Algebraic Cryptography Exam 2 Review


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1 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: Exercise 1. Prove that for any ring R, the ring M n (R is noncommutative whenever n 2. Proof. Note that any ring has the elements 0, 1 which are distinct. Then consider that [ ] [ ] [ ] = and [ ] [ ] = [ ] Therefore M 2 (R is noncommutative. Then it suffices to note that there is a natural monomorphism M 2 (R M n (R whenever n 2 which is given by A A 0. Exercise 2. Prove that for a natural number p, px = 0 for some 0 x F if and only if py = 0 for every y F. Proof. One direction is trivial. For the other direction, suppose that px = 0 for some nonzero x F. Since x 0 and F is a field, x has a multiplicative inverse x 1. Then py = pxx 1 y = 0x 1 y = 0. Exercise 3. Prove that any field either has characteristic zero or characteristic p, where p is prime. (it cannot have composite characteristic Proof. Recall that the characteristic of a field F is defined as the smallest natural number n for which n 1 = 0 (by the last problem, this is equivalent to n y = 0 for every y F. Suppose n were composite, so that n = mk for some 1 < m, k < n. Then m 1, k 1 are nonzero. But (m 1(k 1 = (mk 1 = n 1 = 0. This is a contradiction since a field cannot have zero divisors (for these elements would not have multiplicative inverses. Exercise 4. Prove that if F has characteristic p for some prime, then F contains a copy of F p, and similarly, if F has characteristic zero, then F contains a copy of Q. 1
2 Proof. Suppose F has nonzero characteristic p. By the previous problem, p must be prime. Then the map n n 1 from F p F is a monomorphism, which is easy to check using the definition of n 1 := ( and the definition of the characteristic of a field. Therefore F contains (an isomorphic copy of F p. Now suppose F has characteristic zero. Then there is a natural ring monomorphism ϕ : Z F given by ϕ(n n 1 (here 0 1 := 0, n 1 := ( if n > 0, and n 1 := ( n 1 if n < 0. Then by the universal property of the field of fractions, ϕ extends uniquely to a monomorphism Q F (since Q is the field of fractions of the integral domain Z. Exercise 5. Prove that [R : Q] = ( bonus: more precisely, the degree is 2 ℵ0 = c and [C : R] = 2. Proof. Let F K be an extension of fields and let B be a basis for K as an Fvector space. If B is a finite set {x 1,..., x n }, then there is a natural bijection (F n K given by (c 1,..., c n n j=1 c jx j. This function is surjective because B is a spanning set, and injective because B is linearly independent. Since Q n is countable for every n N and R is uncountable, this proves that Q R cannot be a finite extension and so the degree must be infinite. Now suppose B is infinite. Let F n (B denote the set subsets of B with n elements. Then there is a natural bijection defined by (F n (B (F n K n=1 ( {y1,..., y n }, (c 1,..., c n c j y j Again, this function is a bijection since B is a basis. Let λ, κ, β denote the (infinite cardinalities of F, K, B, respectively. Then the bijection above yields the cardinality equation βλ = κ. This is because when β is infinite the set of finite subsets of B also has cardinality β. Moreover, if λ is infinite then (F n also has cardinality λ. Therefore the countable union on the left has cardinality βλ. Applying this to the case of Q R, we find βℵ 0 = c, and therefore β = c since ℵ 0 < c. To see that [C : R] = 2, notice that {1, i} is a basis for C as an Rvector space. Exercise 6. Suppose R is an integral domain (i.e., has no zero divisors. Note that R is a subring of R[x] (by identifying R with the constant polynomials. Prove (R[x] = R under this identification. Proof. Notice that any unit in R is also a unit in (R[x], so it is the other inclusion which requires proof. Suppose R is an integral domain. Take any two polynomial p, q R[x] and let 0 a, b R be their leading coefficients. Then the polynomial pq R[x] has leading coefficient ab 0 since R is an j=1 2
3 integral domain. Thus, in order for pq to be the identity in R[x] (the constant polynomial 1, it must be the case that both pq are constant, for otherwise pq has positive degree by the argument above. But if p, q are both constant and pq = 1 R, then p, q R. Thus (R[x] R. Exercise 7. Prove that if a polynomial f R[x] has odd degree n > 2, then f is reducible. Proof. Let a n 0 denote the coefficient of the x n term of f. Let sgn(a n be 1 if a n > 0 and 1 if a n < 0. Then and hence f(x lim x ± x n = a n lim f(x = sgn(a n(±. x ± In particular, this entails that there exist a b R so that f(a < 0 and f(b > 0. Therefore, on the interval I joining a, b, by the Intermediate Value Theorem there is some c I for which f(c = 0. In other words, c is a root of f, and so we may factor f(x = (x cg(x for g R[x] of degree strictly less than n. Therefore f is reducible. Exercise 8. Prove the conditions in the definition of algebraically closed are actually equivalent. Exercise 9. Explain why most elements of R are transcendental over Q. Proof. I claim that there are only countably many algebraic elements over Q. Since R is uncountable, most of its elements must be transcendental. Indeed, there are countably many polynomials in Q[x] (identified with n=1 Qn, which is a countable union of countable sets and is therefore countable itself. Each polynomial p Q[x] has finitely many (deg p roots (not necessarily in p, and the countable union of finite sets is countable. Therefore there are only countably many algebraic elements over Q. Since R has cardinality c, most (i.e., all but countably many of its elements must be transcendental over Q. Exercise 10. Suppose F is a field and α is algebraic over F. Prove that the set J = {f F[x] α is a root of f} is an ideal of F[x]. Conclude that α has a minimum polynomial; that is, a polynomial m α F[x] so that m α (α = 0 and whenever f F[x] with f(α = 0, m α divides f. Note: To make the minimum polynomial unique, we also require that it is monic, i.e., the coefficient of the highest degree term is 1. Proof. Let F be a field an α an algebraic element over F. We begin by showing that J is an ideal. Indeed, note that if f, g J, then (f +g(α = f(α+g(α = 0, so f + g J. Moreover, if h F[x], then (fh(α = f(αh(α = 0 h(α = 0, so fh J. Therefore J is an ideal of F[x]. 3
4 Since F is a field, F[x] is a principal ideal domain, and therefore J is principal and generated by some polynomial m α. That is, J = m α, so m α (α = 0 and any element of J has the form m α h for some h F[x], thus m α divides any polynomial for which α is a root. To uniquely identify m α, multiply its leading coefficient (necessarily nonzero by the multiplicative inverse in F to make m α a monic polynomial. Note that m α must be irreducible over F. For if not, then m α would factor as fg for some f, g F[x] of positive degree, and α would necessarily be a root of one of these polynomials; this would contradict the fact that J is generated by m α since m α would not divide f, g. Exercise 11. Prove the equivalence in the previous definition. That is, if α is algebraic over F, prove that [F(α : F] = deg m α. As a corollary, conclude that α F if and only if F(α has degree 1 over F if and only if F(α = F. Proof. This is essentially proven in the solutions to Homework 4. Exercise 12. Prove that the splitting field of x 3 2 is Q(ω, 3 2 and that [Q(ω, 3 2 : Q] = 6. Proof. Notice that the roots of x 3 2 are 3 2, ω 3 2, ω where ω is a primitive cube root of unity. Thus, we clearly have that the splitting field K = Q( 3 2, ω 3 2, ω Q( 3 2, ω. However, at the same time ω = (ω 2 3( K, so the reverse inclusion holds as well. Now, clearly x 3 2 is irreducible over Q (because it is cubic and has no roots in Q, so [Q( 3 2 : Q] = 3. Furthermore, since ω / R, we also have ω / Q( 3 2. The minimum polynomial of ω over this field is x 2 + x + 1 (because it is quadratic and ω, ω 2 are the roots. Therefore [Q( 3 2, ω : Q( 3 2] = 2, and hence [K : Q] = 6. Exercise 13. If f F[x] is irreducible of degree d, what are the minimum and maximum possible degrees of the splitting field of f over F? Proof. Suppose f F[x] is irreducible of degree d. The splitting field K of f over F has degree at least d because we need to adjoin at least one root α 1 of f to F. If the other roots α 2,..., α d are in the field F(α 1 then this is the splitting field and it has degree d. For the upper bound, we prove that the maximal degree of K over F is d! and we proceed by induction on the degree d. The claim is obvious if d = 1. Now suppose that the claim holds for some d 1 N. Then when we adjoin α 1 to F, the polynomial f factors as (x α 1 g over F(α 1 with degg = d 1. The splitting field of g over F(α 1 is the same as the splitting field of f over F, and by induction the former has degree at most (d 1!. Thus, [K : F] = [K : F(α 1 ] [F(α 1 : F] (d 1!d = d!. Technically, we should provide examples of both of these situations occurring. Notice that by our results about finite fields, the minimum degree of the splitting field occurs whenever we take the splitting field of an irreducible polynomial over a finite field. For the maximal degree, constructions over Q are 4
5 doable, but nontrivial (this is a special case of what is called the inverse Galois problem; i.e., given a finite group, find a Galois extension of Q with the specified group as its Galois group. The easier way involves some abstract field theory, which is not really beneficial in my mind, so we will just skip it. Exercise 14. If a polynomial f(x has a root α of multiplicity m 2 (ie., has (x α m, then α is also a root of its derivative f (x. Exercise 15. Suppose F is a finite field with q elements, which we will henceforth denote F q (we will prove uniqueness up to isomorphism later. Prove that q = p n for some n N and some prime p. Proof. Since F q is a finite field, it cannot have characteristic zero (otherwise it would contain Q, which is infinite. Thus F q has prime characteristic p, and therefore contains F p as a subfield. Recall that whenever we have an inclusion of fields, we can view the larger one as a vector space over the smaller one. Thus F q is an F p vector space of some dimension n, which is necessarily finite since F q is finite. Let B = {v 1,..., v n } be a basis for F q over F p. Then the function F n p F q given by (c 1,..., c n n k=1 c kv k is a bijection since B is a basis. Therefore q = p n. Exercise 16. Prove that lcm(j, n = Proof. See the solutions to homework 5 nj gcd(j,n. Exercise 17. Let F K be an extension of fields and τ Aut(K an automorphism which fixes F. If α is a root of an irreducible polynomial f F[x], then τ(α is a conjugate of α over F. That is, τ(α is also a root of f. Proof. Let τ Aut(K/F, and suppose α is a root of an irreducible polynomial f over F. Suppose f(x = n c kx k for some c k F. Note that τ(c k = c k since τ fixes F. Then since τ is an automorphism of K, ( n 0 = τ(0 = τ(f(α = τ c k α k = τ(c k τ(α k = c k τ(α k = f(τ(α. Therefore τ(α is also a root of f, i.e., it is a conjugate of α. distinct monic irre Exercise 18. If f is a prime number, then there are pf p f ducible polynomials of degree f over F p. Proof. Recall that the polynomial x pf x factors over F p into all monic irreducible polynomials over F p of degrees d which divide f. If f is prime, this means it factors into all the monic irreducible polynomials of degrees 1, f. Notice that monic irreducible polynomials of degree 1 are just (x c for some c F p, and there are clearly p of these. Let n denote the number of monic irreducible polynomials over F p of degree d. Then by equating degrees, we must have p f = 1 p + f n. Solving for n we obtain n = pf p f as desired. 5
6 Exercise 19. Provide a formula for the number of distinct monic irreducible polynomials of degree f (not necessarily prime over F p in terms of the divisors of f. (Just try to do this for say, f is a prime power (by induction, or f is a semiprime (products of two primes; the general formula is a bit more complicated. Proof. Suppose that f = q n is a prime power. We will prove by strong induction that the number of distinct monic irreducible polynomials of degree q n over F p is p qn p qn 1 q n. The base case n = 1 follows from the previous exercise. So suppose that n N and the formula holds for all k < n. By a theorem, we know that x pqn x factors into all monic irreducible polynomials of degrees d dividing q n, namely, 1, q, q 2,..., q n. Let m k denote the number of monic irreducible polynomials over F p of degree q k. Then m 0 = p and by the induction hypothesis, m k = pqk p qk 1 for k < n. Then by equating q k degrees, we find that p qn = q k m k = p + (q k pqk p qk 1 + q n m n = p + k=1 k=1 = p qn 1 + q n m n. q k ( p qk p qk 1 + q n m n Solving for m n we obtain m n = pqn p qn 1 q. n Suppose that f = qr is a semiprime. Then notice that x pqr x factors into all monic irreducible polynomials of degrees d dividing qr, i.e., degrees 1, q, r, qr. Let m denote the number of monic irreducible polynomials of degree qr. Then by equating degrees, we find ( p q p p qr = 1 p + q Solving for m, we find q + r ( p r p m = pqr p q p r + p. qr r + qr m. For the completely general case, notice that x pn x factors into all monic irreducible polynomials over F p of degrees d dividing n. Then if m d denotes the number of monic irreducible polynomials over F p of degree d, we can equate degrees to obtain p n = dm d. d n By Möbius inversion, this yields nm n = d n µ(dp n d, 6
7 from which we find m n = 1 µ(dp n d. n d n Above, µ denotes the Möbius function. That is, µ(n is the sum of the primitive nth roots of unity. An alternative characterization is: 1 if n is a squarefree positive integer with an even number of prime factors µ(n := 1 if n is a squarefree positive integer with an odd number of prime factors 0 if n has a squared prime factor The Möbius inversion formula follows from properties of Dirichlet convolution. Notice that the formula for m n given above easily recovers the previous two formulas given for m q n and m qr. We can even do much more complicated ones with the formula for m n given above. For example, the formula above yields m 60 = 1 ( p 60 p 30 p 20 p 12 + p 10 + p 6 + p 4 p In particular, there are monic irreducible polynomials of degree 60 over F 2. By the way, Möbius inversion is a really cool trick for functions involving Dirichlet convolution (i.e., sums where the index ranges over the divisors of a positive integer. 7
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