Zachary Scherr Math 503 HW 5 Due Friay, Feb 26 1 Reaing 1. Rea Chapter 9 of Dummit an Foote 2 Problems 1. 9.1.13 Solution: We alreay know that if R is any commutative ring, then R[x]/(x r = R for any r R. In the present situation we let R = F [y], then F [x, y]/(y 2 x = F [x, y]/(x y 2 = R[x]/(x y 2 = R = F [y]. This shows that F [x, y]/(x y 2 in an integral omain. The ring F [x, y]/(y 2 x 2, on the other han, has zero ivisors. In this ring, 0 = ȳ 2 x 2 = (ȳ x(ȳ + x, an neither ȳ x nor ȳ+ x are zero in this ring. This shows that F [x, y]/(y 2 x 2 an F [x, y]/(y 2 x cannot be isomorphic. 2. 9.1.14 Solution: Consier the map ϕ: R[x, y] R[t] given by x t j y t i where i, j are relatively prime positive integers. Then ϕ(x i y j = (t j i (t i j = 0 showing that x i y j ker(ϕ. The purpose of this problem is to show that ker(ϕ = (x i y j since then the first isomorphism theorem woul yiel that R[x, y]/(x i y j is isomorphic to a subring of R[t], an hence is an integral omain an thus (x i y j is prime. To achieve this, let f(x, y ker(ϕ. The polynomial x i y j is monic in y an hence if we view R[x, y] as (R[x][y] then the results of 7.4.14 tell us that there exists p(x, y R[x, y] having y egree less than j with f(x, y p(x, y (mo x i y j. Thus we may write for some n 0 an a rs R. p(x, y = n j 1 a rs x r y s r=0 s=0 Since both f(x, y an x i y j are containe in ker(ϕ, it follows that p(x, y ker(ϕ as well. Thus 0 = ϕ(p(x, y = = n j 1 a rs ϕ(x r y s r=0 s=0 n j 1 a rs t jr+is. r=0 s=0
I claim that the exponents showing up on the right han sie of the above equation are istinct integers. To prove this, assume there exist non-negative integers r an r, an integers 0 s, s < j with jr + is = jr + is. Rewrite this equation as j(r r = i(s s. Then j i(s s, but gc(i, j = 1 implies that j s s. Since 0 s, s < j this forces s = s. Thus jr = jr an by cancellation (R is an integral omain we get r = r as well. Therefore, since 0 = n j 1 a rs t jr+is r=0 s=0 yet all the egrees of terms on the right han sie are istinct integers, we must have all a rs = 0. This proves that p(x, y = 0 an hence f(x, y 0 (mo x i y j. We have just prove that ker(ϕ (x i y j which by our above remarks proves equality, an therefore the ieal (x i y j is prime 3. 9.2.2 Solution: Let F be a finite fiel of orer q, an let f(x F [x] be a polynomial of egree n. Since (f(x = (cf(x for any c F, we may assume WLOG that f(x is monic. By 7.4.14, we know that every g(x F [x] is congruent moulo f(x to a unique polynomial a n 1 x n 1 + a n 2 x n 2 +... + a 1 x + a 0 where a i F. Since there are q possibilities for each a i, an there are n coefficients, we see that there are q n elements in F [x]/(f(x. 4. 9.2.3 Solution: Let f(x F [x]. The ring F [x] is a PID, hence f(x is irreucible if an only if the ieal (f(x is prime if an only if the ieal (f(x is maximal. Thus f(x is irreucible if an only if F [x]/(f(x is a fiel. 5. 9.2.5 Solution: By the fourth isomorphism theorem, ieals in F [x]/(p(x are in one-to-one corresponence with ieals in F [x] containing (p(x. The ring F [x] is a PID, so every ieal containing (p(x is generate by some element f(x F [x]. Of course (f(x (p(x if an only if p(x ivies f(x, so we see that the ieals containing f(x are precisely those generate by ivisors of f(x. To eal with the fact that associate elements generate the same ieal, we can just stick to monic ivisors of p(x. Thus the ieals in F [x]/(p(x are precisely the principal ieals ( f(x where f(x is a monic ivisor of p(x. Page 2
6. 9.3.1 Solution: We prove this by proving the contrapositive. Let R be a UFD an let p(x R[x] be a monic polynomial. Assume that p(x = a(xb(x where a(x, b(x F [x] are monic polynomials. By Proposition 5, there exist α, β F with p(x = A(xB(x where A(x = αa(x R[x] an B(x = βb(x R[x]. Since a(x an b(x are monic, the leaing coefficient of A(x is α an the leaing coefficient of B(x is β, implying that α, β R. Moreover, the leaing coefficient of A(xB(x is αβ an the leaing coefficient of p(x is 1, so in fact αβ = 1. Thus show that both a(x, b(x R[x]. a(x = βa(x b(x = αb(x Now, by efinition we have 2 Z[2 2], yet 2 = 2 2 2 showing that 2 is an element of the fraction fiel of Z[2 2]. We have proving that Z[2 2] is not a UFD. (x 2(x + 2 = x 2 2 Z[2 2][x] 7. 9.3.2 Solution: Let f(x, g(x Q[x] have the property that f(xg(x Z[x]. Extracting a lowest common enominator, following by factoring out common factors from the remaining coefficients, we may write f(x = a b f(x g(x = c g(x where f(x, g(x Z[x] both have content 1 an gc(a, b = gc(c, = 1. Gauss s Lemma implies that ct( f(x g(x = 1 as well, an since f(xg(x = ac b f(x g(x has integer coefficients, we must have b ac. Any coefficient of f(x is of the form a b m where m Z an similarly any coefficient of g(x is of the form c n where n Z. Their prouct equals ab c mn which is an integer since b ac. 8. 9.4.8 Solution: To prove that both of these rings are fiels, it suffices to show that f(x = x 2 + 1 an g(y = y 2 + 2y + 2 are irreucible in F 11 [x] an F 11 [y] respectively. Since these polynomials have Page 3
egree 2, we can just check to see that they o not have roots. We have a f(a g(a 0 1 2 1 2 5 2 5 10 3 10 6 4 6 4 5 4 4 6 4 6 7 6 10 8 10 5 9 5 2 10 2 1 showing that f(x an g(y are irreucible. Both K 1 an K 2 are thus fiels an have size 121 since f(x an g(y have egree 2. Now, consier the map ψ : F 11 [x] K 2 given by p(x p(ȳ + 1. The evaluation map is a ring homomorphism, an ψ(x 2 + 1 = (ȳ + 1 2 + 1 = ȳ 2 + 2ȳ + 2 = 0. Thus x 2 + 1 ker(ψ so ψ inuces a ring homomorphism F 11 [x]/(x 2 + 1 F 11 [y]/(y 2 + 2y + 2. The map ψ is surjective since ψ(x 1 = ȳ. Thus the inuce map K 1 K 2 is surjective. Since both fiels have the same carinality, this map is necessarily injective an therefore is an isomorphism. REMARK: By the quaratic formula, the polynomial x 2 + 1 has a root in F 11 if an only if 1 F 11, an y 2 + 2y + 2 has a root in F 11 if an only if 4 = 7 F 11. Notice that both polynomials are irreucible if an only if 1 is not a square in F 11 since 4 is a square if an only if 1 is a square. Of course 11 3 (mo 4 so we ve previously proven that 1 is not a square for such primes. 9. 9.4.11 Solution: To prove that x 2 + y 2 1 is irreucible in Q[x, y], it suffices to prove that it is irreucible in (Q[y][x]. Let R = Q[y]. The element y 1 R[y] is irreucible hence prime since R is a UFD. Notice that y 2 1 = (y 1(y + 1 so y 1 ivies y 2 1 yet (y 1 2 oes not ivie y 2 1. Since x 2 + (y 2 1 R[x] is monic, Eisenstein tells us that x 2 + y 2 1 is irreucible. 10. 9.4.12 Solution: The book proves that f(x = x n 1 + x n 2 +... + x + 1 Page 4
is irreucible when n is prime. Thus we shoul prove that f(x is reucible when n is composite. We have x n 1 = (x 1(x n 1 + x n 2 +... + x + 1. By unique factorization in Z[x], it thus suffices to prove that x n 1 has more than two irreucible factors. Since n is composite, we may write n = ab for integers a, b > 1. I claim that x a 1 ivies x n 1. To see this, we work moulo x a 1 to get x n 1 = (x a b 1 1 b 1 (mo x a 1 0 (mo x a 1. Thus there exists g(x Z[x] (this polynomial is easy to write own explicitly with x n 1 = (x a 1g(x = (x 1(x a 1 + x a 2 +... + x + 1g(x showing that x n 1 has at least three irreucible factors. 11. 9.4.13 Solution: The polynomial f(x = x 3 +nx+2 Z[x] is monic an non-constant, hence is irreucible over Z if an only if it is irreucible over Q. Since f(x has egree 3, it is irreucible if an only if it has a root in Q. By the rational root theorem, the only possible roots are ±1, ±2. Evaluating we get f(1 = n + 3 f( 1 = 1 n f(2 = 2n + 10 f( 2 = 6 2n. Setting each of these equal to 0 an solving for n shows that f(x is reucible if an only if n = 1, 3, or 5. 12. Prove that the polynomial f(x = x 2 + x + 1 F 5 [x] is irreucible. Fin an element in F 5 [x]/(x 2 + x + 1 which generates the group of units. Solution: Since f(x has egree 2, to check if f(x is irreucible it suffices to show that f(x has no roots in F 5. We can compute f(0 = 1 f(1 = 3 f(2 = 2 f(3 = 3 f(4 = 1 showing that f(x is irreucible an hence F 5 [x]/(x 2 + x + 1 is a fiel of 25 elements. There are several ways to fin a generator for the multiplicative group of units in this fiel. The most brute Page 5
force way is to just start taking powers of various elements an looking for elements with orer 24. Here is an alternative way. We know that x 3 1 = (x 1(x 2 + x + 1, so in F 5 [x]/(x 2 + x + 1 we have x = 3 (for those of you who in t take 502, this is the notation we use for the orer of an element in a group. It s easy to check that 2 = 4 (in particular, 2 generates the group of units in F 5 so since gc(3, 4 = 1 we get 2 x = 12. Thus we will be one if we can fin some α F 5 [x]/(x 2 + x + 1 with α 2 = 2 x. Write α = a x + b, then α 2 = a 2 x 2 + 2ab x + b 2 = a 2 (4 x + 4 + 2ab x + b 2 = (4a 2 + 2ab x + (4a 2 + b 2. Thus we immeiately get that 0 = 4a 2 + b 2 = b 2 a 2 = (b a(b + a implying that b = ±a. Plugging into the leaing coefficient gives 2 = 4a 2 ± a 2 an since the right han sie better be non-zero we get 2 = 3a 2. This equation has two solutions, one of them is a = 2 an thus b = 2 = 3. The generator we have foun is α = 2 x + 3. You may have foun a ifferent generator since there are ϕ(24 = 8 generators of this group. They are: α = 2 x + 3 α 5 = 3 x + 1 α 7 = 4 x + 1 α 11 = x + 2 α 13 = 3 x + 2 α 17 = 2 x + 4 α 19 = x + 4 α 23 = 4 x + 3. 3 Challenge Problems Challenge Problems ten to be harer than the rest of the problems (an sometimes more interesting. You o not nee to turn these in, but you shoul get something out of thinking about these. 1. Prove that every prime ieal of Z[x] is of one of the following forms 1. (0 2. (f(x where f(x Z[x] is an irreucible non-constant polynomial 3. (p where p Z is prime 4. (p, g(x where p Z is prime, g(x Z[x] is non-constant, an g(x (Z/pZ[x] is irreucible. Which of the above ieals are maximal? Page 6
Solution: Let P Z[x] be a prime ieal. Then we know that I = P Z is a prime ieal of Z. Therefore either I = (0 or I = (p for some prime p Z. In the latter case, we have (p P an so by the fourth isomorphism theorem we see that P correspons to a prime ieal in Z[x]/(p = F p [x]. This ring is a PID an so ALL of its prime ieals are of the form (0 or ḡ(x for some g(x Z[x] in which ḡ(x is irreucible in F p [x]. Thus in this case we see that either P = (p or P = (p, g(x where g(x Z[x] is irreucible in F p [x]. Since the maximal ieals of F p [x] are precisely the non-zero prime ieals, we see that (p, g(x is maximal yet (p is not. Next, we consier the harer case of I = (0, which is to say that P contains no integers. We can obtain Q[x] from Z[x] via Q[x] = D 1 Z[x] where D = Z {0}. In this case, we have that P D =, an so by homework 3 we know that P correspons to a prime ieal in Q[x]. Given the natural embeing ι: Z[x] Q[x], we have that P = ι 1 ( P for some prime ieal P Q[x]. The prime ieals of Q[x] are precisely the ieals of the form (0 an (f(x where f(x is an irreucible element in Q[x]. One possibility in this case is that P = ι 1 ((0 = (0. Otherwise, we must have P = ι 1 ((f(x for some irreucible polynomial f(x Q[x]. We can scale f(x Q[x] by units to assume WLOG that f(x Z[x] an that ct(f(x = 1. Because f(x is irreucible in Q[x], it follows that f(x is irreucible in Z[x] (because it has content 1 an hence it is prime since Z[x] is a UFD. Thus the ieal (f(x is prime in Z[x] an since D 1 (f(x = (f(x in Q[x], the corresponence of prime ieals proves that P = (f(x where f(x Z[x] in an irreucible, non-constant polynomial. Neither of these types of ieals are maximal, although we require an argument to show that (f(x is never maximal. Let f(x Z[x] be any non-constant polynomial. By the infiniteness of Z, there exists n Z with f(n 0, ±1. Thus f(n has some prime factor p. It s clear that we have the ieal containment (f(x (p, f(x. We will be one proving that (f(x is non-maximal precisely when we ve proven that (p, f(x is a proper ieal of Z[x]. To o so, assume by way of contraiction that (p, f(x = Z[x]. Then there exist polynomials h(x, k(x Z[x] with Plug in n to both sies to obtain 1 = p h(x + f(xk(x. 1 = p h(n + f(nk(n. This equation is absur since p ivies both terms on the right han sie of the equality yet p oes not ivie 1. 2. Let R be the set of all functions from the positive integers to the complex numbers: R = {f : Z + C}. Page 7
Consier the operations of + an on R where (f + g(n = f(n + g(n (f g(n = ( n f(g n (a Prove that R is a commutative ring uner + an with ientity element ɛ efine by { 1, n = 1 ɛ(n = 0, n > 1 Solution: That R is an abelian group uner + an that istributes over aition follows easily from the ring properties of C. The operation (calle Dirichlet convolution is commutative since (f g(n = ( n f(g = f(ag(b = (g f(n. n a,b Z + The har axiom to check is that is associative. To o so, let f, g, h R. Then for any n Z + we have (f g h(n = (f g(ah(b = ( f(cg( h(b. We can rewrite this last sum as cb=n f(cg(h(b. c=a This triple sum becomes ( f(c g(h(e = f(c(g h(e = f (g h(n. ce=n b=e ce=n Therefore f (g h = (f g h as claime. Lastly we must check that ɛ is an ientity element for. Of course for any n 1 we have (ɛ f(n = ( n ɛ(f = f(n n showing that ɛ f = f. (b Prove that the units in R are precisely those functions with f(1 0. Solution: One irection is easy. Assume that f R has an inverse g R. Then ɛ = f g so 1 = ɛ(1 = ( n f(g = f(1g(1 n showing that f(1 0. To prove the converse, we will recursively efine an inverse to f R. Assume that f(1 0. Let g R be the function satisfying g(1 = 1 f(1 Page 8
an for n > 1, g(n = 1 f(1 b<n f(ag(b. Notice that this efinition makes sense since the terms on the right han sie only use g(b for b < n. Let h = f g. Then h(1 = f(1g(1 = 1 an for n > 1, h(n = f(ag(b = f(1g(n + f(ag(b = 0 b<n by efinition of g. Therefore f g(n = ɛ(n for all n Z + an so f g = ɛ. (c A multiplicative function f : Z + C is a function satisfying f(ab = f(af(b whenever gc(a, b = 1. Prove that if f, g R are both multiplicative then so is f g. Solution: Let f, g R be multiplicative functions an let h = f g. We want to prove that h(xy = h(xh(y whenever x, y Z + an gc(x, y = 1. By efinition of h, we have h(xh(y = ( x f(g ( x f(eg. e x e y We can istribute this sum as h(xh(y = x e y ( x ( y f(f(eg g. e Since gc(x, y = 1 we also have gc(, e = 1 for any x an e y. multiplicativity of f an g so write h(xh(y = ( xy f(eg. e x e y Thus we can use the Given x an e y we clearly have e xy. Conversely, every k which ivies xy can be written uniquely in the form k = ab where a x an b y. You can prove this by showing that a = gc(k, x an b = gc(k, y, which follows by unique factorization. This observation allows us to conclue that h(xh(y = ( xy f(kg = h(xy. k k xy ( Consier the function 1 R efine by 1(n = 1 for every n 1 an let µ R be the inverse of 1. Then g = f 1 if an only if g µ = f. Use this to prove that if ϕ(n is the Euler phi function, ϕ(n = ( n µ. n (The function µ can be use to prove that there are irreucible polynomials over F p of egree n for any n 1. You can look at exercises 9.5.5 an 9.5.6 to see how alreay this is useful. This formula also gives an alternative proof that ϕ is multiplicative, which we ve previously seen via the Chinese Remainer Theorem. Page 9
Solution: I claim that n = n ϕ(. The easiest way to prove this is to use the fact that ϕ is multiplicative to prove this when n = p m, an then appeal to the previous problem to exten to all n 1. Here is an alternative proof. Consier the n fractions 1 n, 2 n,..., n 1 n, n n. Suppose we write each of these fractions in its lowest terms. Then the enominator of the resulting fraction will be a factor of n. Let be some positive ivisor of n, an let s count how many fractions have enominator in lowest terms. The fraction a n has enominator if an only if gc(a, n = n. Since 1 a n, this occurs if an only if a = n i where 1 i an n ( n = gc(a, n = gc i, n = n gc(i, which happens if an only if gc(i, = 1. Thus the number of such fractions is ϕ(. Since there are n fractions in total, we get that n = n ϕ(. This equation tells us that I = ϕ 1 where I R is the function I(n = n. Convolving both sies with µ gives I µ = ϕ 1 µ = ϕ ɛ = ϕ an hence ϕ(n = n ( n µ. Page 10