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1 THE JOHNS HOPKINS UNIVERSITY Faculty of Arts and Sciences FINAL EXAM - FALL SESSION ADVANCED ALGEBRA I. Examiner: Professor C. Consani Duration: take home final. No calculators allowed. Total Marks = 100 SOLUTIONS

2 1. [10 marks] Consider the ring of the Gaussian integers Z[i] (i = 1). (a) Is 4 + i a prime element in Z[i]? (b) Compute the cardinality of Z[i]/(4 + i). What group is it? (c) Find the G.C.D.(1 + 3i, 5 + i). Sol. (a) N(4 + i) = = 17 is a prime number in Z, and so 4 + i is an irreducible element of Z[i]. Moreover, Z[i] is a Euclidean domain, and so every irreducible element is also a prime element. Therefore 4 + i is a prime element in Z[i]. (b) The cardinality of R = Z[i]/(4 + i) is precisely N(4 + i) = 17. Let I = (4 + X). By the third isomorphism theorem we have: R /(X 2 + 1, 4 + X) /I/(X 2 + 1, 4 + X)/I Z/17Z, where the last isomorphism is obtained by noticing that X = (4 + X)(4 X) + 17 in, so that X = 17 in /I. It follows that R is the cyclic group of order 17. (c) We apply the division algorithm in Z[i]: 5 + i 1 + 3i = i and so we choose the approximate quotient 1 i, to get 5 + i (1 i)(1 + 3i) = 1 i Therefore 5 + i = (1 i)(1 + 3i) + 1 i where N(1 i) = 2 < N(1 + 3i) = 10. Now we repeat the process with 1 + 3i and 1 i: 1 + 3i = 1 + 2i 1 i and so 1 + 3i = ( 1 + 2i)(1 i) and the division algorithm ends. The algorithm tells us that GCD(5+i, 1+3i) = 1 i.

3 2. [20 marks] Give a proof or disprove the following statement: Z[ 3] is an Euclidean domain. Sol. O = Z[ 1+ 3 ] is a Euclidean domain, but Z[ 3] is a proper subring, so we may 2 have some doubts that the division algorithm of O when applied in Z[ 3] holds within Z[ 3]. Similarly we may have some reasonable doubts that the unique factorization in Z[ 3] holds, although O is a UFD, and so we turn our attention to the possibility of finding an element of Z[ 3] with non-unique factorization. We search for possible candidates among elements of Z[ 3] with small norm, the norm itself providing a means to discover possible factorizations. By trying out N(a + bi) = a 2 + 3b 2 for different small integer values of a and b, we soon find that 4 = = So 4 = (1 + i 3)(1 i 3) = 2 2. If α Z[ 3] is a unit, then there is a β Z[ 3] such that αβ = 1, and so N(α)N(β) = 1, which shows that N(α) = 1. Conversely, if N(α) = 1, since N(α) = αᾱ we see that α is a unit in Z[ 3]. Since the only integer solutions to a 2 + 3b 2 = 1 are a = ±1, b = 0, the units of Z[ 3] are ±1. If 2 = αβ in Z[ 3] then 4 = N(2) = N(α)N(β). N(α) = N(β) = 2 is impossible, since no element of Z[ 3] has norm 2. So without loss we have N(α) = 4, N(β) = 1 so β is a unit and hence 2 is irreducible in Z[ 3]. A similar argument shows that both 1 ± i 3 are irreducible since N(1 ± i 3) = 4 also. Therefore 4 has two factorizations into irreducibles in Z[ 3] which are clearly not associate, and thus Z[ 3] is not a UFD, so also not a Euclidean domain.

4 3. [10 marks] Consider the domain R = Z[ 3] := {a + b 3 a, b Z}. (a) Which among the following elements of R are invertible and why? , 2 3, 1 + 3, (b) Does the following equality of ideals hold in R? ( ) = (1 + 3) Explain in details your answer. (c) Is (3 + 3) a prime ideal of R? Explain in details. (d) Determine a maximal ideal M such that X 2 3 M. Sol. (a) We need only compute norms to see which elements in the list have norm ±1, where N(a + b 3) = a 2 3b 2. N( ) = 2, N(2 3) = 1, N(1 + 3) = 2, N( ) = 1. So the second and fourth elements in the list are units, the others are not. (b) Yes, the equality holds since and are associates by part (a) of this exercice: = (2 3)( ). (c) No it is not a prime ideal. N(3 + 3) = 6. If π is a prime element in R, then (π) Z is a prime ideal of Z, hence is pz for some prime p Z. Then p (π) shows that p = ππ in R, and so p 2 = N(p) = N(π)N(π ). Since π is not a unit in R, N(π) ±1, and it follows that p N(π) p 2 in integers. Since 6 is not a prime or the square of a prime (up to sign) in Z, (3 + 3) is not a prime ideal in R. (d) We consider the following ideal of : M = (X + 1) + (X 2 3) = (X + 1, X 2 3). We have M (X+1) M (X+1) by the third isomorphism theorem. Now /(X + 1) Z via the evaluation map X 1, and under this isomorphism, the ideal M/(X + 1) corresponds to the ideal 2Z. Hence (X+1) M (X+1) Z 2Z F 2 and therefore M is indeed a maximal ideal in. Why did we pick M as we did? Since /(X 2 3) Z[ 3] via the evaluation map X 3, the fourth isomorphism theorem tells us that the ideals of containing (X 2 3) are in one-to-one correspondence with the ideals of Z[ 3], and in particular that maximal ideals correspond to maximal ideals. The third isomorphism theorem tells us also that for any ideal I of containing (X 2 3) we have I (X 2 3) I (X 2 3) Z[ 3] Ī where Ī is the ideal of Z[ 3] corresponding to I under the isomorphism induced by evaluation at 3 described above. We don t need to know whether Z[ 3] is a Euclidean domain, or a PID or even a UFD. But we do know by part (a) that is an irreducible factor of 2 in Z[ 3], and this makes it a good candidate, since a

5 first guess to form a maximal ideal of containing (X 2 3) is simply to add in a prime element of Z, forming for example (2, X 2 3). (Note however that this ideal is not maximal, and in fact is not even prime, in. Arguments similar to the isomorphism arguments above show that /(2, X 2 3) F 2 [X]/(X + 1) 2, or also similarly, that /(2, X 2 3) F 2 [ 3], which is not an integral domain since (1 + 3) 2 = = 0 mod 2.) Since 2 does not remain prime in Z[ 3] we instead choose a (hopefully) prime (but certainly irreducible) factor of 2 such as 1+ 3, and consider the pre-image of the ideal (1+ 3) in under the evaluation map X 3, which is precisely M. (M = (1 + 3) in the notation above.) That s how we came upon our particular M as a candidate. (Note also that maximality of M shows that (1 + 3) is a maximal ideal of Z[ 3], and hence is a prime factor of 2.) Given the discussion above, we could also try to choose an integer prime p which remains prime in Z[ 3]. Say we have a prime p Z which remains prime in Z[ 3]. This occurs if and only if the reduction of X 2 3 modulo p is irreducible in F p [X]. Let M = (p, X 2 3). Then M p M p F p [X] (X 2 3) since the homomorphism reduction of coefficients modulo p which induces the isomorphism /p (Z/pZ)[X] F p [X] takes M to the ideal (X 2 3) of F p [X], where the bar indicates reduction modulo p. But since p remains prime in Z[ 3], X 2 3 is irreducible and hence prime in F p [X], so the ideal (X 2 3) is prime and hence maximal in the PID F p [X]. Therefore /M is a field and M is a maximal ideal. To find a particular p in order to answer the question, we note that we have already seen that 2 does not remain irreducible in Z[ 3], and obviously 3 becomes reducible also. However the reduction of X 2 3 mod 5 remains irreducible in F 5 [X] and so 5 is prime in Z[ 3]. It follows that taking M = (5, X 2 3) would also work.

6 4. [5 marks] Do the equations 3X 10Y = 2, 2X + 6Y = 5 have solutions in Z? If yes, determine for each equation a complete set of solutions. Sol. 2X + 6Y = 5 certainly has no solutions in Z since the left hand side of the equation is always even, the right hand side is odd. (A more formal way of saying this is that 5 is not a multiple of the GCD of 2 and 6, which is 2.) Since the GCD of 3 and 10 is 1, by the division algorithm in Z there exist integers A and B such that 3A + 10B = 1, and then certainly 3(2A) + 10(2B) = 2, so the first equation has solutions in Z. In particular one solution (found by observation) to 3X 10Y = 2 is given by X 0 = 14, Y 0 = 4. But then given this one particular solution we may find all solutions: for any m Z. X = X 0 + m 10 (3, 10) = 14 10m Y = Y 0 m 3 (3, 10) = 4 3m

7 5. [10 marks] Consider the quotient ring R = /(X 4 + 3X 3 + 1). (a) Is ( 2) R a maximal ideal of R? Why? (b) Is R a domain? Is R a field? Explain. (c) Does R have any further unit besides ±1? If yes, give an example of such unit. Sol. (a) Yes it is a maximal ideal. Let p(x) = X 4 + 3X 3 + 1, I = p(x) = (x 4 + 3X 3 + 1). We have ( 2) = (2 + I)/I, and the third isomorphism theorem yields I ( 2) = I 2+I I 2 + I 2 2+I 2 F 2 [X] (X 4 + X 3 + 1) where the last isomorphism is induced by reduction of coefficients modulo 2, which sends p(x) to q(x) = X 4 + X Now q(x) has no roots in F 2, so has no linear factors. Suppose q(x) = (X 2 +ax +b)(x 2 +cx +d) factors into quadratics over F 2, with a, b, c, d F 2. Multiplying out, we find q(x) = X 4 +(a+c)x 3 +(b+d+ac)x 2 + (bc + ad)x + bd. Comparing coefficients we see bd = 1 b = d = 1 and a + c = 1 a = 1, c = 0, without loss of generality. But then 0 = bc + ad = 1, a contradiction, and so q(x) is irreducible over F 2. Since q(x) is irreducible, F 2 [X]/(q(X)) is a field, which proves that ( 2) is a maximal ideal in R. (b) R is a domain but not a field. Since q(x) is the reduction of p(x) modulo 2 and q(x) is irreducible in F 2 [X], this proves that p(x) is irreducible in. Since is a UFD, I is a prime ideal and so R = /I is a domain. ( 2) is a nonzero maximal ideal in R, hence R cannot be a field. (The only ideals of a field are the zero ideal and the field itself.) (c) Yes, R has units besides ±1. For example, (X 3 + 3X 2 + I)( X + I) = X 4 3X 3 + I = X 4 3X 3 + p(x) + I = 1 + I so X + I is a unit in R which is not equal to ±1 + I, since X ± 1 / I.

8 6. [15 marks] Let H be a subgroup of a group G and write Cl(H) = {g 1 Hg : g G} for the conjugacy class of H in G. Show that (N G (H) = the normalizer of H in G). Cl(H) = G : N G (H) Assume that G is a finite group and prove that G cannot be the set-union of its conjugate subgroups ( G). Sol. The group G acts on the set of its subgroups by conjugation. The orbit of H under this action is exactly Cl(H). If G is finite, we know that Orb(H) = G / Stab(H). The stabilizer of H is N G (H). Therefore, Cl(H) = G : N G (H). If G is infinite, one can always argue that the map of sets Cl(H) {gn G (H) g G}, T = g 1 Hg1 1 g 1 N G (G) is (well defined) and bijective. Now, consider H < G (i.e. a proper subgroup of G). Call G : H = h (note that h > 1). Because H < N G (H), it follows that G : N G (H) h. Therefore, H has a most h conjugate subgroups. All together they contain at most elements. ( H 1)h + 1 = G (h 1) < G

9 7. [10 marks] Show that a group G cannot be described as a product of two conjugate subgroups different from G. Sol. We prove the contrapositive. Suppose G = HgHg 1 for some H G and g G. Since multiplication on the right by g is a bijection of G with itself, we have G = Gg = HgHg 1 g = HgH. Then we must have 1 = hgh for some h, h H, and so g = h 1 h 1 H. Hence ghg 1 = H, and so G = H 2 = H.

10 8. [20 marks] Show that if a group G has two normal, proper, distinct subgroups H, K of index p > 1, p prime number, s.t. H K = {1}, then: G = p 2 and G is not cyclic. Sol. Let H and K be two distinct subgroups satisfying the hypothesis. Then neither subgroup can contain the other, since in that case the contained subgroup would then have index strictly greater than p. We have G = G : {1} = G : H K < since both indexes G : H and G : K are finite. Thus G cannot be cyclic, since H = G / G : H = G / G : K = K, and cyclic groups have unique subgroups of any given (allowable) order. Since K is a normal subgroup of G, HK = KH is a subgroup and H < G = N G (K), so we can apply the second isomorphism theorem to conclude that H/(H K) HK/K. Now K HK G, and since K is normal in G, HK/K G/K. Moreover, G/K = p is prime, so either HK/K = {K} (the trivial group in G/K) and hence HK = K, or HK/K = G/K and hence HK = G (by the fourth isomorphism theorem, if you like). If HK = K then we have H HK = K, which we have already noted is not possible; hence HK = G. Since H K = {1} we find that H is isomorphic to G/K, so H = p. But then G = G H / H = G : H H = p 2. Note that without the assumption that H and K are distinct subgroups there is a counterexample to the question as literally stated: take G = Z/pZ and H = K = {1}. Then G = p and G is cyclic.

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