# x 3 2x = (x 2) (x 2 2x + 1) + (x 2) x 2 2x + 1 = (x 4) (x + 2) + 9 (x + 2) = ( 1 9 x ) (9) + 0

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1 1. (a) i. State and prove Wilson's Theorem. ii. Show that, if p is a prime number congruent to 1 modulo 4, then there exists a solution to the congruence x 2 1 mod p. (b) i. Let p(x), q(x) be polynomials in Z3[x], given by p(x) = x 6 + 2x 5 + 2x 4 + 2x 2 + x + 1 and q(x) = x 5 + 2x 3. Find all gcds of p(x) and q(x) in Z3[x]. Solution: Using long division of polynomials, the Euclidean algorithm gives: x 6 + 2x 5 + 2x 4 + 2x 2 + x + 1 = (x + 2) (x 5 + 2x 3 ) + (2x 3 + 2x 2 + x + 1) (x 5 + 2x 3 ) = (2x 2 + x + 2) (2x 3 + 2x 2 + x + 1) + (2x 2 + 1) (2x 3 + 2x 2 + x + 1) = (x + 1) (2x 2 + 1) + 0 Hence the gcds are 2x 2 + 1, x ii. Is Z3[x] a principal ideal domain? Is Z3[x]= hp(x)i a eld? Justify your answers. Solution: Since Z3 is a eld, Z3[x] is a Euclidean ring and therefore a PID. Since (x 2 + 2)jp(x), p(x) is not irreducible; hence Z3[x]= hp(x)i is not a eld. iii. Give a single generator for the ideal hp(x); q(x)i. Solution: x iv. Give a brief explanation as to why the Euclidean algorithm is guaranteed to terminate. Solution: The norm of the remainder, which is non-negative integer, is strictly decreasing at every step. 2. (a) Find a gcd of x 3 2x and x x + 1 in Q[x]. Solution: Using long division of polynomials, the Euclidean algorithm gives: x 3 2x = (x 2) (x x + 1) + (x 2) x x + 1 = (x 4) (x + 2) + 9 (x + 2) = ( 1 9 x ) (9) + 0 Hence 9 (or 1) is a gcd of the two polynomials. In other words, the two polynomials are relatively prime. 1

2 Alternatively, one can spot that x x + 1 = (x 1) 2 and then it is sucient to show that (x 1) 6 j (x ). (b) State and prove one of the following: Gauss' Lemma Eisenstein Criterion Note: In the literature \Gauss' Lemma" can refer to various (related) results. Here we mean that, for a UFD R, if f; g are primitive in R[x], then so is fg. This is the simplest version. The version of the Eisenstein Criterion as we saw it in class was the following: Suppose R is a unique factorization domain with quotient eld F. Let f(x) = P n i=0 a ix i be a polynomial of degree at least 1 with coecients in R. Suppose there exists an irreducible element p of R such that: 8i = 0; 1; : : : n 1; pja i ; p 6 ja n ; p 2 6 ja0. Then f[x] is irreducible in F [x]. (c) State what it means for a eld F to be a eld of fractions of an integral domain R. (d) Show that the polynomial x is irreducible in Q[x]. Solution: We have x 2 Z[x], and Q is the eld of quotients of the UFD Z. Clearly 2 is a prime integer which divides all the coecients (0,-2) except the leading coecient (1), and does not divide 2. Hence by the Eisenstein criterion x is irreducible in Q[x]. (e) Show that i. the polynomial (p(x; y)) 2 is primitive in (Q[x])[y] ii. the polynomial p(x; y) is irreducible in Q[x; y] where p(x; y) = (x x + 1)y 4 + (x 3 2x)y 3 + (x )y 2 + (x )y (x) Solution: (i): The rst two coecients of p(x; y) are the polynomials of part 2a. Since they are coprime, the polynomial is primitive. Hence by Gauss' Lemma, so is p(x; y) 2 (since Q is a eld, Q[x] is a Euclidean ring, and therefore a UFD, so Gauss' Lemma indeed does apply). (ii) We apply the Eisenstein Criterion to p(x; y), seen as an element of (Q[x])[y], with R = Q[x] (which is a UFD, as above). We take the 2

3 irreducible p = x (from part 2d). This clearly divides all the coecients apart from the leading coecient. From part 2a, it does not divide the leading coecient (otherwise the gcd in part 2a would not be 1). Moreover, since deg ((x ) 2 ) = 4 > 2 = deg 1 2 (x) and in Q[x] the degree of a product is the sum of the degrees (Q being an integral domain), (x ) 2 does not divide the constant term. Thus we may apply the Eisenstein Criterion to obtain that p(x; y) is irreducible in F [y], where F is the eld of fractions of Q[x]. In class we saw a Lemma stating that Lemma: Let F be the eld of fractions of the unique factorization domain R, and suppose f(x) 2 R[x] is primitive. Then f(x) is irreducible in R[x] i f(x) is irreducible in F [x]. Since the polynomial is primitive, the Lemma applies; hence p(x; y) is irreducible in R[y] = (Q[x])[y] = Q[x; y]. Note: In class (as here) we identied, without proof, R[x; y] with (R[x])[y]. We also saw in an exercise an application of the Eisenstein Criterion to (Q[x])[y]. 3. (a) Let D be an integral domain with multiplicative identity, and suppose the element a 2 D has a factorization into irreducibles. Prove that this factorization is unique up to associates, assuming one of the following: D is a Euclidean Ring; D is principal ideal domain; D = R[x] for some unique factorization domain R. (b) Consider the subring E of the eld of complex numbers given by 1 E = 2 (a + bi) j a; b 2 Z; a b is even : In other words, E consists of the complex numbers x + p y 3 such that x and y are either both integers or both odd multiples of 2 1. For example, 1 3p p 3 and are both in E, but 1 + 1p 2 3 is not. i. Find the units of E. Solution: Suppose (x + p y 3) (z + p t 3) = 1 3

4 where x; y; z; t are integers such that x y and z t are both even. Then (x + p y 3)(z + p t 3) = 4 p p whence x + y 3 z + t 3 = j4j p x + y 3 2 p z + t 3 2 = 16 (x 2 + 3y 2 )(z 2 + 3t 2 ) = 16: The positive factors of 16 are 1,2,4,8,16. However, x 2 + 3y 2 = 1 can only be achieved with jxj = 1; jyj = 0, which is to be discarded because x y should be even. Also, x 2 +3y 2 = 2 has no solutions at all for integer values of x; y. Similarly z 2 +3t 2 = 1 and z 2 +3t 2 = 2 can be discarded. The only way of factorizing 16 avoiding 1 and 2 is 16 = 4 4. Now x 2 + 3y 2 = 4 is has 6 solutions in integers: (x; y) = (1; 1); ( 1; 1); (1; 1); ( 1; 1); (2; 0); ( 2; 0). Note that in all cases x y is even. Thus the units are to be found among 1 2 ( 1 p 3); 1. Simple computations show that (1 + p p 3) (1 3) = 1 p ( 1 3) ( 1 + p 3) = 1 and 1 and -1 are clearly their own inverses. Hence the units are precisely the 6 elements 1 2 ( 1 p 3); 1. ii. State whether the following argument is correct: In E, we have that 2:2 = 4 = (1 + p p 3)(1 3). Therefore, E is not a unique factorization domain. If correct, ll in the missing details. If not, explain why. Solution: The reasoning is incorrect, because multiplying the factors 1 p 3) on the right hand side by the units 1 2 ( 1 p 3) we obtain the factors on the left hand side, 2 and 2. Thus the factorizations are the same up to units. iii. State whether 13 is irreducible in E or not, and justify your answer (if irreducible, give a proof; if not, give a non-trivial factorization). Solution: Suppose (x + p y 3) (z + p t 3) = 13 4

5 Then (x + p y 3)(z + p t 3) = 4 13 p p whence x + y 3 z + t 3 = j4 13j p x + y 3 2 p z + t 3 2 = (x 2 + 3y 2 )(z 2 + 3t 2 ) = : One way of achieving the above is jxj = 2 = jzj; jyj = jtj = 4. Note x y and z t are both even. Moreover, taking x = 2 = z, y = 4 = t, we obtain the factorization (1 + 2 p 3)(1 2 p 3) = 13: Since 1 2 p 3 are non-units (by part 3(b)i), this is a non-trivial factorization. This shows that 13 is not irreducible. 4. (a) Let D be an integral domain with multiplicative identity, and suppose the element a 2 D is not 0 and is not a unit. Show that a can be expressed as a nite product of irreducibles, assuming one of the following. D is a Euclidean Ring; D is principal ideal domain; D = R[x] for some unique factorization domain R. (b) Let a; b be Gaussian integers given by a = i and b = 3 + 6i. Find a gcd of a; b in the ring of Gaussian integers, and Gaussian integers x; y such that this gcd can be expressed in the form xa + yb. Solution: i = q0 (3 + 6i) + (1 3i) (3 + 6i) = q1 (1 3i) + (1 + 2i) (1 3i) = q2 (1 + 2i) + 0 where q0 = (2 + i); q1 = ( 1 + i); q2 = ( 1 i). Thus 1+2i is a gcd and we may take x = q1 = (1 i) and y = 1 + q0q1 = ( 2 + i). (c) Show that, if m; n are relatively prime positive integers, then m is invertible in Z n. (You may assume m < n.) Solution: There exist integers a; b such that am + bn = 1. Hence am 1 mod n. (d) Give an example of a unique factorization domain which is not a Euclidean ring. (You do not need to prove that your example is as claimed). Solution: We saw in class that Z[x] is a UFD but not a PID. Since PID ) ER, Z[x] is an example of a UFD which is not an ER. 5

6 (e) Find a gcd of x 4 + 4x 3 + 6x 2 + 4x + 1 and x 4 4x 3 + 6x 2 4x + 1 in Z[x]. Solution: The numbers constitute the fth line of Pascal's Triangle, that is, they are the binomial coecients 4 i, i = 0; 1; : : : ; 4. Hence the two polynomials are (1 + x) 4 and (1 x) 4. We saw that, for an integral domain R with 1 (such as Z) and a unit a of R, any polynomial ax+b is irreducible in R[x]. Thus 1 x are both irreducible. Since they are coprime, the two polynomials are coprime; one gcd is (a) Let R be a commutative ring with multiplicative identity. Show that R is a eld if and only if it has no proper ideals. (b) State the Second Isomorphism Theorem. (c) Let S be a commutative ring with multiplicative identity. Let a; b 2 S be such that hbi hai. Show that bja. (d) Let T be the ring Z24. Apart from the empty ideal and T itself, how many ideals are there in T? Solution: By the Second Isomorphism Theorem, there is a bijection between the ideals of T and the ideals of Z containing h24i. Thus it is sucient to count the ideals in Z which are strictly in between T and Z. Since Z is a PID, these ideals are of the form hbi for some b 2 Z. By part 5c, bj24. Thus jbj 2 f2; 3; 4; 6; 8; 12g. Moreover, by an exercise from class (essentially the converse of part 5c), for any such b we have h24i ( hbi ( Z. Clearly if b1 = b2, then hb1i = hb2i while if jb1j 6= jb2j then hb1i 6= hb2i. Thus there are 6 proper ideals in T. (e) Give a positive integer n such that Z n is not a eld, but has precisely one non-trivial factor ring (that is, only one factor ring apart from Z n itself and f0g). Show that this factor ring is a eld. (Note that there are various possibilities for n). Solution: The factor rings are in one-to-one correspondence with the ideals of Z n. From part 5d it should be clear that the proper ideals of Z n are in one-to-one correspondence with the non-trivial positive integer divisors of n. Looking at the factorization of n into primes, we see that this can only be achieved by taking n = p 2 for some prime p. The simplest example is thus n = 4. Whatever the choice of p, the unique ideal in Z n is maximal, hence the factor ring is a eld. (f) Let J be a proper ideal in a principal ideal domain Y. Show that J is contained in a maximal ideal. Hence, or otherwise, show that there 6

7 exists a factor ring F which is a eld and such that the elements of Y =J (seen as cosets of Y ) partition the elements of F. Solution: If J is maximal, we have the required maximal ideal containing J. Otherwise, there exists an ideal J1 such that J ( J1 ( Y. Again, we may assume J1 is not maximal (otherwise we are done). In this way we obtain a chain of ideals J ( J1 ( J2 ( By the following Lemma seen in class: Lemma: Suppose I1 ( I2 ( I2 ( is a strictly increasing chain of ideals in a PID. Then the chain is nite at some point we must obtain a maximal ideal J k containing J. We take the eld Y =J k for F. If x; y belong to the same coset with respect to J, then x y 2 J. Since J J k, x y 2 J k, and x; y belong to the same coset with respect to J k. Thus the partition of Y into cosets with respect to J is a renement of the partition into cosets with respect to J k, which are the elements of F. 7

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