ○ It follows immediately that r^′=r](https://shawnzhong.com/wp-content/uploads/2018/05/img_5aee501817d08.png)
![Ideals Generated by Subset • Definition ○ Let R be a commutative ring ○ If A is a subset of R, then the ideal generated by A is ○ (A)≔{r_1 a_1+…+r_n a_n│n∈Z(≥1),r_i∈R,a_i∈A}⊆R ○ If A is finite, then we write (A) as (a_1,…,a_n ) • Notice: when |A|=1, (A) is a principal ideal • Example: (2,x)⊆Z[x] ○ Suppose, by way of contradiction, that (2,x)=(p) for some p∈Z[x] ○ Since 2∈(p) § 2=pq for some q∈Z[x] § 0=deg2=degp+degq § degp=degq=0 ○ Since x∈(p) § Choose r∈R[x] s.t. pr=x, then deg〖r=1〗 § Write r=ax+b,where a,b∈Z § Then pr=p(ax+b)=x § So pa=1, by comparing coefficients § Since p∈Z[x] and a∈Z, p∈{±1} ○ Therefore (2,x)=Z[x] ○ So, 1=2p^′+xq′, where p^′,q^′∈Z[x] ○ Evaluating both side at 0, we get 1=2p^′ (0)=0 ○ This is a contradiction, so (2,x)⊆Z[x] • Example: Z[x]\/(2,x)≅Z\/(2) ○ Define F:Z[x]→Z\/2Z given by a_0 x^n+…+a_1 x+a_0↦(a_0 ) ̅ ○ F is a ring homomorphism § F factors as Z[x]→Z→Z\/2Z, where p↦p(0)↦p((0) ) ̅ § Composition of homomorphisms is still a homomorphism ○ F is certainly surjective ○ (2,x)⊆kerF § Let p∈(2,x) § Then p=2g+xh for some g,h∈Z[x] § Since xh has no constant term, and 2g has even constant term § F(p)=F(2g)=F(g)=0 ̅∈Z\/2Z ○ kerF⊆(2,x) § Let p∈kerF § Write p=a_n x^n+…+a_1 x+a_0 § Write a_0=2b,b∈Z § Then p=x(a_n x^(n−1)+…+a_1 )+2b∈(2,x) ○ Therefore, kerF=(2,x) ○ By the First Isomorphism Theorem of , Z[x]\/(2,x)≅Z\/2Z≅Z\/(2) ○ Note: Z[x]\/(x,n)≅Z\/(n) Maximal Ideal • An ideal M in a ring R is maximal if • M≠R, and the only ideals containing M are M and R Proposition 71: Criterion for Maximal Ideal • Statement ○ If R is a commutative ring, and M⊆R is an ideal ○ Then M is maximal ⟺R\/M is a field • Proof (⟹) ○ The only ideals containing M are R and M ○ Thus, R\/M has exactly 2 idals, by the Correspondence Theorem ○ Namely, the zero ideal, and the entire ring ○ Let x+M∈R\/M ○ Suppose x∉M i.e. x+M≠0 ○ Then (x+M)=R\/M ○ So 1+M∈(x+M) ○ Choose y+M∈R\/M s.t. (x+M)(y+M)=1+M ○ This shows x+M is a unit ○ Therefore R\/M is a field • Proof (⟸) ○ Suppose R\/M is a field ○ Then R\/M has exactly two ideals, 0 and R\/M ○ By the Correspondence Theorem, ○ There are exactly two ideals containing M: R and M ○ By definition of maximal ideal, M is maximal Examples of Maximal Ideals • What are the maximal ideals in Z? ○ (n)∈Z is maximal ⟺Z\/(n) is a field ⟺ n is prime • Is (x)⊆Z[x] maximal? ○ No, (x)⊊(2,x)≠Z[x] ○ Also, by First Isomorphism Theorem, Z[x]\/(x)≅Z, but Z is not a field § Define a ring map Z[x]→Z given by p→p(0) § F is surjective, and kerF=(x) • Is (x^2+1)⊆R[x] maximal? ○ R[x]\/(x^2+1)≅ℂ is a field • Is (x^2−1)⊆R[x] maximal ○ R[x]\/(x^2−1)≅R×R is not a field, since (1,0) is not a unit ○ Another way to see (x^2−1) is not maximal § (x^2−1)⊊(x−1)⊊R[x] § (x^2−1)⊊(x+1)⊊R[x]](https://shawnzhong.com/wp-content/uploads/2018/05/img_5aee4f89b32e7.png)
![Theorem 70: The First Isomorphism Theorem for Rings • Statement ○ If f:R→S is a ring homomorphism, then there is an induced isomorphism ○ f ̅:R\/kerf→im(f), given by r+kerf↦f(r) • Proof ○ We need only check f ̅(1_(R/kerf ) )=1_S, and f ̅ preserves multiplication ○ f ̅(1_(R/kerf ) )=f ̅(1+kerf )=f(1_R )=1_S ○ f ̅((r_1+I)(r_2+I))=f ̅(r_1 r_2+I)=f(r_1 r_2 )=f(r_1 )f(r_2 )=f ̅(r_1+I) f ̅(r_2+I) • Example: R[x]\/(x^2+1)≅ℂ ○ Let F:R[x]→ℂ given by p↦p(i) ○ F is a ring homomorphism § In fact, if R is a subgring of some ring S, and s∈S, then § The function R[x]→S given by p↦p(s) is a ring homomorphism ○ F is surjective § If a+bi∈ℂ, then F(a+bx)=a+bi ○ (x^2+1)⊆kerf § If p(x^2+1)∈(x^2+1), then § F(p(x^2+1))=F(p)F(x^2+1)=p(i)p(i^2+1)=0 ○ kerf⊆(x^2+1) § Let p∈kerf § Using polynomial division, we can find q,r∈R[x] s.t. § p=q(x^2+1)+r where degr deg(x^2+1)=2 § Write r=ax+b for some a,b∈R § Since p∈kerf, p(i)=0 § 0=p(i)=q(i)×(i^2+1)+r(i)=r(i)=ai+b § So a=b=0 § Therefore p=q(x^2+1), and p∈(x^2+1) ○ Therefore, kerf=(x^2+1) ○ By the First Isomorphism Theorem of Rings, R[x]\/(x^2+1)≅ℂ • Example: R[x]\/(x−a)≅R, where a∈R ○ Let F:R[x]→R given by p↦p(a) ○ F is surjective § F(b)=b,∀b∈R ○ F is a ring homomorphism ○ (x−a)⊆kerf § If p(x−a)∈(x−a), then § F(p(x−a))=F(p)F(x−a)=p(a)p(a−a)=0 ○ kerf⊆(x−a) § Let p∈kerf § Divide x−a into p to obtain q,r∈R[x] s.t. § p=q(x−a)+r, where degr 1 § Since p∈kerf, 0=p(a)=q(a)(a−a)+r=r § Thus r=0, so p=q(x−a)∈(x−a) ○ Therefore, kerf=(x−a) ○ By the First Isomorphism Theorem of Rings, R[x]\/(x−a)≅R • Example: R[x]\/(x^2−1)≅R×R ○ Recall: Chinese Remainder Theorem § If I,J are ideals in a commutative ring R s.t. I+J=R § Then R\/IJ≅R\/I×R\/J, where § I+J={x+y│x∈I,y∈J} § IJ={x_1 y_1+…+x_n y_n│n∈Z(≥1),x_i∈I,y_i∈J} ○ (x^2−1)⊆(x+1)(x−1) § This is obvious, since x^2−1∈(x+1)(x−1) ○ (x+1)(x−1)⊆(x^2−1) § Let p_1 q_1+…+p_n q_n∈(x−1)(x+1), where p_i∈(x−1),q_i∈(x+1) § Each term p_i q_i is of form □ f_i (x−1)⋅g_i (x+1)=f_i g_i (x^2−1) for some f_i,g_i∈R § Thus p_i q_i∈(x^2−1)⇒p_1 q_1+…+p_n q_n∈(x^2−1) ○ Thus (x^2−1)=(x+1)(x−1) ○ R[x]\/(x+1)(x−1)≅R×R § 1/2 (x+1)−1/2 (x−1)=1∈R[x] § ⇒(x+1)+(x−1)=R[x] § ⇒1∈(x+1)+(x−1) § Chinese Remainder Theorem implies R[x]\/(x+1)(x−1)≅R×R ○ Therefore, R[x]\/(x^2−1)≅R×R Other Isomorphism Theorems for Rings • The Second Isomorphism Theorem for Rings ○ If I is an ideal of a ring R, and S is a subring of R ○ Then S+I is also a subring of R, where ○ I is an ideal of S+I, and (S+I)\/I≅S\/(I∩S) • The Third Isomorphism Theorem for Rings ○ If I⊆J are ideals of a ring R, then (R\/I)⁄(J\/I)≅R\/J • Correspondence Theorem ○ If R is a ring, and I is an ideal of R ○ Then there is a bijection {ideals of R\/I}⟷{ideals of R containing I}](https://shawnzhong.com/wp-content/uploads/2018/05/img_5aec790d90b90.png)
