# 1 Rings 1 RINGS 1. Theorem 1.1 (Substitution Principle). Let ϕ : R R be a ring homomorphism

Size: px
Start display at page:

Download "1 Rings 1 RINGS 1. Theorem 1.1 (Substitution Principle). Let ϕ : R R be a ring homomorphism"

Transcription

1 1 RINGS 1 1 Rings Theorem 1.1 (Substitution Principle). Let ϕ : R R be a ring homomorphism (a) Given an element α R there is a unique homomorphism Φ : R[x] R which agrees with the map ϕ on constant polynomials and sends x α. (b) Given elements α 1,, α n R there is a unique ring homomorphism Φ : R[x 1,..., x n ] such that Φ R = φ and Φ[x i ] = α i. Lemma 1.2. For every ring R there is a unique ring homomorphism Z R. Lemma 1.3. If R is a ring and a R then {ra : r R} = (a) is an ideal. Theorem 1.4. A ring R is a field if and only if it has exactly two ideals. Corollary 1.5. Let F be a field and R a non-zero ring. Then every homomorphism ϕ : F R is injective. Lemma 1.6. Every ideal in Z is principle. Theorem 1.7. Let g(x) be a monic polynomial in R[x] and let α be an element of R such that g(α) = 0. Then x α divides g(x). Theorem 1.8. If F is a field then every ideal of F [x] is principle. Corollary 1.9. Let F be a field and let f, g F [x] which are both non-zero. Then there is a unique monic d(x) F [x] called the greatest common divisor of f, g such that (a) d generates the ideal (f, g) of F [x] generated by f, g. (b) d divides f and g (c) If h is any divisor of f and g then h divides d. (d) There are p, q F [x] such that d = pf + qg Theorem Let I be an ideal of a ring R. (a) There is a unique ring structure on the set R/I such that the canonical map π : R R/I sending a a + I is a homomorphism.

2 1 RINGS 2 (b) The kernel of π is I. Theorem 1.11 (Mapping Property of Quotient Rings). Let f : R R be a ring homomorphism with kernel I and let J be an ideal which is contained in I. Denote R/J by R. (a) There is a unique homomorphism f : R R such that fπ = f (b) First Isomorphism Theorem: If J = I then f maps R isomorphically to the image of f. Theorem 1.12 (Correspondence Theorem). Let R = R/J and let π denote the canonical map R R (a) There is a bijective correspondence between the set of ideals of R which contain J and the set of all ideals of R given by I π(i) and I π 1 (I) (b) (Third Isomorphism Theorem) If I R corresponds to I R then R/I and R/I are isomorphic rings. Definition Let R be a ring extension of R and let α R. WE define R[α] = {Σr i α i : r i R} Theorem Let R R and let α R. Then there is a unique map ϕ : R[x] R such that ϕ is the identity on R and takes x α. Further F [α] = im[α] Definition C = R[i] Theorem Let R be a ring and let f(x) be a monic polynomial of positive degree with coefficients in R. Let R[α] be the ring obtained by adjoining an element satisfying f(α) = 0. The elements of R[α] are in bijective correspondence with vectors (r 0,, r n 1 ) R n via a map µ where µ(r 0,, r n 1 ) = r 0 + r 1 α + r 2 α r n 1 α n 1 Theorem Let R be a ring and let a, b R such that ab = 0. Further let c R[α] be such that ψ(a)c = 1 where ψ : R R[α]. Then if R[α] 0 we must have ψ(b) = 0. Definition We say b R is a Zero Divisor if there is a non-zero a R such that ab = 0

3 1 RINGS Integral Domains Definition A ring R is an Integral Domain if it has no zero divisors. I.e. 0 1 and if ab = 0 then a = 0 or b = 0. Theorem If R is an integral domain then it satisfies the cancelation law ( a, b, c)(a 0) (ab = ac) (b = c) Theorem If R is an integral domain then so is R[x]. Theorem Let R be an integral domain with finitely many elements is a field. Theorem Let R be an integral domain. Then there exists an embedding φ : R F into a field F. Definition Let R be an integral domain. A fraction in R will be a pair a/b where a, b R and b 0. Two fractions a 1 /b 1, a 2 /b 2 are called equivalent a 1 /b 1 a 2 /b 2 if a 1 b 2 = a 2 b 1. There is a ring structure on F = {[a/b] equivalence classes of fractions} such that F is a field. It is called the Field of Fractions of R. Theorem Let R be an integral domain with field of fractions F and let ϕ : R K be an injective homomorphism from R into a field K. Then Φ(a/b) = ϕ(a)ϕ(b) 1 Definition An ideal M is Maximal if M R but M is not contained in any other ideals other than M, R. Theorem If R is an integral domain then M is a maximal ideal if and only if R/M is a field. Theorem The maximal ideals of the ring Z of integers are the principle ideals generated by the prime integers. Theorem The maximal ideals of the polynomial ring C[x] are the principle ideals generated by the linear polynomials x a.

4 1 RINGS 4 Theorem 1.30 (Hilbert s Nullstellensatz). The maximal ideals of the polynomial ring C[x 1,..., x n ] are in a bijective correspondence with points of complex n-dimensional space. a = a 1,, a n M a = (x 1 a 1, x 2 a 2,..., x n a n ). Theorem If a, b Z have no factor in common other than ±1 then there are c, d such that ac + bd = 1. Theorem Let p be a prime integer and let a, b be integers. Then if p divides ab p divides a or p divides b. Theorem 1.33 (Fundamental Theorem of Arithmetic). Every integer a 0 can be written as a product a = cp 1 p k where c is ±1 and each p i is prime. And further, up to the ordering this product is unique. Theorem Let F be a field (a) If two polynomials f, g F [x] have no common non-constant factors then there are polynomials r, s F [x] such that rf + sg = 1 (b) If an irreducible polynomial p F [x] divides a product fg then p divides one of the factors. (c) Every nonzero polynomial f F [x] can be written as a product cp 1 p n where c F [x] and the p i are monic irreducible polynomials and n 0. This factorization is unique except for the ordering of terms. Theorem Let F be a field and let f(x) be a polynomial of degree n with coefficients in F. Then f has at most n roots in F. Definition Let R be an integral domain. If a, b R we say a divides b if ( r R)ar = b We say that a is a proper divisor of b if b = qa for some q R and neither q R and neither q nor a is a unit.

5 1 RINGS 5 We say a non-zero element a R is irreducible if it is not a unit and if it has no proper divisor. We say that a, a R are associates if a divides a and a divides a. It is easy to show that if a, a are associates then a = ua for some unit u R. Theorem Let R be an integral domain. u is a unit (u) = (1) a, a are associates (a) = (a ) a divides b (a) (b) a is a proper divisor of b (1) (a) (b) Theorem Let R be an integral domain. Then the following are equivalent. (a) For every a R, a 0 if a is not a unit then where each b i is irreducible. a = b 1 b n (b) R does not contain an infinite increasing chain of principle ideals (a 1 ) (a 2 ) (a 3 ) Definition We say that an integral domain R is a Unique Factorization Domain (UFD) if (i) Existence of factors is true for R (ii) If a R and a = p 1 p n and a = q 1 q m where p i, q j are irreducible. Then m = n and after reordering p i, q i are associates for each i. Definition Let R be an integral domian. p R is prime if p 0 and ( a, b R) if p divides ab then p divides a or p divides b. Theorem Let R be an integral domain such that existence of factorization holds. Then R is a UFD if and only if ever irreducible element is prime.

6 2 FIELDS 6 Theorem Let R be a UFD and let a = p 1... p n, b = q 1... q m be prime factorizations. Then a divides b if and only if m n and after reordering p i, q i are associates for all i n. Theorem Let R be a UFD and a, b R such that at least one of a 0 or b 0 holds. Then there exists a greatest common divisor d of a, b such that (i) d divides a and b (ii) if an element e divides a and b then e divides d Definition Let R be an integral domain. We say R is a principle ideal domain (PID) if every ideal is principle. Theorem In an integral domain all prime elements are irreducible. In a PID all irreducible elements are prime. Theorem Every PID is a UFD Lemma Let R be any ring. Then the union of an increasing chain of ideals is an ideal. Theorem Let R be a PID (a) Let 0 p R. Then R/(p) is a field if and only if p is irreducible. (b) The maximal ideals of R are those generated by irreducible elements. 2 Fields Definition 2.1. Let F be a field. And let α K such that K F. We say that α is algebraic over F if there is a polynomial over F which is satisfied by α. I.e. if α n + a n 1 α n a 1 α + α = 0 for some a 0,..., a n 1 F We say that α is transcendental over F if it is not algebraic over F.

7 2 FIELDS 7 Lemma 2.2. Let F K be fields with α K. Then if ϕ α : F [x] K f(x) f(α) we have α is transcendental if and only if ϕ is injective. Or more specifically if the kernel of ϕ is 0. Definition 2.3. Let F K be fields with α K. Further let ϕ α : F [x] K f(x) f(α) We then know that ker(ϕ α ) is principle as F [x] is a principle ideal domain. So in particular it is generated by a single element f α (x) F [x]. But because K is a field we must have f α (x) is irreducible (because otherwise K would have a zero divisor) Hence f α (x) is the only irreducible polynomial in (f α (x)) (because every element of the ideal is a multiple of f α (x)) and we call f α the Irreducible Polynomial for α over F. Definition 2.4. Let F (α) be the smallest field containing both α and F. Similarly let F (α 1,..., α n ) be the smallest field containing α 1,..., α n and F. Lemma 2.5. Recall that F [α] is the ring {Σa n α n : a n F } and is the smallest ring containing both F and α. We then have F (α) is isomorphic to the field of fractions of F [α] In particular we have that if α is transcendental then F [x] F [α] is an isomorphism and hence F (α) is isomorphic to the field F (x) of rational functions. Theorem 2.6. (a) Suppose that α is algebraic over F and let f(x) be its irreducible polynomial over F. The map F [x]/(f) F [α] is an isomorphism and F [α] is a field. Thus F [α] = F (α) (b) More generally let α 1,..., α n be algebraic elements of a field extension K of F. Then F [α 1,..., α n ] = F (α 1,..., α n ). Theorem 2.7. Let α be an algebraic over F and let f(x) be its irreducible polynomial. Suppose f(x) has degree n. Then (1, α,..., α n 1 ) is a basis for F [α] as a vector space over F

8 2 FIELDS 8 Theorem 2.8. Let α K and β L be algebraic elements of two extensions of F. There is an isomorphism of fields σ : F (α) F (β) which is the identity on F and which sends α β if and only the irreducible polynomials for α and β over F are equal. Definition 2.9. Let K, K be field extensions of F. An isomorphism ϕ : K K which restricts to the identity on F is called an Isomorphism of field extensions of an F -isomorphism Theorem Let ϕ : K K be an isomorphism of field extensions of F and let f(x) be a polynomial with coefficients in F. Let α be a root of f in K and let α = ϕ(α) be its image in K. Then α is also a root of f. Definition Let K be a field extension of a field F. We can always regard K as a vector space over F where addition is field addition and multiplication by F is simply multiplication. We say that the degree of K as an extension of F is the dimension of the vector space (denoted [K : F ]). Extensions of degree 2 are called quadratic, of degree are called cubic, ect. The term degree comes from the case when K = F (α) for an algebraic α over F and so (1, α,..., α n 1 ) form a basis for the vector space (where n is the degree of the irreducible polynomial). In this case we also call the degree the degree of α over F Theorem If α is algebraic over F then [F (α) : F ] is the degree of the irreducible polynomial of α. Theorem Let F K L be fields. Then [L : F ] = [L : K][K : F ]. These are called towers of field extensions. Corollary Let K be an extension of F of finite degree n and let α K. Then α is algebraic over F and it s degree divides n.

9 2 FIELDS 9 Corollary Every irreducible polynomial in R[x] has degree 1 or 2 Theorem Let K be an extension of F. The elements of K which are algebraic over F form a subfield of K. Definition We say that an extension K of F is an algebraic extensions (and K is algebraic over F ) if every element of K is algebraic over F. Theorem Let F K L be fields. If L is algebraic over K and K is algebraic over F, then L is algebraic over F. Theorem Let F be a field. Let F = F [x]/i. Then I = (f) for some f F [x] and F is a field if and only if the (f) is irreducible. Corollary Let F be a field and let f F [x] be irreducible. Then K = F [x]/(f) is a field extension of F and the residue of x is a root of f in K. Theorem Let F be a field and let f(x) be a monic polynomial in F [x] of positive degree. Then there is a field extension K of F such that f(x) factors into linear factors over K. Theorem Let f, g F [x] and let K be a field extension of F. (a) Division with remainder of g by f gives the same answer whether carried out in F [x] or in K[x]. (b) f divides g in K[x] if and only if f divides g in F [x]. (c) The monic greatest common divisor d of f, g is the same wether computed in K[x] or in F [x]. (d) If f and g have a common root in K, then they are not relatively prime in F [x]. Conversely if f and g are not relatively prime in F [x] then there exists an extension field L in which they have a common root. (e) If f is irreducible in F [x] and f and g have a common root in K then f divides g in F [x]. Definition Let F be a field and let f F [x]. If f(x) = a n x n + a n 1 x n a 1 x + a 0

10 2 FIELDS 10 then we define f (x) = na n x n 1 + (n 1)a n 1 x n 2 a 1 Where we interpret n as the image of n Z under the unique ring homomorphism Z F. It can be shown that things like the product rule hold for these formal derivatives. Lemma Let F be a field and let f(x) F [x]. Let α F be a root of f(x). Then α is a multiple root, i.e. that (x α) 2 divides f(x) if and only if α is a root of f(x) and of f (x). Theorem Let f(x) F [x] where F is a field. Then there exists a field extension K of F in which f has a multiple root if and only if f and f are not relatively prime. Theorem Let f be an irreducible polynomial in F [x]. Then f has no multiple roots in any field extension of F unless the derivative f is the zero polynomial. In particular if F has characteristic 0, then f has no multiple root. Definition Let K be a field extension of F. Let α 1,..., α n be a sequence of elements of K. We say that α 1,..., α n are Algebraically Dependent if there is a polynomial f F [x 1,..., x n ] such that f(α 1,..., α n ) = 0. We say α 1,..., α n are algebraically independent otherwise. Lemma Let F K. α 1,..., α n are algebraically independent if and only if the substitution map ϕ : F [x 1,..., x n ] K which takes f(x 1,..., x n ) to f(α 1,..., α n ) has ker(ϕ) = 0. Corollary If α 1,..., α n are algebraically independent over F then F (α 1,..., α n ) is isomorphic to F (x 1,..., x n ) the field of rational functions in x 1,..., x n. Definition An extension o the form F (α 1,..., α n ) where α 1,..., α n are algebraically independent is called a Pure Transcendental extension. Definition A transcendence basis for a field K of F is a set of elements α 1,..., α n such that K is algebraic over F (α 1,..., α n )

11 2 FIELDS 11 Theorem Let (α 1,..., α m ) and (β 1,..., β n ) be elements in a field extension K of F which are algebraically independent. If K is algebraic over F (β 1,..., β n ) then m n and (α 1,..., α m ) can be completed to a transcendence basis for K by adding n m many of the β i. Corollary Any two transcendence basis for a field extension F K have the same number of elements. 2.1 Finite Fields Definition We say that q = p r = K is the order of a field K. When dealing with finite fields p will always be a prime and q will be the order of the field we are talking about. Fields with q = p r elements are often denoted F q. Theorem Let p be a prime and let q = p r be a power of p with r 1. Let K be a field with order q. (a) There exists a field of order q (b) Any two fields of order q are isomorphic. (c) Let K be a field of order q. The multiplicative group K of nonzero elements of K is a cyclic group of order q 1. (d) The elements of K are roots of the polynomial x q x. This polynomial has distinct roots and it factors into linear factors in K (e) Every irreducible polynomial of degree r in F p [x] is a factor of x q x. The irreducible factors of x q x in F p [x] are precisely the irreducible polynomials in F p [x] whose degree divides r. (f) A field K of order q contains a subfield of order q = p k if and only if k divides r. Corollary Let K be a finite field. Then there is an element a K such that for all b K, b 0 there is an n ω such that a n = b. Definition A generator for the cyclic group F p is called an primitive element modulo p.

12 2 FIELDS 12 Theorem Let p be a prime and let q = p r. (a) The polynomial x q x has no multiple root in any field L of characteristic p. (b) Let L be a field of characteristic p and let K be the set of roots of x q x in L. Then K is a subfield of L. Theorem Let L be a field of characteristic p, and let q = p r. Then in the polynomial ring L[x, y], we have (x + y) q = x q + y q Lemma Let k be an integer dividing r, say r = ks, and let q = p r, q = p k. Then x q x divides x q x. Definition A field F is algebraically closed if every polynomial f(x) F [x] has a root in F. Theorem 2.42 (Fundamental Theorem of Algebra). Every nonconstant polynomial with complex coefficients has a complex root. Definition Let F be a field. F is an algebraic closure of F if F is algebraically closed F is algebraic over F. Corollary Let F be a subfield of C. Then the subset F of C consisting of all numbers algebraic over F is an algebraic closure of F. Theorem Every field F has an algebraic closure and if K 1, K 2 are algebraic closures of F there is an isomorphism ϕ : K 1 K 2 which is the identity map on F. Corollary Let F be an algebraic closure of F, and let K be any algebraic extension of F. Then there is a subextension K F which is isomorphic to K. Lemma Let f(x) be a complex polynomial. Then f(x) takes on a minimum value at some point x 0 C.

13 2 FIELDS Field Extensions Definition If K is a field extension of F we say K/F. Definition An F -automorphism of K is an automorphism of K which is the identity on F. Definition The group of all F -automorphisms of K is called the Galois Group of the field extension (G(K/F )) Theorem For any finite extension K/F the order of G(K/F ) divides the degree [K : F ] of the field extension. Definition A finite field extension K/F is called a Galois Extension if G(K/F ) = [K : F ] Definition Let G be a group of automorphisms of K. The set of elements fixed by every element of G is called the fixed field of G K G = {α K : ϕ(α) = α for all ϕ G} Corollary Let K/F be a Galois extension with Galois group G = G(K/F ). The fixed field of G is F. Definition Let f(x) F [x] be a nonconstant monic polynomial. A splitting field for f(x) over F is an extension K of F such that (i) f(x) factors into linear factors in K : f(x) = (x α 1 ) (x α n ) with α i K (ii) K is generated by the roots of f(x) : K = F (α 1,..., α n ) Theorem If K is a splitting field of a polynomial f(x) over F then K is a Galois extension of F. Conversely, every Galois extension is a splitting field of some polynomial f(x) F [x]. Corollary Every finite extension is contained in a Galois extension. Corollary Let K/F be a Galois extension and let L be an intermediate field: F L K. Then K/L is a Galois extension too.

14 2 FIELDS 14 Theorem (a) Let K be an extension of a field F, let f(x) be a polynomial with coefficients in F and let σ be an F -automorphism of K. If α is a root of f(x) in K then σ(α) is also a root. (b) Let K be a field extension generated over F by elements α 1,..., α r and let σ be an F -automorphism of K. If σ fixes each of the generators α i then σ is the identity automorphism. (c) Let K be a splitting field of a polynomial f(x) over F. group G(K/F ) operates faithfully on the set {α 1,..., α r }. The Galois Theorem 2.60 (The Main Theorem). Let K be a Galois extension of a field F and let G = G(K/F ) be its Galois group. The function H K H is a bijective map from the set of subgroups of G to the set of intermediate fields F L K. It s inverse is L G(K/L) This correspondence has the property that if H = G(K/L) then [K : L] = H hence [L : F ] = [G : H] Theorem 2.61 (Existence of a primitive element). Let K be a finite extension of a field F of characteristic 0. there is an element γ K such that K = F (γ). Definition We call an element γ K such that F (γ) = K a primitive element for K over F. Theorem Let G be a finite group of automorphisms of a field K and let F be its fixed field. Let {β 1,..., β r } be the orbit of an element β = β 1 K under the action of G. Then β is algebraic over F, it s degree over F is r and its irreducible polynomial over F is g(x) = (x β 1 ) (x β r ). Further note that r divides G. Corollary Let K/F be a Galois extension. Let g(x) be an irreducible polynomial in F [x]. If g has one root in K then it factors into linear factors in K[x].

15 2 FIELDS 15 Theorem Let G be a group of order n of automorphisms of a field K and let F be its fixed field. Then [K : F ] = n. Corollary Let G be a finite group of automorphisms of a field K and let F be its fixed field. Then K is a Galois extension of F and its Galois group is G. Definition Let ϕ : F F be an isomorphism. Then ϕ extends to an isomorphism F [x] F [x] by an x n a n x n where a n = ϕ(a n ). We denote by f(x) the image of f(x) under this map. Lemma Let f(x) F [x] be an irreducible polynomial. Let α is a root of f(x) in an extension field K of F and let α be a root of f(x) in an extension field K of F. Then there is a unique isomorphism ϕ 1 : F (α) F (α) which restricts to ϕ on a the subfield F and which sends α to α. Theorem Let ϕ : F F be an isomorphism of fields. Let f(x) F [x] be nonconstant and let f(x) be the corresponding polynomial in F [x]. Let K and K be the splitting fields for f(x) and f(x). Then there is an isomorphism ψ : K K which restricts to ϕ on F. Corollary Any two splitting fields of f(x) F [x] are isomorphic. Theorem Let K be the splitting field of a polynomial f(x) F [x]. Then K is a Galois extension of F. That is G(K/F ) = [K : F ] Lemma With the notation of the previous lemma, the number of isomorphisms ψ : K K extending ϕ is equal to [K : F ] Definition Since any two splitting fields K of f(x) F [x] are isomorphic, the Galois group G(K/F ) depends, up to isomorphism, only on f. It is often referred to as the Galois group of the polynomial over F. Corollary Let K/F be a finite field extension. the following are equivalent.

16 2 FIELDS 16 (i) K is a Galois extension of F. (ii) K is the splitting field of an irreducible polynomial f(x) F [x]. (ii ) K is the splitting field of a polynomial f(x) F [x]. (iii) F is the fixed field for the action of the Galois group G(K/F ) on K. (iii ) F is the fixed field for an action of a finite group of automorphisms of K. Theorem 2.75 (Main Theorem). Let K be a Galois extension of a field F and let G = G(K/F ) be its Galois group. the function H K H is a bijective map form the set of subgroups of G to the set of intermediate fields F L K. Its inverse function is L G(K/L) This correspondence has the property that if H = G(K/L) then [K : L] = H and [L : F ] = [G : H] Theorem Let K/F be a Galois extension and let L be an intermediate field. Let H = G(K/L) be the corresponding subgroup of G = G(K/F ). Then (a) Let σ be an element of G. The subgroup of G which corresponds to the conjugate subfield σl is the conjugate subgroup σhσ 1. In other words G(K/σL) = σhσ 1. (b) L is a Galois extension of F if and only if H is a normal subgroup of G. When this is so, then G(L/F ) is isomorphic to the quotient group G/H Definition Let F C be a subfield of C which contains a primitive pth root of unity ζ p = e 2πi/p. Lemma If α is a root of f(x) = x p a then α, ζ p α, ζ 2 pα,..., ζ p 1 p α are the roots of f(x). So the splitting field of x p a is generated by a single root K = F [α]

17 2 FIELDS 17 Theorem Let F C and let F contain a pth root of unity. Further let a F be an element which is not a pth power in F. Then the splitting field of f(x) = x p a has degree p over F and its Galois group is a cyclic group of order p. Theorem Let F be a subfield of C which contains a pth root of unity ζ p and let K/F be a Galois extension of degree p. Then K is obtained by adjoining a pth root to F. Theorem Let p be a prime integer and let ζ p = e 2πi/p. For any subfield F of C the Galois group of F (ζ p ) over F is a cyclic group. Lets consider the Galois group of a product of polynomials f(x)g(x) over F. Let K be a splitting field of fg. Then K contains a splitting field of K of f and F of g. So we have the following diagram. K K F Theorem With the above notation, let G = G(K/F ) and H = G(F /F ) and G = G(K /F ). (i) G and H are quotients of G (ii) G is isomorphic to a subgroup of the product G H.

### RINGS: SUMMARY OF MATERIAL

RINGS: SUMMARY OF MATERIAL BRIAN OSSERMAN This is a summary of terms used and main results proved in the subject of rings, from Chapters 11-13 of Artin. Definitions not included here may be considered

### Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman

Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 31, 2006 TALK SLOWLY AND WRITE NEATLY!! 1 0.1 Symbolic Adjunction of Roots When dealing with subfields of C it is easy to

### Fields and Galois Theory. Below are some results dealing with fields, up to and including the fundamental theorem of Galois theory.

Fields and Galois Theory Below are some results dealing with fields, up to and including the fundamental theorem of Galois theory. This should be a reasonably logical ordering, so that a result here should

### Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman

Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 17, 2006 TALK SLOWLY AND WRITE NEATLY!! 1 0.1 Integral Domains and Fraction Fields 0.1.1 Theorems Now what we are going

### Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman

Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 17, 2006 TALK SLOWLY AND WRITE NEATLY!! 1 0.1 Factorization 0.1.1 Factorization of Integers and Polynomials Now we are going

### Notes on graduate algebra. Robert Harron

Notes on graduate algebra Robert Harron Department of Mathematics, Keller Hall, University of Hawai i at Mānoa, Honolulu, HI 96822, USA E-mail address: rharron@math.hawaii.edu Abstract. Graduate algebra

### RUDIMENTARY GALOIS THEORY

RUDIMENTARY GALOIS THEORY JACK LIANG Abstract. This paper introduces basic Galois Theory, primarily over fields with characteristic 0, beginning with polynomials and fields and ultimately relating the

### Math 429/581 (Advanced) Group Theory. Summary of Definitions, Examples, and Theorems by Stefan Gille

Math 429/581 (Advanced) Group Theory Summary of Definitions, Examples, and Theorems by Stefan Gille 1 2 0. Group Operations 0.1. Definition. Let G be a group and X a set. A (left) operation of G on X is

### Factorization in Polynomial Rings

Factorization in Polynomial Rings Throughout these notes, F denotes a field. 1 Long division with remainder We begin with some basic definitions. Definition 1.1. Let f, g F [x]. We say that f divides g,

### FIELD THEORY. Contents

FIELD THEORY MATH 552 Contents 1. Algebraic Extensions 1 1.1. Finite and Algebraic Extensions 1 1.2. Algebraic Closure 5 1.3. Splitting Fields 7 1.4. Separable Extensions 8 1.5. Inseparable Extensions

### Supplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.

Glossary 1 Supplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.23 Abelian Group. A group G, (or just G for short) is

### Math Introduction to Modern Algebra

Math 343 - Introduction to Modern Algebra Notes Field Theory Basics Let R be a ring. M is called a maximal ideal of R if M is a proper ideal of R and there is no proper ideal of R that properly contains

### School of Mathematics and Statistics. MT5836 Galois Theory. Handout 0: Course Information

MRQ 2017 School of Mathematics and Statistics MT5836 Galois Theory Handout 0: Course Information Lecturer: Martyn Quick, Room 326. Prerequisite: MT3505 (or MT4517) Rings & Fields Lectures: Tutorials: Mon

### Finite Fields. Sophie Huczynska. Semester 2, Academic Year

Finite Fields Sophie Huczynska Semester 2, Academic Year 2005-06 2 Chapter 1. Introduction Finite fields is a branch of mathematics which has come to the fore in the last 50 years due to its numerous applications,

### Contradiction. Theorem 1.9. (Artin) Let G be a finite group of automorphisms of E and F = E G the fixed field of G. Then [E : F ] G.

1. Galois Theory 1.1. A homomorphism of fields F F is simply a homomorphism of rings. Such a homomorphism is always injective, because its kernel is a proper ideal (it doesnt contain 1), which must therefore

### Extension fields II. Sergei Silvestrov. Spring term 2011, Lecture 13

Extension fields II Sergei Silvestrov Spring term 2011, Lecture 13 Abstract Contents of the lecture. Algebraic extensions. Finite fields. Automorphisms of fields. The isomorphism extension theorem. Splitting

### GALOIS THEORY. Contents

GALOIS THEORY MARIUS VAN DER PUT & JAAP TOP Contents 1. Basic definitions 1 1.1. Exercises 2 2. Solving polynomial equations 2 2.1. Exercises 4 3. Galois extensions and examples 4 3.1. Exercises. 6 4.

### 1. Group Theory Permutations.

1.1. Permutations. 1. Group Theory Problem 1.1. Let G be a subgroup of S n of index 2. Show that G = A n. Problem 1.2. Find two elements of S 7 that have the same order but are not conjugate. Let π S 7

### Algebra Prelim Notes

Algebra Prelim Notes Eric Staron Summer 2007 1 Groups Define C G (A) = {g G gag 1 = a for all a A} to be the centralizer of A in G. In particular, this is the subset of G which commuted with every element

### CSIR - Algebra Problems

CSIR - Algebra Problems N. Annamalai DST - INSPIRE Fellow (SRF) Department of Mathematics Bharathidasan University Tiruchirappalli -620024 E-mail: algebra.annamalai@gmail.com Website: https://annamalaimaths.wordpress.com

### AN INTRODUCTION TO GALOIS THEORY

AN INTRODUCTION TO GALOIS THEORY STEVEN DALE CUTKOSKY In these notes we consider the problem of constructing the roots of a polynomial. Suppose that F is a subfield of the complex numbers, and f(x) is

### Finite Fields. Sophie Huczynska (with changes by Max Neunhöffer) Semester 2, Academic Year 2012/13

Finite Fields Sophie Huczynska (with changes by Max Neunhöffer) Semester 2, Academic Year 2012/13 Contents 1 Introduction 3 1 Group theory: a brief summary............................ 3 2 Rings and fields....................................

### Sample algebra qualifying exam

Sample algebra qualifying exam University of Hawai i at Mānoa Spring 2016 2 Part I 1. Group theory In this section, D n and C n denote, respectively, the symmetry group of the regular n-gon (of order 2n)

### NOTES FOR DRAGOS: MATH 210 CLASS 12, THURS. FEB. 22

NOTES FOR DRAGOS: MATH 210 CLASS 12, THURS. FEB. 22 RAVI VAKIL Hi Dragos The class is in 381-T, 1:15 2:30. This is the very end of Galois theory; you ll also start commutative ring theory. Tell them: midterm

### Galois theory (Part II)( ) Example Sheet 1

Galois theory (Part II)(2015 2016) Example Sheet 1 c.birkar@dpmms.cam.ac.uk (1) Find the minimal polynomial of 2 + 3 over Q. (2) Let K L be a finite field extension such that [L : K] is prime. Show that

### Math 201C Homework. Edward Burkard. g 1 (u) v + f 2(u) g 2 (u) v2 + + f n(u) a 2,k u k v a 1,k u k v + k=0. k=0 d

Math 201C Homework Edward Burkard 5.1. Field Extensions. 5. Fields and Galois Theory Exercise 5.1.7. If v is algebraic over K(u) for some u F and v is transcendental over K, then u is algebraic over K(v).

### Galois Theory of Cyclotomic Extensions

Galois Theory of Cyclotomic Extensions Winter School 2014, IISER Bhopal Romie Banerjee, Prahlad Vaidyanathan I. Introduction 1. Course Description The goal of the course is to provide an introduction to

### Galois Theory, summary

Galois Theory, summary Chapter 11 11.1. UFD, definition. Any two elements have gcd 11.2 PID. Every PID is a UFD. There are UFD s which are not PID s (example F [x, y]). 11.3 ED. Every ED is a PID (and

Graduate Preliminary Examination Algebra II 18.2.2005: 3 hours Problem 1. Prove or give a counter-example to the following statement: If M/L and L/K are algebraic extensions of fields, then M/K is algebraic.

### Course 311: Abstract Algebra Academic year

Course 311: Abstract Algebra Academic year 2007-08 D. R. Wilkins Copyright c David R. Wilkins 1997 2007 Contents 3 Introduction to Galois Theory 41 3.1 Field Extensions and the Tower Law..............

### ALGEBRA QUALIFYING EXAM SPRING 2012

ALGEBRA QUALIFYING EXAM SPRING 2012 Work all of the problems. Justify the statements in your solutions by reference to specific results, as appropriate. Partial credit is awarded for partial solutions.

### SUMMARY ALGEBRA I LOUIS-PHILIPPE THIBAULT

SUMMARY ALGEBRA I LOUIS-PHILIPPE THIBAULT Contents 1. Group Theory 1 1.1. Basic Notions 1 1.2. Isomorphism Theorems 2 1.3. Jordan- Holder Theorem 2 1.4. Symmetric Group 3 1.5. Group action on Sets 3 1.6.

### GALOIS THEORY AT WORK: CONCRETE EXAMPLES

GALOIS THEORY AT WORK: CONCRETE EXAMPLES KEITH CONRAD 1. Examples Example 1.1. The field extension Q(, 3)/Q is Galois of degree 4, so its Galois group has order 4. The elements of the Galois group are

### but no smaller power is equal to one. polynomial is defined to be

13. Radical and Cyclic Extensions The main purpose of this section is to look at the Galois groups of x n a. The first case to consider is a = 1. Definition 13.1. Let K be a field. An element ω K is said

### Ph.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018

Ph.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018 Do 6 problems with at least 2 in each section. Group theory problems: (1) Suppose G is a group. The

### Page Points Possible Points. Total 200

Instructions: 1. The point value of each exercise occurs adjacent to the problem. 2. No books or notes or calculators are allowed. Page Points Possible Points 2 20 3 20 4 18 5 18 6 24 7 18 8 24 9 20 10

### IUPUI Qualifying Exam Abstract Algebra

IUPUI Qualifying Exam Abstract Algebra January 2017 Daniel Ramras (1) a) Prove that if G is a group of order 2 2 5 2 11, then G contains either a normal subgroup of order 11, or a normal subgroup of order

### Math 121 Homework 2 Solutions

Math 121 Homework 2 Solutions Problem 13.2 #16. Let K/F be an algebraic extension and let R be a ring contained in K that contains F. Prove that R is a subfield of K containing F. We will give two proofs.

### Part IX. Factorization

IX.45. Unique Factorization Domains 1 Part IX. Factorization Section IX.45. Unique Factorization Domains Note. In this section we return to integral domains and concern ourselves with factoring (with respect

### CHAPTER I. Rings. Definition A ring R is a set with two binary operations, addition + and

CHAPTER I Rings 1.1 Definitions and Examples Definition 1.1.1. A ring R is a set with two binary operations, addition + and multiplication satisfying the following conditions for all a, b, c in R : (i)

### The Galois group of a polynomial f(x) K[x] is the Galois group of E over K where E is a splitting field for f(x) over K.

The third exam will be on Monday, April 9, 013. The syllabus for Exam III is sections 1 3 of Chapter 10. Some of the main examples and facts from this material are listed below. If F is an extension field

### Section VI.33. Finite Fields

VI.33 Finite Fields 1 Section VI.33. Finite Fields Note. In this section, finite fields are completely classified. For every prime p and n N, there is exactly one (up to isomorphism) field of order p n,

### THROUGH THE FIELDS AND FAR AWAY

THROUGH THE FIELDS AND FAR AWAY JONATHAN TAYLOR I d like to thank Prof. Stephen Donkin for helping me come up with the topic of my project and also guiding me through its various complications. Contents

### Algebra Exam, Spring 2017

Algebra Exam, Spring 2017 There are 5 problems, some with several parts. Easier parts count for less than harder ones, but each part counts. Each part may be assumed in later parts and problems. Unjustified

### Chapter 3. Rings. The basic commutative rings in mathematics are the integers Z, the. Examples

Chapter 3 Rings Rings are additive abelian groups with a second operation called multiplication. The connection between the two operations is provided by the distributive law. Assuming the results of Chapter

### Theorem 5.3. Let E/F, E = F (u), be a simple field extension. Then u is algebraic if and only if E/F is finite. In this case, [E : F ] = deg f u.

5. Fields 5.1. Field extensions. Let F E be a subfield of the field E. We also describe this situation by saying that E is an extension field of F, and we write E/F to express this fact. If E/F is a field

### Fields and Galois Theory

Fields and Galois Theory Rachel Epstein September 12, 2006 All proofs are omitted here. They may be found in Fraleigh s A First Course in Abstract Algebra as well as many other algebra and Galois theory

### Finite Fields. [Parts from Chapter 16. Also applications of FTGT]

Finite Fields [Parts from Chapter 16. Also applications of FTGT] Lemma [Ch 16, 4.6] Assume F is a finite field. Then the multiplicative group F := F \ {0} is cyclic. Proof Recall from basic group theory

### Course 311: Hilary Term 2006 Part IV: Introduction to Galois Theory

Course 311: Hilary Term 2006 Part IV: Introduction to Galois Theory D. R. Wilkins Copyright c David R. Wilkins 1997 2006 Contents 4 Introduction to Galois Theory 2 4.1 Polynomial Rings.........................

### φ(xy) = (xy) n = x n y n = φ(x)φ(y)

Groups 1. (Algebra Comp S03) Let A, B and C be normal subgroups of a group G with A B. If A C = B C and AC = BC then prove that A = B. Let b B. Since b = b1 BC = AC, there are a A and c C such that b =

### MATH 3030, Abstract Algebra Winter 2012 Toby Kenney Sample Midterm Examination Model Solutions

MATH 3030, Abstract Algebra Winter 2012 Toby Kenney Sample Midterm Examination Model Solutions Basic Questions 1. Give an example of a prime ideal which is not maximal. In the ring Z Z, the ideal {(0,

### Practice problems for first midterm, Spring 98

Practice problems for first midterm, Spring 98 midterm to be held Wednesday, February 25, 1998, in class Dave Bayer, Modern Algebra All rings are assumed to be commutative with identity, as in our text.

### MATH 431 PART 2: POLYNOMIAL RINGS AND FACTORIZATION

MATH 431 PART 2: POLYNOMIAL RINGS AND FACTORIZATION 1. Polynomial rings (review) Definition 1. A polynomial f(x) with coefficients in a ring R is n f(x) = a i x i = a 0 + a 1 x + a 2 x 2 + + a n x n i=0

### Notes on Field Extensions

Notes on Field Extensions Ryan C. Reich 16 June 2006 1 Definitions Throughout, F K is a finite field extension. We fix once and for all an algebraic closure M for both and an embedding of F in M. When

### Field Theory Qual Review

Field Theory Qual Review Robert Won Prof. Rogalski 1 (Some) qual problems ˆ (Fall 2007, 5) Let F be a field of characteristic p and f F [x] a polynomial f(x) = i f ix i. Give necessary and sufficient conditions

### Algebra Exam Fall Alexander J. Wertheim Last Updated: October 26, Groups Problem Problem Problem 3...

Algebra Exam Fall 2006 Alexander J. Wertheim Last Updated: October 26, 2017 Contents 1 Groups 2 1.1 Problem 1..................................... 2 1.2 Problem 2..................................... 2

### Fields. Victoria Noquez. March 19, 2009

Fields Victoria Noquez March 19, 2009 5.1 Basics Definition 1. A field K is a commutative non-zero ring (0 1) such that any x K, x 0, has a unique inverse x 1 such that xx 1 = x 1 x = 1. Definition 2.

### MATH 361: NUMBER THEORY TENTH LECTURE

MATH 361: NUMBER THEORY TENTH LECTURE The subject of this lecture is finite fields. 1. Root Fields Let k be any field, and let f(x) k[x] be irreducible and have positive degree. We want to construct a

### Polynomial Rings. i=0. i=0. n+m. i=0. k=0

Polynomial Rings 1. Definitions and Basic Properties For convenience, the ring will always be a commutative ring with identity. Basic Properties The polynomial ring R[x] in the indeterminate x with coefficients

### Galois Theory. This material is review from Linear Algebra but we include it for completeness.

Galois Theory Galois Theory has its origins in the study of polynomial equations and their solutions. What is has revealed is a deep connection between the theory of fields and that of groups. We first

### 4.5 Hilbert s Nullstellensatz (Zeros Theorem)

4.5 Hilbert s Nullstellensatz (Zeros Theorem) We develop a deep result of Hilbert s, relating solutions of polynomial equations to ideals of polynomial rings in many variables. Notation: Put A = F[x 1,...,x

### ALGEBRA PH.D. QUALIFYING EXAM September 27, 2008

ALGEBRA PH.D. QUALIFYING EXAM September 27, 2008 A passing paper consists of four problems solved completely plus significant progress on two other problems; moreover, the set of problems solved completely

### ABSTRACT ALGEBRA 2 SOLUTIONS TO THE PRACTICE EXAM AND HOMEWORK

ABSTRACT ALGEBRA 2 SOLUTIONS TO THE PRACTICE EXAM AND HOMEWORK 1. Practice exam problems Problem A. Find α C such that Q(i, 3 2) = Q(α). Solution to A. Either one can use the proof of the primitive element

### MATH 101A: ALGEBRA I, PART D: GALOIS THEORY 11

MATH 101A: ALGEBRA I, PART D: GALOIS THEORY 11 3. Examples I did some examples and explained the theory at the same time. 3.1. roots of unity. Let L = Q(ζ) where ζ = e 2πi/5 is a primitive 5th root of

### NOTES ON FINITE FIELDS

NOTES ON FINITE FIELDS AARON LANDESMAN CONTENTS 1. Introduction to finite fields 2 2. Definition and constructions of fields 3 2.1. The definition of a field 3 2.2. Constructing field extensions by adjoining

### Algebra Ph.D. Entrance Exam Fall 2009 September 3, 2009

Algebra Ph.D. Entrance Exam Fall 2009 September 3, 2009 Directions: Solve 10 of the following problems. Mark which of the problems are to be graded. Without clear indication which problems are to be graded

### Real Analysis Prelim Questions Day 1 August 27, 2013

Real Analysis Prelim Questions Day 1 August 27, 2013 are 5 questions. TIME LIMIT: 3 hours Instructions: Measure and measurable refer to Lebesgue measure µ n on R n, and M(R n ) is the collection of measurable

### A Harvard Sampler. Evan Chen. February 23, I crashed a few math classes at Harvard on February 21, Here are notes from the classes.

A Harvard Sampler Evan Chen February 23, 2014 I crashed a few math classes at Harvard on February 21, 2014. Here are notes from the classes. 1 MATH 123: Algebra II In this lecture we will make two assumptions.

### Homework 10 M 373K by Mark Lindberg (mal4549)

Homework 10 M 373K by Mark Lindberg (mal4549) 1. Artin, Chapter 11, Exercise 1.1. Prove that 7 + 3 2 and 3 + 5 are algebraic numbers. To do this, we must provide a polynomial with integer coefficients

### Algebraic Cryptography Exam 2 Review

Algebraic Cryptography Exam 2 Review You should be able to do the problems assigned as homework, as well as problems from Chapter 3 2 and 3. You should also be able to complete the following exercises:

### MAT 535 Problem Set 5 Solutions

Final Exam, Tues 5/11, :15pm-4:45pm Spring 010 MAT 535 Problem Set 5 Solutions Selected Problems (1) Exercise 9, p 617 Determine the Galois group of the splitting field E over F = Q of the polynomial f(x)

### AN INTRODUCTION TO THE THEORY OF FIELD EXTENSIONS

AN INTRODUCTION TO THE THEORY OF FIELD EXTENSIONS SAMUEL MOY Abstract. Assuming some basic knowledge of groups, rings, and fields, the following investigation will introduce the reader to the theory of

### GEOMETRIC CONSTRUCTIONS AND ALGEBRAIC FIELD EXTENSIONS

GEOMETRIC CONSTRUCTIONS AND ALGEBRAIC FIELD EXTENSIONS JENNY WANG Abstract. In this paper, we study field extensions obtained by polynomial rings and maximal ideals in order to determine whether solutions

### 22M: 121 Final Exam. Answer any three in this section. Each question is worth 10 points.

22M: 121 Final Exam This is 2 hour exam. Begin each question on a new sheet of paper. All notations are standard and the ones used in class. Please write clearly and provide all details of your work. Good

### Keywords and phrases: Fundamental theorem of algebra, constructible

Lecture 16 : Applications and Illustrations of the FTGT Objectives (1) Fundamental theorem of algebra via FTGT. (2) Gauss criterion for constructible regular polygons. (3) Symmetric rational functions.

### Section X.55. Cyclotomic Extensions

X.55 Cyclotomic Extensions 1 Section X.55. Cyclotomic Extensions Note. In this section we return to a consideration of roots of unity and consider again the cyclic group of roots of unity as encountered

### Public-key Cryptography: Theory and Practice

Public-key Cryptography Theory and Practice Department of Computer Science and Engineering Indian Institute of Technology Kharagpur Chapter 2: Mathematical Concepts Divisibility Congruence Quadratic Residues

### Math 547, Exam 1 Information.

Math 547, Exam 1 Information. 2/10/10, LC 303B, 10:10-11:00. Exam 1 will be based on: Sections 5.1, 5.2, 5.3, 9.1; The corresponding assigned homework problems (see http://www.math.sc.edu/ boylan/sccourses/547sp10/547.html)

Algebra Qualifying Exam August 2001 Do all 5 problems. 1. Let G be afinite group of order 504 = 23 32 7. a. Show that G cannot be isomorphic to a subgroup of the alternating group Alt 7. (5 points) b.

### Extension theorems for homomorphisms

Algebraic Geometry Fall 2009 Extension theorems for homomorphisms In this note, we prove some extension theorems for homomorphisms from rings to algebraically closed fields. The prototype is the following

### Algebra Qualifying Exam, Fall 2018

Algebra Qualifying Exam, Fall 2018 Name: Student ID: Instructions: Show all work clearly and in order. Use full sentences in your proofs and solutions. All answers count. In this exam, you may use the

### Algebraic Varieties. Chapter Algebraic Varieties

Chapter 12 Algebraic Varieties 12.1 Algebraic Varieties Let K be a field, n 1 a natural number, and let f 1,..., f m K[X 1,..., X n ] be polynomials with coefficients in K. Then V = {(a 1,..., a n ) :

### Algebra Review. Instructor: Laszlo Babai Notes by Vincent Lucarelli and the instructor. June 15, 2001

Algebra Review Instructor: Laszlo Babai Notes by Vincent Lucarelli and the instructor June 15, 2001 1 Groups Definition 1.1 A semigroup (G, ) is a set G with a binary operation such that: Axiom 1 ( a,

### Computations/Applications

Computations/Applications 1. Find the inverse of x + 1 in the ring F 5 [x]/(x 3 1). Solution: We use the Euclidean Algorithm: x 3 1 (x + 1)(x + 4x + 1) + 3 (x + 1) 3(x + ) + 0. Thus 3 (x 3 1) + (x + 1)(4x

### COURSE SUMMARY FOR MATH 504, FALL QUARTER : MODERN ALGEBRA

COURSE SUMMARY FOR MATH 504, FALL QUARTER 2017-8: MODERN ALGEBRA JAROD ALPER Week 1, Sept 27, 29: Introduction to Groups Lecture 1: Introduction to groups. Defined a group and discussed basic properties

### Math 547, Exam 2 Information.

Math 547, Exam 2 Information. 3/19/10, LC 303B, 10:10-11:00. Exam 2 will be based on: Homework and textbook sections covered by lectures 2/3-3/5. (see http://www.math.sc.edu/ boylan/sccourses/547sp10/547.html)

### Rings and groups. Ya. Sysak

Rings and groups. Ya. Sysak 1 Noetherian rings Let R be a ring. A (right) R -module M is called noetherian if it satisfies the maximum condition for its submodules. In other words, if M 1... M i M i+1...

### Rings. Chapter Homomorphisms and ideals

Chapter 2 Rings This chapter should be at least in part a review of stuff you ve seen before. Roughly it is covered in Rotman chapter 3 and sections 6.1 and 6.2. You should *know* well all the material

### Section 33 Finite fields

Section 33 Finite fields Instructor: Yifan Yang Spring 2007 Review Corollary (23.6) Let G be a finite subgroup of the multiplicative group of nonzero elements in a field F, then G is cyclic. Theorem (27.19)

### Abstract Algebra, Second Edition, by John A. Beachy and William D. Blair. Corrections and clarifications

1 Abstract Algebra, Second Edition, by John A. Beachy and William D. Blair Corrections and clarifications Note: Some corrections were made after the first printing of the text. page 9, line 8 For of the

### 2 (17) Find non-trivial left and right ideals of the ring of 22 matrices over R. Show that there are no nontrivial two sided ideals. (18) State and pr

MATHEMATICS Introduction to Modern Algebra II Review. (1) Give an example of a non-commutative ring; a ring without unit; a division ring which is not a eld and a ring which is not a domain. (2) Show that

### QUALIFYING EXAM IN ALGEBRA August 2011

QUALIFYING EXAM IN ALGEBRA August 2011 1. There are 18 problems on the exam. Work and turn in 10 problems, in the following categories. I. Linear Algebra 1 problem II. Group Theory 3 problems III. Ring

### Math 553 Qualifying Exam. In this test, you may assume all theorems proved in the lectures. All other claims must be proved.

Math 553 Qualifying Exam January, 2019 Ron Ji In this test, you may assume all theorems proved in the lectures. All other claims must be proved. 1. Let G be a group of order 3825 = 5 2 3 2 17. Show that

### Module MA3411: Abstract Algebra Galois Theory Michaelmas Term 2013

Module MA3411: Abstract Algebra Galois Theory Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents 1 Basic Principles of Group Theory 1 1.1 Groups...............................

### Math 121 Homework 3 Solutions

Math 121 Homework 3 Solutions Problem 13.4 #6. Let K 1 and K 2 be finite extensions of F in the field K, and assume that both are splitting fields over F. (a) Prove that their composite K 1 K 2 is a splitting

### A BRIEF INTRODUCTION TO LOCAL FIELDS

A BRIEF INTRODUCTION TO LOCAL FIELDS TOM WESTON The purpose of these notes is to give a survey of the basic Galois theory of local fields and number fields. We cover much of the same material as [2, Chapters

### Galois Theory and the Insolvability of the Quintic Equation

Galois Theory and the Insolvability of the Quintic Equation Daniel Franz 1. Introduction Polynomial equations and their solutions have long fascinated mathematicians. The solution to the general quadratic

### Profinite Groups. Hendrik Lenstra. 1. Introduction

Profinite Groups Hendrik Lenstra 1. Introduction We begin informally with a motivation, relating profinite groups to the p-adic numbers. Let p be a prime number, and let Z p denote the ring of p-adic integers,