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## MAT 441, Studyguide

1 review
by: Deepthi Notetaker

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# MAT 441, Studyguide MAT 441

Deepthi Notetaker
ASU

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## About this Document

This is review of all lemmas, definitions for exam-2
COURSE
Ring Theory
PROF.
Dr. Andrew Bremner
TYPE
Study Guide
PAGES
5
WORDS
CONCEPTS
Math, Algebra, ringtheory
KARMA
50 ?

## 2

1 review
"Almost no time left on the clock and my grade on the line. Where else would I go? Deepthi has the best notes period!"
Claude Stracke

## Popular in Mathematics (M)

This 5 page Study Guide was uploaded by Deepthi Notetaker on Tuesday April 19, 2016. The Study Guide belongs to MAT 441 at Arizona State University taught by Dr. Andrew Bremner in Winter 2016. Since its upload, it has received 20 views. For similar materials see Ring Theory in Mathematics (M) at Arizona State University.

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Date Created: 04/19/16
(Mid­term 2) Theory​   Lemmas, theorems and definitions    Polynomials over a ring R  Definition:  Polynomials over a ring R (with coefficients in R) are expressions of type  r0​ 1​+r2​+........n​​ where, x is referred to an indeterminate (n≥0 an1​ 2​…..,n​are  coefficients) subject to certain conventions.  Lemma: ​ All polynomials together with ‘+’, ‘.’ forms a ring called polynomial ring over  R; denoted as R[x] (contains R)  Lemma: Suppose R is an integral domain . Let f(x), g(x) ∈ R[x]. Then  deg(f(x).g(x))=deg f(x) + deg g(x)  Theorem: ​ Suppose R is an integral domain, then the ring R[x] is an integral domain.  Theorem: Let R be an integral domain. Let f(x)∈R[x]  Then f(x) is a unit in R[x]  Division algorithm for polynomials  Let F be a field  Let a(x), b(x) ∈ F[x], then there exists polynomials q(x), r(x) satisfying   1. a(x)=b(x)q(x)+r(x)  2. r(x)=0 or deg r(x)<deg b(x)  Moreover, q(x), r(x) are unique  Let f:R→S be a homomorphism of rings. Let K⊆R, K={r∈R|f(r)=0​ s​(K is the kernel of  the homomorphism). Show that K is the subring of R  Def:​ Two polynomials a(x), b(x) are said to be associates if they differ only by a unit i.e.  there exists λ∈F, λ being non­zero, such that b(x)=λa(x)  Thm: ​ If b(x)|a(x), then λb(x)|a(x)  Thm:​  If b(x)|a(x), then deg(b(x))≤deg(a(x))  Def:​ A polynomial a(x) is monic if its lead coefficient is 1.  Any non­zero a(x)∈F(x) is an associate of a monic polynomial  Def:​ d(x) is a greatest common divisor(gcd) of a(x), b(x) in F[x] if   1. d(x)|a(x); d(x)|b(x)  2. If e(x)|a(x) and e(x)|b(x) then deg(e(x))≤deg(d(x))  Def:​ Let a(x), b(x)∈F[x], then there exists a unique monic gcd of a(x), b(x)  Def: A polynomial f(x)∈F[x] is irreducible if and only if it’s divisors are the units and  the associates of f(x)  Lemma:​  Every polynomial of degree 1 is irreducible in F[x]  Theorem:​  f(x) is irreducible ⇔f(x) cannot be factored into two polynomials each with  degree less than deg(f)  Thm: ​If f(x) is irreducible and f(x)|g(x)h(x) then f(x)|g(x) or f(x)|h(x)  Thm:​ Let f(x)∈F[x] be a non­constant polynomial. Then f(x) is the product in an  essentially unique way of a finite number of irreducible polynomials  The remainder theorem  Let F be a field, with f(x)∈F[x]. Let a∈F. The remainder when f(x) is divided by x­a is  equal to f(a)  Congruence in polynomial rings    Def: Let F[x] be a ring of polynomials over a Field F, we define what it means for two  polynomials f(x), g(x) to be congruent mod m(x)  f(x≡g​(x) mod m(x)) means m(x) divides f(x)­g(x)    Lemma:​  Congruence in polynomials is an equivalence relation  Lemma: ​ If≡k mod m, g≡l mod m , then f+g≡k+l mod m and fg≡kl mod m    n​ n​ Corollary:Suppose f≡k mod m. Then f=k​ mod m for n=1, 2, 3, …….    Def: Fix m(x)∈F[x]. The congruence class [f] of f∈F[x] is defined by [f]≡f ∈F[x] | g​ mod m​}  Theorem: ​[f(x)]=[g(x)] ⇔ ≡g(x) mod m(x)  In F[x], we call the set of all possible classes of remainder polynomials upon division by  m(x) as “F[x] modulo m(x)” i.e. the ring F[x]/m(x) of classes of remainder polynomials  mod m(x)  Theorem:​  F[x]/m(x) is a field ⇔ m(x) is an irreducible polynomial  Ideals  Def: I is said to be an ideal of R if for all x∈I, for all r∈R we have rx∈I, xr∈I (say I has  absorption property)  Note: see for left ideal and right ideal  Theorem: ​I is an ideaℤ⇔ I=n​ℤ for some∈ℤ​  Theorem: ​A non­empty subset I of the ring R is a⇔ 1. a­b∈I for all a∈I; b∈I                                                                                                   2. For all a∈I, r∈R, ra∈I; ar∈I  Let c, , c, …………, c∈R. Consider R is commutative, I=c+r​c​+.....c​| ∈R}.  1​ 2​ 3​ n​ 1​1​ 2​2​ n​n ​ i​ Then I is an ideal of R     Def: An ideal that is generated by a single element ‘a’, namely, I=aR={ar | r∈R} is  called the principal ideal generated by ‘a’  Def: Let R be a ring. Let I be an ideal of R. We say r, s∈R are congruent modulo I ⇔ r­s  ∈ I  Cosets  Def:  R/I is the set of all cosets of I in R = {a+I | a∈R}  Lemma: ​ Congruence modulo I is an equivalence relation  Lemma: ​ If ≡b mod I, c≡d mod I then a+c≡b+d mod I and ac≡bd mod I  Fundamental theorem:​  [x]=[y] (or x+I = y⇔ x≡y mod I  Corollary: x]=[0]⇔ x∈I   Theorem: ​R/I, +, . is a ring  Lemma: ​ If R is commutative, then R/I is commutative ring. If R has a multiplicative  identity , then R/I has a multiplicative identity   R​ Def: LetØ:R→S be ring homomorphism. The Kernel of Ø, ker(Ø) is defined by ker(Ø) =  {r∈R | Ø(r)=0S​ Lemma: ​ ker(Ø) is an ideal of R  Lemma: ​ Let Ø:R→S be a ring homomorphism, then Ø is inje⇔ ker(Ø​)={0}  Lemma:​  Let R be a ring, with ideal I. Then, we have an associated quotient ring R/I.   Defineπ : R→R/I                    a→a+I  Then π is a surjective ring homomorphism and ker(π)=I  Lemma:​  Im(Ø)={s∈S | s=Ø(r) for some∈R​}, then Im(Ø) is a subring of S  First Isomorphism theorem: ​  Let Ø:R→S be a surjective ring homomorphism. Then K=Ker(Ø) is an ideal of R and  R/K≅​S  Def: An ideal≠R is said to be prime idea∈I ⇒ a∈I or b∈I  Def: An ideal≠R is said to be maximal if there is no ideal other than R containing I i.e.  If I⊆J⊆R for some ideal J, then J=I or J=R  Theorem:​  R/I is an integral domain iff I is prime ideal                        I is maximal ideal iff R/I is a field  Corollary: Suppose I is a maximal ideal, then I is a prime ideal  Def: Let a,∈R we say a divides b in R if there exists c∈R satisfying ac=b  Theorem: ​Let u∈R be a unit. Let r∈R, then u divides r                        Any associate of r divides r  Theorem: ​Let p∈R be non­zero and not a unit. Then p is irreducible in R ⇔ (p=rs ⇒ r  is a unit or s is a unit)  Euclidean domain  Def: An integral domain R is said to be Euclidean domain if there ▯ from nction ​ non­zero elements to non­negative integers   ▯:R­{0}→ℤ+​which satisfies the following  1.▯(a​​(ab) for all nonzero∈R  b​ 2. Foa, ∈R, b​≠0 then there exists∈R satisfying a=bq+r with r=0▯(r)<▯(b)  Theorem:​  Gvenα​,β ∈ [i]β≠0, there exists q, [i] satisfyα=​βq+r and  0≤▯(r)<▯(β)  Let R be an Euclidean domain with Euclidean function ▯. Then the following conditions  are equivalent:  1. u is a unit  2. ▯(u)=▯(1)  3. ▯(c)=▯(uc) for so∈R, c≠0  Def: R is a Principal Ideal Domain (PID) if R is an integral domain in which every ideal  is principal  Theorem: ​Let R be a n Euclidean domain, with Euclidean f▯:R​→​ℕ, then R is a  PID  Lemma: ​ Let R be an integral domain. Then  1. (a)⊆(b⇔ b divides a  2. (a)=(b) ⇔ a|b and b|a  3. (a(b) ⇔ b|a and b is not an associate of a  Sub lemma: ​a|b and b|a ⇔ a and b are not associates  Corollary: The set of all the Euclidean domains is contained in the set of all principal  ideal domains  Def: Let R be an integral domain, R is said to satisfy the ascending chain  condition(a.c.c) if whenever )⊆(a​)⊆…………. Then the chain “stops” in the sense that  1​ 2​ there is an intege0​​ℕ such that (1​(a​2​⊆…………(a 0 −1⊆(a 0=(a n0+1)=(an0+2 )=........  Theorem: ​ Let R be a principal ideal domain. Then R satisfies the ascending chain  condition  Theorem: ​ Let R be a principal ideal domain. Let p be an irreducible element of R. Then  p|bc ⇒ p|b or p|c  Theorem: ​ In a principal ideal domain, every non­zero non­unit element can be written  essentially uniquely as a product of irreducibles  Unique Factorization Domain (UFD)  Def:​ UFD is defined to be an integral domain in which every non­zero unit can be  written essentially uniquely as a product of irreducibles. From previous theorem, we  already proved that every PID is a UFD  Def:​ In the ring R define a norm function →ℤ​ by  N(a+b√D)=(a+b√D)(a­b√D)=a​ 2­Db​

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