Eighth Homework Solutions

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1 Math 4124 Wednesday, April 20 Eighth Homework Solutions 1. Exercise 5.2.1(e). Determine the number of nonisomorphic abelian groups of order First we write 2704 as a product of prime powers, namely To find the number of abelian groups of order 16, we find the number of partitions of 4. The partitions are (4), (3,1), (2,2), (2,1,1), (1,1,1,1) (so for example (3,1) corresponds to the group Z/8Z Z/2Z). Thus there are 5 abelian groups of order 16. Also there are 2 abelian groups of order 9. Therefore the total number of abelian groups of order 2704 is 5 2 = Exercises 5.2.2(e) and 5.2.3(e). List the elementary divisors and invariant factors for the abelian groups of order = Thus the number of abelian groups of order is 2 4 = 16. In the table below n will indicate Z/nZ; thus using this notation there are two abelian groups of order p 2, namely p 2 and p, p. Elementary Divisors Invariant Factors 2 2,3 2,5 2, ,3 2,5 2,7,7 6300,7 2 2,3 2,5,5, ,5 2 2,3,3,5 2, ,3 2,2,3 2,5 2, ,2 2 2,3 2,5,5,7,7 1260,35 2 2,3,3,5,5, ,15 2,2,3,3,5 2, ,6 2 2,3,3,5 2,7,7 2100,21 2,2,3 2,5,5, ,10 2,2,3 2,5 2,7,7 3150,14 2 2,3,3,5,5,7,7 420,105 2,2,3 2,5,5,7,7 630,70 2,2,3,3,5 2,7,7 1050,42 2,2,3,3,5,5, ,30 2,2,3,3,5,5,7,7 210, Exercise Let R be a commutative ring with 1. Prove that the center of the ring M n (R) is the set of scalar matrices. First we show that the scalar matrices are in the center of M n (R), so let A be a scalar matrix, say A = ri where r R. Then the i j entry of A is rδ i j, where δ is the Kronecker δ. If B M n (R) has entries b i j, then the i j entry of AB is k rδ ik b k j = rb i j, whereas the i j entry of BA is k b ik rδ k j = b i j r. Since R is commutative, rb i j = b i j r for all i, j, hence A commutes with B. It follows that A is in the center of M n (R).

2 Conversely suppose A is in the center of M n (R) and write A = (a i j ). Then by Exercise 7.2.6, we have E pq AE rs = a qr E ps. Also because A is in the center, this is E pq E rs A = δ qr E ps A. Therefore a qr E ps = δ qr E ps A. By considering the case q r, we now see that a qr = 0 if q r. Also a qq E 11 = E 11 A, so a qq is independent of q and we deduce that A is a scalar matrix. 4. Exercise Prove that the rings 2Z and 3Z are not isomorphic. Suppose θ : 2Z 3Z is an isomorphism. Set x = θ2. Then x 2 = θ(2 2 ) = θ4 and 2x = θ2 + θ2 = θ4. Therefore x 2 2x = 0. The only solutions to this are x = 0 and x = 2. But 2 / 3Z. Therefore θ2 = 0. But this is not possible because θ is one-to-one, so the result is proven.

3 Math 4124 Wednesday, April 27 Ninth Homework Solutions 1. Exercise Decide which of the following are ideals of the ring Z[x]. Let I be the relevant subset of Z[x]. (a) The set of all polynomials whose constant term is a multiple of 3. Yes. Obviously 0 I and I is an abelian group under addition. Finally suppose f = a 0 +a 1 x+ +a m x m I and g = b 0 +b 1 x+ +b n x n Z[x]. Then 3 divides a 0 and f g = a 0 b 0 + (a 1 b 0 + a 0 b 1 )x + + a m b n x m+n. Since 3 divides a 0 b 0, we see that f g I and we have proven that I is an ideal. (b) The set of all polynomials whose coefficient of x 2 is a multiple of 3. No. x I yet x 2 / I, so I is not even a subring. (c) The set of all polynomials whose constant term, coefficient of x and coefficient of x 2 are zero. Yes. This is just the ideal x 3 Z[x]. (d) Z[x 2 ]. No. 1 I but 1x / I (on the other hand, I is a subring). 2. Exercise Let I and J be ideals of the ring R with a 1. (a) Prove that I + J is the smallest ideal of R containing both I and J. (b) Prove that IJ is an ideal contained in I J. (c) Give an example where IJ I J. (d) Prove that if R is commutative and if I + J = R, then IJ = I J. (a) Obviously I +J is an ideal containing both I and J. Suppose K is an ideal containing both I and J. Since K is closed under addition, it must contain I +J. Therefore I + J is the smallest ideal containing I and J. (b) Since I R, we have IJ IR I. Similarly IJ J and it follows that IJ I J. (c) Let R = Z and I = J = 2Z. Then IJ = 4Z and I J = 2Z. (d) In view of (b), we need to prove that if x I J, then x IJ. Since R = I + J, we may write 1 = i + j where i I and j J. Then x = ix + x j IJ + IJ = IJ as required. 3. Define a ring homomorphism θ : R[x] C by θ(r) = r for r R and θ(x) = i. Then θ is onto, because if a + bi C where a,b R, then a + bi = θ(a + bx). By the fundamental homomorphism theorem R[x]/kerθ = C. Set M = kerθ. Then M R[x] (because kerθ is an ideal). Finally M is a maximal ideal because R[x]/M is a field.

4 4. There are several ways to do this. One way is to use the isomorphism (R/I)/(M/I) = R/M. Then M/I is a maximal ideal of R/I tells us that (R/I)/(M/I) is a field, hence R/M is a field and consequently M is a maximal ideal of R. 5. Consider the ring R := Z/6Z. Then P := 3Z/6Z is a maximal ideal of R, because R/P = Z/3Z which is a field, so certainly P is a prime ideal of R. It consists of two elements, namely 0 and 3. Thus if P 2 P, then P 2 has strictly less than two elements and hence P 2 = 0. This in particular would tell us that 3 2 = 0. But 3 2 = 9 = 3 0 and we conclude that P 2 = P.

5 Math 4124 Wednesday, May 2 Tenth Homework Solutions 1. (a) Let s,t S and suppose st / S. Then stx = 0 for some nonzero x R. Then tx 0 because t is a nonzero divisor, and hence stx because s is a nonzero divisor and tx 0. It follows that S is a multiplicatively closed subset of R. Obviously it contains 1 but not 0. (b) Let r/s S 1 R, where r R and s S. If r is a zero divisor, then rx = 0 for some nonzero x R, hence (r/s)(x/1) = 0 and x/1 0. This shows that r/s is a zero divisor. On the other hand if r is not a zero divisor, then r S by definition of S and thus s/r S 1 R. Since (r/s)(s/r) = 1, we see that r/s is a unit and we re finished. 2. Let K = kerα and J = kerβ. Then R/K = Z/2Z, a field, and we see that K is a maximal ideal of R. Similarly J is a maximal ideal of R. Since R/J = R/K, we see that J K and hence J + K = R. By the Chinese Remainder theorem, we see that R/(I J) = R/I R/J = Z/2Z Q. It follows that there is a ring epimorphism θ : R Z/2Z Q. 3. We need to show that if p/1 divides (a/s)(b/t) in S 1 R, then p/1 divides either (a/s) or (b/t). Since divisibility is not affected by multiplying by a unit, we may assume that p/1 divides ab/1, so (p/1)(r/x) = ab/1, for some r/x S 1 R. Since R is an integral domain, this tells us that pr = abx. Since p is prime, we see that p divides a or b or x. If p divides x, then p is a unit which contradicts the hypothesis that p is a prime, so without loss of generality we may assume that p divides a. We conclude that p/1 divides a/1 and the result follows. 4. Exercise on p Prove that R[x]/(x 2 + 1) = C. As in homework 9 problem 3, define θ : R[x] C by θ(r) = r for r R and θ(x) = i. Then θ is onto. Also θ(x 2 + 1) = (θx) 2 + θ(1) = 0, so (x 2 + 1) kerθ. Suppose f R[x] and θ( f ) = 0. By the division algorithm, we may write f = (x 2 + 1)q + r where q,r R[x] and degr 1. Then θ(r) = 0. Write r = ax+b where a,b R. Then θ(r) = a + ib; this can be 0 only when a = b = 0. This shows that f (x 2 + 1) and hence kerθ = (x 2 +1). The result now follows from the fundamental homomorphism theorem.

6 Math 4124 Wednesday, April 6 April 6, Ungraded Homework Exercise on page 165. In each of parts (a) to (d), give the number of nonisomorphic abelian groups of the specified order. (a) 100 (b) 576 (c) 1155 (d) (a) First write 100 as a product of prime factors; thus 100 = The number of partitions of 2 is 2, namely {2} and {1,1}, thus the number of isomorphism classes of abelian groups of order p 2 is 2. It follows that the number of isomorphism classes of groups of order 100 is 2 2 = 4. (b) 576 = The partitions of 6 are {6}, {5,1}, {4,2}, {4,1,1}, {3,3}, {3,2,1}, {3,1,1,1}, {2,2,2}, {2,2,1,1}, {2,1,1,1,1}, {1,1,1,1,1,1}; thus there are 11 partitions of 6. Also there are 2 partitions of 2. Therefore the number of isomorphism classes of abelian groups of order 576 is 2 11 = 22. (c) 1155 = The number of partitions of 1 is 1. Therefore the number of isomorphism classes of abelian groups of order 1155 is = 1. (d) = The partitions of 3 are {3}, {2,1}, {1,1,1}, so there are 3 partitions of 3. Therefore the number of isomorphism classes of abelian groups of order is 3 3 = 9. Exercise on page 165. In each of parts (a) to (d), give the lists of elementary divisors for all abelian groups the specified order and then match each list with the corresponding list of invariant factors give the number of nonisomorphic abelian groups of the specified order. (a) 270 (b) 9801 (c) 320 (d) 105 (a) 270 = Elementary Divisors Invariant Factors 2,5, ,5,3 2,3 3,90 2,5,3,3,3 3,3,30 (b) 9801 = Elementary Divisors Invariant Factors 3 4, ,3,11 2 3, ,3 2,11 2 9, ,3,3,11 2 3,3,1089 3,3,3,3,11 2 3,3,3, ,11,11 11, ,3,11,11 33, ,3 2,11,11 99,99 3 2,3,3,11,11 3,33,99 3,3,3,3,11,11 3,3,33,33

7 (c) 320 = Elementary Divisors Invariant Factors 2 6, ,2,5 2, ,2 2,5 4,80 2 4,2,2,5 2,2,80 2 3,2 3,5 8,40 2 3,2 2,2,5 2,4,40 2 3,2,2,2,5 2,2,2,40 2 2,2 2,2 2,5 4,4,20 2 2,2 2,2,2,5 2,2,4,20 2 2,2,2,2,2,5 2,2,2,2,20 2,2,2,2,2,2,5 2,2,2,2,2,10 (d) 105 = Elementary Divisors Invariant Factors 3,5,7 105 Exercise on page 166. In each of parts (a) to (d), determine which pairs of abelian groups listed are isomorphic (here the expression {a 1,a 2,...,a k } denotes the abelian groups Z a1 Z a2 Z ak ). Remarks Obviously if two groups are isomorphic, they have the same order (but not conversely). Also, we shall use the elementary divisor form of the structure theorem for finitely generated abelian groups. (a) {4,9}, {6,6}, {8,3}, {9,4}, {6,4}, {64}. The orders of the groups are 36,36,24,36,24,64. In terms of elementary divisors, they are Z 4 Z 9, Z 2 2 Z2 3, Z 3 Z 8, Z 4 Z 9, Z 2 Z 3 Z 4, Z 64. We conclude that the first and fourth groups are isomorphic, but there are no other isomorphisms. (b) {2 2,2 3 2 }, {2 2 3,2 3}, { }, { ,2}. The order of the groups are 72,72,72,72 (thus the orders of the groups are no help in determining isomorphism classes here). In terms of elementary divisors, they are Z 2 Z 4 Z 9, Z 2 Z 4 Z 2 3, Z 8 Z 9, Z 2 Z 4 Z 9. Thus the first and fourth are isomorphic, but there are no other isomorphisms. (c) { , }, { ,5 7 2 }, {3 5 2,7 2,3 5 7}, {5 2 7,3 2 5,7 2 }. The orders of the groups are all The elementary divisor decompositions are Z 9 Z 5 Z 25 Z 7 Z 49, Z 9 Z 5 Z 25 Z 7 Z 49, Z 3 Z 3 Z 5 Z 25 Z 7 Z 49, Z 9 Z 5 Z 25 Z 7 Z 49. The first, second and fourth groups are isomorphic, but there are no other isomorphisms.

8 (d) { , ,2 5 2 }, { , }, {2 2,2 7,2 3,5 3,5 3 }, {2 5 3, ,2 3,7}. The order of all the groups is The elementary divisor decompositions are Z 2 Z 4 Z 8 Z 5 Z 25 Z 125 Z 7, Z 8 Z 8 Z 125 Z 125 Z 7, Z 2 Z 4 Z 8 Z 125 Z 125 Z 7, Z 2 Z 4 Z 8 Z 125 Z 125 Z 7. The third and fourth are isomorphic, but there are no other isomorphisms.

9 Math 4124 Monday, April 11 April 11, Ungraded Homework Exercise on page 230. Decide which of the following (a) (f) are subrings of Q: (a) the set of all rational numbers with odd denominators (when written in lowest terms) (b) the set of all rational numbers with even denominators (when written in lowest in terms) (c) the set of nonnegative rational numbers (d) the set of squares of rational numbers (e) the set of all rational numbers with odd numerators (when written in lowest in terms) (f) the set of all rational numbers with even numerators (when written in lowest terms) Let R be the described set, so we need to decide whether R is a subring of Q. (a) Yes. (i) 0 R because 0 = 0/1. (ii) Suppose a/r,c/s R where r,s are odd. Then obviously a/r R, and a/r + b/s = (as + br)/rs. This is in R because rs is odd and when we reduce to lowest terms, the denominator will still be odd. (iii) Suppose a/r,c/s R. Then (a/r)(c/s) = (ac)/(rs); since rs is odd and the denominator will remain odd when reducing to lowest terms, it follows that R is closed under multiplication. (b) No, because R is not closed under addition. For example 1/2,1/2 R, but 1/2 + 1/2 = 1/1 / R. (c) No, because R is not closed under inverses for addition. For example 1 R but 1 / R. (d) No, because R is not closed under addition. For example 1,1 R but / R. (e) No, because R is not closed under addition. For example 1,1 R but / R. (f) Yes. (i) 0 R because 0 = 0/1. (ii) Suppose a/r,c/s R where a,c are even and the fractions are in lowest terms. Then obviously a/r R. Also we may write a = 2x and b = 2y, and r,s are odd. Then a/r + c/s = 2(xs + yr)/rs R. (iii) Suppose a/r,c/s R where a,c are even and the fractions are in lowest terms. Then we may write a = 2x and b = 2y, and r,s are odd. Thus (a/r)(c/s) = 4(xy)/rs R.

10 Exercise on page 231. For a fixed element a R, define C(a) = {r R ra = ar}. Prove that C(a) is a subring of R containing a. Prove that the center of R is the intersection of the subrings C(a) over all a R. (i) 0 C(a) because a0 = 0a = 0. (ii) Suppose x,y C(a). Then ax = xa and ay = ya. Thus so x + y C(a). Also so x C(a). a(x + y) = ax + ay = xa + ya = (x + y)a a( x) = (ax) = (xa) = ( x)a (iii) Suppose x,y C(a). Then ax = xa and ay = ya. Thus a(xy) = xay = (xy)a, so xy C(a) and we have proved that C(a) is a subring of R. The final part, that R is the intersection of all C(a), is just set theory. Exercise on page 231 (last two sentences). Let R and S be rings. Prove that R S is commutative if and only if both R and S are commutative. Prove that R S has an identity if and only if both R and S have identities. Let (a, p),(b,q) R S. Then (a, p)(b,q) = (b,q)(a, p) if and only if (ab, pq) = (ba,qp); that is if and only if ab = ba and pq = qp. It follows that R S is commutative if and only if R and S are commutative. Suppose R and S have an identity; let us call the identity of R e, and the identity of S f. Then re = er = r for all r R and s f = f s = s for all s S. It follows that (e, f ) is an identity for R S, because (r,s)(e, f ) = (re,s f ) = (r,s) = (re,s f ) = (r,s)(e, f ). Conversely suppose R S has an identity; let us call it (e, f ) where e R and f S. Then (r,s)(e, f ) = (e, f )(r,s) = (r,s) for all (r,s) R S. By looking at the first coordinate, we see that re = er = r, consequently e is an identity for R. Similarly by looking at the second coordinate, we see that f is an identity for S. Exercise on page 237. Let p(x) = 2x 3 3x 2 + 4x 5 and let q(x) = 7x x 4. In each of parts (a),(b) and (c) compute p(x) + q(x) and p(x)q(x) under the assumption that the coefficients of the two given polynomials are taken from the specified ring (where the integer coefficients are taken mod n in parts (b) and (c)): (a) R = Z, (b) R = Z/2Z, (c) R = Z/3Z.

11 (a) p(x) + q(x) = 9x 3 3x x 9 p(x)q(x) = 14x 6 21x x 4 142x x 2 181x + 20 (b) p(x) + q(x) = x 3 + x 2 + x + 1, p(x)q(x) = x 5 + x (c) p(x) + q(x) = x, p(x)q(x) = 2x 6 + x 4 + 2x 3 + 2x + 2 Exercise on page 238. Let S be a ring with an identity 1 0. Let n Z + and let A be an n n matrix with entries from S whose i, j entry is a i j. Let E i j be the element of M n (S) whose i, j entry is 1 and whose other entries are all 0. (a) Prove that E i j A is the matrix whose ith row equals the jth row of A and all other rows are zero. (b) Prove that AE i j is the matrix whose jth column equals the ith column of A and all other columns are zero. (c) Deduce that E pq AE rs is the matrix whose p,s entry is a qr and all other entries are zero. Note that A = a i j E i j. Also E i j E pq = δ jp E iq, where δ indicates the Kronecker δ; that is i, j δ ii = 1 and δ i j = 0 if i j. (a) (b) E i j A = p,q a pq E i j E pq = p,q a pq E iq δ jp = a jq E iq q This matrix clearly has 0 s in the rows other than the ith. On the other hand the qth entry of the ith row is a jq, that is the qth entry of the jth row of A. AE i j = p,q a pq E pq E i j = p,q a pq E p j δ qi = a pi E p j p This matrix clearly has 0 s in the columns other than the jth. On the other hand the pth entry of the jth column is a p j, that is the pth entry of the jth column of A. (c) From (a) we see that E pq A is the matrix whose pth row is the jth row of A, and has zeros elsewhere. Now apply (b).

12 Math 4124 Wednesday, April 13 April 13, Ungraded Homework Exercise on page 248. Decide which of the following are ring homomorphisms from M 2 (Z) to Z. ( ) a b (a) a c d ( ) a b (b) a + d c d ( ) a b (c) ad bc c d Let us call the relevant homomorphism θ. (a) No, because multiplication is not respected If x = θ(x)θ(y) = 0 0 = 0, but θ(xy) = 1. ( ) and y = ( ) 0 0, then 1 0 (b) No, because multiplication is not respected. In fact taking x,y as above, we get θ(x)θ(y) = 0 0 = 0, but θ(xy) = 1. (c) No, because addition is not respected. In fact taking x,y as above, we get θ(x)+θ(y) = = 0, but θ(x + y) = 1. Exercise on page 248. Decide which of the following are ideals of the ring Z Z. (a) {(a,a) a Z} (b) {(2a,2b) a,b Z} (c) {(2a,0) a Z} (d) {(a, a) a Z} Let the relevant set of elements be I. (a) No. For example (1,1) I, but (1,1)(1,0) / I. (b) Yes. 0 I and I is closed under +. Finally if (x,y) Z Z, then (2a,2b)(x,y) = (2ax,2by) I. (c) Yes. 0 I and I is closed under +. Finally if (x,y) Z Z, then (2a,0)(x,y) = (2ax,0) I. (d) No. For example (1, 1) I, but (1, 1)(1, 1) = (1,1) / I.

13 Math 4124 Monday, April 18 April 18, Ungraded Homework Exercise on page 256 Let R be a commutative ring with identity 1 0. Prove that the principal ideal generated by x in the polynomial ring R[x] is a prime ideal if and only if R is an integral domain. Prove that (x) is a maximal ideal if and only if R is a field. Define a ring homomorphism θ : R[x] R by θ(r) = r for r R and θ(x) = 0; another way to describe this homomorphism is θ( f ) = f (0). Then θ is onto and kerθ = (x). Therefore R[x]/(x) = R by the fundamental isomorphism theorem. Since (x) is a prime ideal if and only if R[x]/(x) is an integral domain, and (x) is a maximal ideal if and only if R[x]/(x) is a field, the result is proven. Exercise on page 256 Let R be an integral domain. Prove that (a) = (b) for some elements a,b R if and only if a = ub for some unit u of R. Suppose a = ub for some unit u of R. Then a br, so (a) (b). Also if uv = 1, then b = va and we deduce that (b) (a). Therefore (a) = (b). Conversely suppose (a) = (b). The result is obvious if a = 0, because then b = 0 and we can take u = 1. Therefore we may assume that a 0. Then we may write a = br and b = as for some r,s R and we have a = ars. Since a 0 and R is an integral domain, we see that 1 = rs, so r is a unit and the result follows.

14 Math 4124 Wednesday, April 20 April 20, Ungraded Homework Exercise on page 256 Let R be a commutative ring with identity 1 0. Prove that R is a field if and only if 0 is a maximal ideal. First suppose R is a field. To prove that 0 is a maximal ideal, we must show that there are no ideals strictly between 0 and R, equivalently 0 and R are the only ideals of R. Let I be a nonzero ideal of R. Let 0 x I. Then xx 1 = 1 I and we deduce that 1r I for all r R. This shows that I = R and it follows that 0 is a maximal ideal of R. Conversely suppose 0 is a maximal ideal of R. To show that R is a field, we need to prove that every nonzero element of R has a multiplicative inverse, so let 0 x R. Since xr is a nonzero ideal of R (remember that R is commutative), we see that xr = R and hence there exists y R such that xy = 1. This proves that x is invertible as required. Exercise on page 257 Let R be a commutative ring with identity 1 0. Prove that if P is a prime ideal of R and P contains no zero divisors, then R is an integral domain. Let 0 a,b R. We want to prove that ab 0. If a or b P, this is true because P contains no (nonzero) zero divisors. Therefore we may assume that a,b / P. But then ab / P because P is a prime ideal and the result follows. Let R be a commutative ring identity 1 0 and let M be a maximal ideal of R. Suppose M is nilpotent, that is there exists n N such that M n = 0. Prove that M is the only prime ideal of R. Let P be a prime ideal of R. If M is not contained in P, we may choose x M \ P. Then x n = 0, hence x n P and since P is prime, we deduce that x P. This is a contradiction. Therefore M P and since M is maximal and P R, we deduce that P = M and the result follows.

15 Math 4124 Monday, April 25 April 25, Ungraded Homework Exercise on page 264 Let R be an integral domain and let D be a multiplicatively closed subset of R which contains 1 but not 0. Prove that the ring of fractions D 1 R is isomorphic to a subring of the quotient field of R (hence is also an integral domain). Let S = R 0 and let φ : R S 1 R denote the natural homomorphism defined by φ(r) = r/1. Note that S 1 R is the quotient field of the integral domain R, and φ is a monomorphism. In this situation we usually identify R with φ(r), but we won t do so here. Since φ(d) 0 for all d D, we see that φ(d) is invertible for all d D. By the universal property for D 1 R, we may extend φ to a map α : D 1 R S 1 R. Specifically if θ : R D 1 R is the natural homomorphism defined by θ(r) = r/1, then φ = αθ. All that remains to prove is that α is a monomorphism, equivalently kerα = 0, so suppose r/d kerα where r R and d D. Since kerα D 1 R, we see that (r/d)(d/1) = r/1 kerα and hence θ(r) kerα. Therefore φ(r) = αθ(r) = 0 and since φ is a monomorphism, we deduce that r = 0. We conclude that r/d = 0 and the result follows. Let R,S be commutative rings with 1 0. Prove that R S is never an integral domain. Furthermore prove there exists I R S with I 0,R S and I 2 = I. Since (1, 0)(0, 1) = (0, 0), we see that R S is not an integral domain. Now set I = {(r, 0) r R}. It is routine to check that I R S. Finally if x I, then we may write x = (r,0) where r R, and then x = (r,0)(1,0) = x(1,0) I 2. Thus I I 2, and since I 2 I because I R S, we conclude that I = I 2. Give an example of a commutative ring with a 1 0 that has a nonzero prime ideal which is not maximal. One possibility is the polynomial ring R := Z[x] and the ideal P := xz[x]. We can define a homomorphism θ : Z[x] Z by θ(a 0 +a 1 x+ +a n x n ) = a 0. It is easy to check that θ is a homomorphism, and it is clearly onto. Its kernel consists of all those polynomials with zero constant term (that is, a 0 = 0), which is precisely P. By the fundamental homomorphism for rings, R/P = Z. Since Z is an integral domain but not a field, the same is true for R/P, and we deduce that P is a prime ideal but not a maximal ideal. Exercise on page 311 Determine whether the following polynomials are irreducible in the rings indicated. For those that are reducible, determine their factorization into irreducibles. The notation F p denotes the finite field Z/pZ, p a prime. (a) x 2 + x + 1 in F 2 [x] (b) x 3 + x + 1 in F 3 [x] (c) x in F 5 [x] (d) x x in Z[x] In each case, let f (x) indicate the relevant polynomial.

16 (a) Since f (0) = f (1) = 1 0, we see that f has no roots in F 2. This means that f has no degree 1 factor. If f is not irreducible, then f must be the product of two degree 1 polynomials, and we have just shown that this is not the case. Therefore f is irreducible. (b) Here f (1) = 0, so x+2 is a factor and we have f (x) = (x+2)(x 2 +x+2). Since x 2 +x+ 2 has no root in F 3, we see that x 2 +x+2 is irreducible and therefore (x+2)(x 2 +x+2) is the factorization of f into irreducibles. (c) Since f (0) = 1 and f (1) = f (2) = f (3) = f (4) = 1, we see that f has no root, so the only possibility is that f factors into two degree 2 irreducible factors. By inspection (or apply the root theorem to y 2 + 1) f = (x 2 + 2)(x 2 + 3), and x and x are irreducible. (d) Suppose x x is not irreducible. Then we can write x x = f (x)g(x) where f,g Z[x] are monic polynomials each of degree at least 1. Taking this mod 5, we get a nontrivial factorization of x in F 5 [x], which contradicts (c).

17 Math 4124 Wednesday, April 27 April 27, Ungraded Homework Let k = Z/2Z, the field with 2 elements. Prove that k[x]/(x 5 + x 2 + 1) is a field with 32 elements. First we show that x 2 + x + 1 is the only irreducible polynomial of degree 2 in k[x]. Indeed there are only 4 polynomials of degree 2, namely x 2 + x + 1, x 2 + 1, x 2 + x, x 2. However the last 3 are not irreducible, since x is a factor of the last 2, and if we plug in x = 1 in x 2 + 1, we get 0 and so x 1 (= x + 1) is a factor of x Next note that x 5 + x has no degree 1 factor, because if we plug in x = 0 or 1, we get 1. It follows that if x 5 + x is not irreducible, it must have an irreducible degree 2 factor. Since the only irreducible degree two polynomial is x 2 + x + 1, we deduce that x 5 + x = (x 2 + x + 1) f for some f k[x]. but x 5 + x = (x 2 + x + 1)(x 3 + x 2 ) + 1, which shows that x 2 + x + 1 does not divide x 5 + x 2 + 1, and we have shown that x 5 + x is irreducible. Since k[x] is a PID, it follows that k[x]/(x 5 + x 2 + 1) is a field, with 2 5 = 32 elements.

18 Math 4124 Monday, May 2 May 2, Ungraded Homework Prove that 2 is irreducible but not prime in Z[ 13]. Here Z[ 13] = {a+b 13 a,b Z}. Define N : Z[ 13] Z by N(a+b 13) = a 2 13b 2 for a,b Z. Then N(αβ) = N(α)N(β) for α,β Z[ 13]; to see this write α = a + b 13 and β = c+d 13. Then N(αβ) = N(ac+13bd +(ad +bc) 13) = (ac+13bd) 2 13(ad + bc) 2, N(α)N(β) = (a 2 13b 2 )(c 2 13d 2 ) and (ac + 13bd) 2 13(ad + bc) 2 = (a 2 13b 2 )(c 2 13d 2 ). Since N is a multiplicative homomorphism (but not a ring homomorphism), we see that if α is a unit in Z[ 13], then N(α) is a unit in Z, i.e. ±1. Conversely suppose N(a + b 13) = ±1. Then (a + b 13)(a b 13) = ±1 and we see that a + b 13 is a unit in Z[ 13]. Next we note that there is no α Z[ 13] such that N(α) = ±2. Indeed if N(a + b 13) = ±2, then a 2 13b 2 = ±2, so a 2 = ±2 mod 13. However there is no such a (just go through the 13 possibilities for a). Suppose 2 = αβ in Z[ 13]. Then 4 = N(α)N(β). Since N(α),N(β) ±2, we see that either N(α) = ±1 or N(β) = ±1. We deduce that α or β is a unit. Also N(2) = 4 ±1, so N(2) is not a unit. Finally 2 0 and we conclude that 2 is irreducible. Now 2 6 = ( )(1 + 13). Thus if 2 is prime, we have 2 divides or This is not the case, because the resulting quotients would be 1/2 + 13/2 and 1/2 + 13/2, neither of which are in Z[ 13]. The result follows.