Quizzes for Math 401


 Gervais Evans
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1 Quizzes for Math 401 QUIZ 1. a) Let a,b be integers such that λa+µb = 1 for some inetegrs λ,µ. Prove that gcd(a,b) = 1. b) Use Euclid s algorithm to compute gcd(803, 154) and find integers λ,µ such that gcd(803, 154) = λ µ 154. Show all your work. Solution: a) If d is the greatest common divisor of a and b then d a and d b. It follows that d λa and d µb, and therefore d λa + µb = 1. The only positive divisor of 1 is 1 itself, so d = 1. b) We have 803 = = = = Thus gcd(803, 154) = 11 and 11 = = 33 ( ) = = 5( ) 154 = QUIZ 2. a) Define Euler phifunction φ and list the results we proved about it. Compute φ(175). b) Let p be a prime number and let n be a positive integer such that p 4 n 3. Prove that p 2 n. Solution: a) φ(n) is the number of positive integers which are N and relatively prime to N. We proved that φ is multiplicative, i.e. φ(mn) = φ(m)φ(n) for any 1
2 relatively prime natural numbers m,n. Furthemore, φ(p) = p 1 and φ(p k ) = (p 1)p k 1 for any k > 0 and any prime number p. If m = p a 1 1 p a p a k k, where p 1,...,p k are prime numbers and a 1 > 0,...,a k > 0 then φ(m) = m(1 1/p 1 )(1 1/p 2 )...(1 1/p k ) = (p 1 1)p a (p 2 1)p a (p k 1)p a k 1 k Note now that 175 = Thus φ(175) = φ(5 2 )φ(7) = = 120. b) Suppose that p k n (i.e. p k is the highest power of p which divides n). Then p 3k n 3 and therefore 3k 4. Since k is an integer, we have k 2 and therefore p 2 n. Second solution: Clearly p 4 n 3 implies that p n 3. Since p is prime we have p n, i.e. n = pm for some integer m. Now p 4 n 3 = p 3 m 3 implies that p m 3. Again, since p is a prime number, we get p m. It follows that m = pk for some integer k, so n = pm = p 2 k, i.e. p 2 n. QUIZ 3. a) Define a pseudoprime relative to the base a and a Carmichael number. What are the pseudoprimes relative to the base 1? Conclude that Carmichael numbers are odd. b) Prove that if gcd(n, 6) = 1 then n 2 1 (mod 12) (Extra credit: prove that n 2 1 (mod 24) ). Solution: a) A positive number n ic called a pseudoprime relative to the base a if n is composite and a n 1 1 (mod n). A composite positive integer n is a pseudoprime relative to 1 iff ( 1) n 1 1 (mod n). If n is even this means that 1 1 (mod 2), i.e n 2, which is not possible since n is composite. On the other hand, if n is odd then the congruence ( 1) n 1 1 (mod n) holds. Thus pseudoprimes relative to base 1 are exactly the composit odd numbers. A Carmichael number is any integer n which is a pseudoprime relative to every base a such that gcd(a,n) = 1. Since gcd( 1,n) = 1 for any n, a Carmichael number is pseudoprime relative to 1, so it must be odd. b) From gcd(n, 6) = 1 we conclude that n is relatively prime to 3 and 4. Since 2
3 φ(3) = 2 = φ(4), we see by Euler s theorem that n 2 1 (mod 3) and n 2 1 (mod 4). Thus n 2 1 is divisible by both 3 and 4 and since these numbers are relatively prime, their product also divides n 2 1, i.e. n 2 1 (mod 12). To prove the congruence n 2 1 (mod 24) it suffices to show that n 2 1 (mod 8), i.e. that 8 n 2 1 = (n 1)(n + 1). Since n is odd, n 1, n + 1 are consecutive even numbers, so one of them must be divisible by 4. Thus (n 1)(n + 1) is a product of an even number and a number divisible by 4, so it is divisible by 8. QUIZ 4. a) Let R be a unital ring and let a R be invertible. 1) Prove that if b R and ab = 0 then a = 0. 2) Prove that if x,y R are such that ax = ay then x = y. b) Define a field by listing all the axioms. Solution: a) Let c be the inverse of a so that ac = ca = 1. If ab = 0 then c(ab) = c 0 = 0. On the other hand, c(ab) = (ca)b = 1 b = b so b = 0. This proves 1). Now if ax = ay then a(x y) = 0 and therefore x y = 0 by 1). In other words, x = y, which proves 2). b) A field is a set F with two binary operations (functions) +, : F F F such that: A1. (a + b) + c = a + (b + c) for all a,b,c F; A2. There exists 0 F such that a + 0 = a = 0 + a for all a F; A3. For every a F there exists b F such that a + b = b + a = 0; A4. a + b = b + a for every a,b F; M1. (a b) c = a (b c) for all a,b,c F; M2. There exists 1 F such that a 1 = a = 1 a for all a F; M3. For every a F {0} there exists b F such that a b = b a = 1; M4. a b = b a for every a,b F; 3
4 D. a (b + c) = (a b) + (a c) and (b + c) a = (b a) + (c a) for every a,b,c F. QUIZ 5. a) Let R be a ring and I an ideal of R. Define a coset of I in R; the ring R/I; the canonical homomorphism. b) 1. Let F be a division ring and I an ideal of F. Prove that I = {0} or I = F. 2. Let F be a commutative unital ring such that {0} and F are the only ideals of F. Prove that F is a field. Hint. Look at principal ideals. Solution: a) A coset of I in R is a set of the form r + I = {r + i : i I} for some r R. R/I is the set of all cosets of I in R. It is a ring with addition and multiplication given by (t+i)+(r+i) = (t+r)+i and (t+i)(r+i) = tr+i. The canonical homomorphism is the homomorphism f : R R/I given by f(r) = r+i. b) Suppose that F is a division ring and I is an ideal of F. If I contains a nonzero element a then 1 = (a 1 )a I and therefore r = r 1 I for all r F. Thus I = F. This proves that either I = {0} or I = F. Suppose now that F is commuative, unital and the only ideals of F are {0} and F. If a F and a 0 then the principal ideal af in not {0} (it contains a) hence it is F. In particular, 1 af, so 1 = ab for some b F. This proves that all nonzero elements are invertible, i.e. F is a field. QUIZ 6. a) State the correspondence theorem. b) Let S = C[ 1, 1] be the ring of all continuous functions from [ 1, 1] to R and let I = {f S : f(0) = 0}. 1. Prove that the map Φ : S R given by Φ(f) = f(0) is a surjective homomorphism. 2. Prove hat S/I and R are isomorphic. 4
5 Solution: a) Correspondence Theorem. Let f : R S be a surjective homomorphism. There is a bijective correspondence between subrings of R which contain kerf and subrings of S. In this correspondence ideals correspond to ideals. This correspondence respects (or preserves) inclusions. Explicitely, a subring T of R which contains kerf corresponds to the subring f(t) of S and a subring K of S corresponds to the subring f 1 (K) of R. In particular, if S = R/I and f is the canonical homomorphism then the correspondence theorem states that an ideal J of R which contains I corresponds to the ideal J/I of R/I. b) For 1) note that Φ(f + g) = (f + g)(0) = f(0) + g(0) = Φ(f) + Φ(g) and Φ(f g) = (f g)(0) = f(0) g(0) = Φ(f) Φ(g). This shows that Φ is a homomorphism. For a given c R the constant function f(x) = c for all x [ 1, 1] belongs to S and Φ(f) = f(0) = c. This proves that Φ is surjective. For 2) note that by definition, I is the kernel of Φ. Thus it is an ideal. Since 1) shows that f is a surjective homomorphism, the First Isomorphism Theorem implies that R and S/ ker f = S/I are isomorphic. Quiz 7. a) Define irreducible elements and prime elements in an integral domain R. b) Let R be a PID. Prove that every nonzero prime ideal in R is maximal. c) Prove that 3 is irreducible in the ring R = Z[ω] of Eisenstein integers (recall ω = ( 1 + 3)/2). Conclude that 3R is a maximal ideal. Solution: a) An nonzero element a R is called irreducible if a is noninvertible and whenever a = xy for some x,y R, either x or y is invertible in R. Equivalently, a 0 is irredcucible if ar is maximal among all proper principal ideals. 5
6 An nonzero element a R is called prime if a is noninvertible and whenever a xy then a x or a y. Equivalently, a 0 is prime iff ar is a prime ideal. b) Let I be a nonzero prime ideal. Since R is a PID, I = ar is principal. Thus a is a prime element, hence irreducible. Suppose that J is an ideal containing I. Then J = br for some b. Since a J we have a = bx for some x R. But a is irreducible so either b is invertible or x is invertible. In the former case we have br = R and in the latter case we have br = ar. This proves that ar is maximal. Second argument: if ar is prime and nonzero then a is irreducible and therefore ar is maximal among all proper principal ideals of R. Since every ideal is principal, ar is maximal. c) Recall that N(a + bω) = a 2 ab + b 2 has the property that N(xy) = N(x)N(y). Thus if neitehr x nor y is invertible, then N(x) > 1 N(y) > 1 and therefore N(xy) is a composite integer. In other words, if N(z) is prime then z is irreducible. But N( 3) = 3 is a prime, so 3 is irreducible. Since R is PID, 3 is prime so 3R is a nonzero prime ideal and by b) it is maximal. Quiz 8. a) Define primitive polynomials and state Gauss Lemma. b) Prove that the polynomial f = 5x 7 + 6x 5 18x is irreducible in Q[x]. Carefully explain your reasoning. Solution: a) Let R be a UFD. A polynomial f = f 0 + f 1 x f m x m R[x] is primitive if it is a nonconstatnt polynomail and its coefficients do not have any noninvertible common divisors, i.e. gcd(f 0,f 1,...,f m ) = 1. Gauss Lemma: Let R be a UFD and let f,g R[x] be primitive polynomials. Then fg is also primitive. b) Clearly the polynomial f(x) = 5x 7 +6x 5 18x Z[x] is primitive. Note that Eisenstein criterion applies to this polynomial with prime ideal 3Z (since 5 3Z, each of 6, 18, 12, 0 belong to 3Z and 12 (3Z) 2 = 9Z). Thus f is irreducible in Z[x] (Eisentsetin criterion says that in any factorization of f in Z[x] one of the facrtors is constant. Since f is primitive, this constant is invertible in Z). A result from class says that nonconstant polynomials in Z[x] remain irreducible in Q[x]. 6
7 Quiz 9 a) Define the group D 6. Choose a subgroup H of D 6 consiting of 2 elements. List all left and right cosets of H in G. What is [G : H]? Is H a normal subgroup? b) Define a left coset of H in G. Define a normal subgroup. Solution: a) The group D 6 is the dihedral group of order 6, i.e. it is the group of isometries of an equilateral triangle A 1 A 2 A 3. If T is the rotation by 2π/3 counterclockwise about the center of A 1 A 2 A 3 and S is the reflection at the symmetry axis passing by A 1 then D 6 = {I,T,T 2,S,ST,ST 2 }. We have T 3 = I = S 2 and TS = ST 2. Note that D 6 has three subgroups of order 2, namely {I,S}, {I,ST } and {I,ST 2 } (note that a subgroup of order 2 is of the form < g >= {e,g} for some element g of order 2). Let us choose H = {I,S}. Then H = {I,S}, TH = {T,ST 2 } and T 2 H = {T 2,ST } are the left cosets of H in G and the right cosets of H in G are H = {I,S}, HT = {T,ST } and HT 2 = {T 2,ST 2 }. In other words, G/H = {H,TH,T 2 H} and H\G = {H,HT,HT 2 }. Recall that [G : H] is the number of left (or right) cosets of H in G so [G : H] = 3 (this can be also seen from Lagrange s theorem). Finally, H is not normal since TH HT. b) A left coset of H in G is a subset of G of the form ah = {ah : h H} for some a H. A subgroup H of G is called normal (notation: H G) if ghg 1 H for every g G and every h H. Equivalently, H is normal if c g (H) = H for every g G, where c g is the conjugation by g. Quiz 10 a) Satae the Third Isomorphism Theorem. ( ) b) Let σ =. Write σ as a product of disjoint cycles Find the order of σ. Write σ 1 as a 2 9 matrix. Solution: a) Third Isomorphism Theorem: Let G be a group, H a subgroup of G and N a normal subgroup of G. Then 1. HN = {hn : h H,n N} is a subgroup of G and N HN; 7
8 2. H N H; 3. the groups H/(H N) and HN/N are isomorphic by the map h(h N) hn. ( ) b) We have σ = = (1, 8, 4, 6)(2, 5, 9)(3, 7). Thus the order of σ is lcm(4, 3, 2) = 12. The inverse ( ) σ 1 = QUIZ 11. a) What is an action of a group G on a set X? b) A group G of order 49 acts without fixed points on a finite set X. Prove that 7 X. Solution: a) An action of a group G on a set X is a function G X X which to a pair (g,x) G X assignes the element g x X and such that: 1. g (h x) = (gh) x for any g,h G and any x X; 2. e x = x for all x X (e is the identity element of G). b) There are several ways of solving this part. Recall the three principles we established for a finite group G acting on a finite set S: Rule 1. If the action of G on S is transitive, then S = G / St(s) for any s S. Rule 2. Let p be a prime number which does not divide S. There is an element s S such that O(s) is not divisible by p. Rule 3. Suppose that a p group G acts on a set S. Let r denote the number of fixed points for this action. Then S r (mod p). In particular, (i) if S is not divisible by p then r > 0, i.e. there is at least one fixed point. (ii) if p S and r > 0 then r p, i.e. if there is a fixed point then there are at least p of them. We can apply Rule 3 to our problem, since G is a 7 group. We are told that r = 0, so X 0 (mod 7), i.e. 7 X. 8
9 Alternateively, we may apply Rule 2. Suppose that 7 X. Then there is an orbit O such that 7 O. On the other hand, the size of each orbit divides the order of the group, so O 49 = 7 2. This two conditions imply that O = 1, which means that there is a fixed point, a contradiction. Finally, we may just give a direct argument. We know that each orbit has size which divides G = 7 2, so it has 1, 7 or 7 2 elements. But orbits of size 1 are excluded (no fixed points) so each orbit has size divisible by 7. Since orbits partition X, X is divisible by 7. 9
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