A few exercises. 1. Show that f(x) = x 4 x 2 +1 is irreducible in Q[x]. Find its irreducible factorization in

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1 A few exercises 1. Show that f(x) = x 4 x 2 +1 is irreducible in Q[x]. Find its irreducible factorization in F 2 [x]. solution. Since f(x) is a primitive polynomial in Z[x], by Gauss lemma it is enough to show that f(x) is irreducible in Z[x]. Observe that f(0) 1 mod 2 and f(1) 1 mod 2, so f(x) mod 2 does not have a root in F 2. So f(x) does not have a root in Z, and by the corollary to Gauss lemma it follows that f(x) does not have a root in Q, so f cannot have a degree one factor in Q[x]. So it remains to show that f does not factor as a product of two quadratic polynomials in Z[x]. If possible, suppose f(x) = (ax 2 +bx+c)(dx 2 +ex+f) is a factorization of f in Z[x]. Then ad = 1, so without loss we may assume that a = d = 1. So f(x) = (x 2 +bx+c)(x 2 +ex+f) = x 4 +(b+e)x 3 +(be+c+f)x 2 +(bf +ce)x+cf. Comparing coefficients we get b+e = 0, (be+c+f) = 1, bf +ce = 0, cf = 1. Now cf = 1 implies either c = f = 1 or c = f = 1, so c+f = 2 or 2, so c+f 2 mod 4. From the first two equations, we get b 2 = 1+c+f 3 mod 4 which is not possible. In F 2 [x], we have f(x) = x 4 +2x 2 +1 x 2 = (x 2 +1) 2 x 2 = (x 2 +1+x)(x 2 +1 x) = (x 2 +x+1) 2 and note that x 2 +x+1 is irreducible in F 2 [x]. 2. Let R be a PID and a,b,c be elements of R such that a and b are relatively prime. If a divides bc, then show that a divides b. solution. Since a and b are relatively prime, gcd(a,b) = 1, on other words, the ideal (a,b) is equal to R. In other words, there exists r,s R such that ra+sb = 1. So rac+sbc = c. Since a divides both rac and sbc, it follows that a divides c. 3. Find all integers x such that x 2 mod 5, x 6 mod 8. solution. Let (8 1 mod 5) denote any integer which represents the inverse of 8 modulo 5. Let e 1 = (8 1 mod 5).8. Then e 1 0 mod 8 and e 1 1 mod 5. Similarly let e 2 = (5 1 mod 8)5. We can pick (8 1 mod 5) = 2 and (5 1 mod 8 = 5. We calculate (e 1,e 2 ) = (2.8,5.5) = (16,25). As in the proof of Chinese remainder theorem, we compute 2e 1 +6e 2 = mod 40. So the general solution of the given congreunces is x = 22+40k where k is any integer. 4. Find the greatest common divisor of (1 + 13i) and (10+11i) in Z[i]. Find a generator for the ideal generated by (1+13i) and (10+11i) in Z[i]. solution. Let a = 1+13i, b = 10+11iand c = gcd(a,b). Since Z[i] is an Euclidean domain, c is a generator for the ideal (a,b). Write N(z) = z z. Since c divides a and b, N(c) must divide gcd(n(a),n(b)) = gcd(170,221) = 17. The integer 17 factors in Z[i] as 17 = (4+i)(4 i). This suggests that (4+i) and (4 i) are possible factors of a and b. One can verify directly that (4+i) divides both a and b. The quotients a/(4+i) and b/(4+i) have norm 10 and 13 respectively. Since 10 and 13 are relatively prime, so are a/(4 + i) and b/(4 + i). So (4+i) = gcd(a,b). 1

2 5. Let R be a commutative ring. Given r R, define ann(r) = {x R: xr = 0}. (a) Verify that ann(r) is an ideal in R. (b) Let C = {ann(r): r R,r 0}. Let Z be a maximal element of C with respect to inclusion. Show that Z is a prime ideal of R. solution. Let x,y ann(r) and s R. Then xr = 0 = yr, so (sx+y)r = 0, so (sx+y) ann(r). Hence ann(r) is an ideal in R. (b) Let s R such that Z = ann(s). Suppose x,y R such that xy Z and y / Z. Then ys 0 and xys = 0. Now ann(ys) ann(s), so by maximality of ann(s), we must have ann(ys) = ann(s). But x ann(ys). So x ann(s). 6. Let R be the ring of real valued continuous functions on [0,1]. For each x [0,1] define M x = {f R: f(x) = 0}. Show that M x is a maximal ideal in R for each x R. Show that these are the only maximal ideals in R. solution. For each x [0,1], we have a ring homomorphism ev x : C R given by ev x (f) = f(x). The image of ev x is equal to the field R, so ker(ev x ) = M x is a maximal ideal in C. Conversely, suppose M is any maximal ideal in C. If possible, suppose that for each x [0,1], there exists f x M such that f x (x) 0. Since f x is continuous, there exists a subset V x, relatively open in [0,1], such that x V x and f x is non-vanishing on V x. The open sets {V x : x [0,1]} cover [0,1]. Since [0,1] is compact, we can find x 1,,x n [0,1] such that V x1 V xn = [0,1]. Then g = fx fx 2 n is an element of M such that g(x) > 0 for each x [0,1], so g 1 C, that is g is a unit in C. Since M contains a unit, it follows that M = C, which is a contradiction. So given any maximal ideal M, there exists x [0,1] such that f(x) = 0 for all x M. In other words M M x. Since M x is maximal, it follows that M = M x. 7. Let a = Consider the homomorphism φ : Z[x] R given by φ(f(x)) = f(a). Show that ker(φ) is a maximal ideal and find a generator for this maximal ideal. solution. The homomorphism extends to a homomorphism Φ : Q[x] R given by Φ(f) = f( 2 + 1). Under the map Φ, the polynomial (x 1) 2 maps to (a 1) 2 = 2. So g(x) = (x 1) 2 2 = x 2 2x 1 is in the kernel of Φ. The quadratic polynomial g does not have any rational roots, so it is irreducible in Q[x]. Since Q[x] is a principal ideal domain, ker(φ) is the principal ideal generated by g. Suppose f ker(φ). We can write f = nh where n is some integer and h Z[x] is a primitive polynomial. Then nh(a) = 0, so h(a) = 0, so h ker(φ). So g(x) divides h(x) in Q[x]. Since both g and h are primitive polynomials in Z[x], Gauss lemma implies that g(x) divides h(x) in Z[x]. So h(x) is in the principal ideal g(x)z[x]. So f(x) g(x)z[x] too. 8. Let R be a commutative ring. Let M be an ideal in R. (a) Suppose every element in R \ M is an unit in R. Then show that M is the unique maximal ideal in R. (b) Suppose M is a maximal ideal such that for each m M, the element 1 + m is an unit in R. Show that M is the unique maximal ideal in R. solution. (a) Let A be any ideal in R. If A contains an unit u of R, then A would contain u 1 u = 1, so it would follow that A = R. So if A is any proper ideal in R, then A does not 2

3 contain any units, so A M. In other words, M contains all the proper ideals of R. So M is the unique maximal ideal of R. (b) Let a / M. Consider the ideal M + (a) = {m + ra: r R,m M} generated by M and a. This ideal contains M and a, so it properly contains M. Since M is maximal it follows that M + (a) = R. In other words, there exists m M and r R such that m+ra = 1. So ra = (1 m) with m M, so ra is a unit, which implies a is a unit. So each element a / M is a unit. Now part (a) implies that M is the unique maximal ideal of R. 9. Show that x p x has p distinct zeroes in Z/pZ = F p for any prime p. Conclude that x p x = x(x 1)(x 2) (x (p 1) in F p. solution. Notethatthemultiplicative groupf p hasorder(p 1), so foreachnon-zeroelement a of F p, we have a p 1 = 1. It follows that a p = a for all a F p. Since this identity ap = a also holdsfora = 0, wegetthata p = aforalla F p. Soeacha F p isarootof(x p x)andhence (x a) isafactor of(x p a) foreach a F p. Since F p (x) isanufd, let f(x) = g 1 (x) g k (x) be the irreducible factorization of f(x). For each a F p we can write f(x) = (x a)g(x) for some g(x). So (x a)g(x) = g 1 (x) g k (x). By the uniqueness of the factorization it follows that (x a) is equal to one of the irreducible factors g j (x) (upto a scalar). So for each a F p, the factor (x a) appears in the list {g 1 (x),,g k (x)} and since the factors (x a) are distinct for distinct values of a, it follows that f(x) = a F p (x a)h(x) for some polynomial h(x). Comparing the degrees, it follows that h(x) is a constant. Comparing the coefficients of x p, it follows that h(x) = Let A and B be two ideals in a commutative ring R such that A+B = R. Then show that A k +B r = 1 for any strictly positive integer k,r. solution. Since A+B = R, there exists a A and b B such that a+b = 1. So a = 1 b, so a k = 1 b where b B, so a k + b = 1 with a k A k and b B. So A k + B = R. Now repeat the argument with the B playing the role of A and A k playing the role of B to conclude that M k +N r = R. 11. (a) Let F be a field and f(x) F[x] be a polynomial of degree n. Say that a F is a root of f with multiplicity m if (x a) m is a factor of f(x) and (x a) m+1 is not a factor of f(x), for some m > 0. Show that f(x) has atmost n roots in F counted with multiplicity. (b) Give example of a commutative ring R and a polynomial f(x) R[x] of degree n for some n 1 that has more than n roots in R. solution. (a) Suppose a 1,,a k be the distinct roots of f with multiplicity m 1,,m k. Let A j be the ideal generated by (x a j ). Note that M j is the kernel of the homomorphism ev aj : F[x] F given by ev aj (f) = f(a j ), so each A j is a maximal ideal. Further, unique factorizationimpliesthatifi j,thena i anda j aredistinct, soa i +A j = 1. Bytheprevious exercise, we find that A m i i +A m j j = 1 for each i j. It follows that k j=1a m j j = k j=1 Am j j. By definition of root multiplicity, we have that f belongs to k j=1a m j j. So f belongs to k j=1 Am j j which is the ideal generated by k j=1 (x a j) m j. So f(x) = (x a 1 ) m1 (x a k ) m k g(x). Looking at degrees of both sides we obtain m 1 + +m k n. (b) Let R = Z/4Z. Let f(x) = 2x 2 2x. Verify that f(a) = 0 for each a R. 12. Find all integer x such that x 3 mod 7, x 0 mod 8, x 5 mod 15. 3

4 solution. Let p 1 = 7, p 2 = 8, p 3 = 15. Note that p 1,p 2,p 3 are relatively prime. As in proof of the Chinese remainder theorem, we find e 1,e 2,e 3 such that e j 1 mod p j and e j 0 mod p k for k j. Then a solution of the given congruences is given by c = 3e e 2 +5e 3. And the general solution is given by c+p 1 p 2 p 3 k = c+840k where k is any integer. For example: we want e 1 1 mod 7, e 1 0 mod 8 and e 1 0 mod 15. The second and the third congruence that e 1 needs to satisfy give us e 1 = 120k for some k, so we want 120k 1 mod 7, or in other words, k should be chosen to be an inverse of 120 modulo 7. We denote by (120 1 mod 7) any such integer k. With similar notation, we have (e 1,e 2,e 3 ) = ((120 1 mod 7) 120,(105 1 mod 8) 105,(56 1 mod 15) 56) = ((1 1 mod 7) 120,((1) 1 mod 8) 105,(11 1 mod 15) 56) = (1 120,1 105,( 4) 56) = (120, 105, 224). Now we compute (3 e e e 3 ) = = mod 840. So x = 80 is a solution to our given congruences (easy to verify directly once we have found it). The general solution is given by x = k where k is any integer. 13. Let R be the set of all rational numbers of the form a/b where a,b are integers with b 0 b relatively prime to 3. Show that R has a unique maximal ideal which is generated by 3. solution. Let M be the ideal of R generated by 3. Elements of M have the form 3q where q is any element of R. If 1 M, then 1 = 3q for some q R, which would mean that q = 1/3 R, which we know is not true. So 1 / M and thus M is a proper ideal. Let a R\M. Then a = m/n with m,n Z and n relatively prime to 3. If m is a multiple of 3, then we would have a = 3(m/3)/n which would mean that a 3R, so m must be relatively prime to 3. So a 1 = n/m belongs to R, hence a is a unit. Thus we have argued that every element of R\M is a unit. So M is the unique maximal ideal of R (by problem 8 (a)) 14. An element x of a commutative ring R is called nilpotent if x n = 0 for some n 0. Let R be a ring and let N be the set of all nilpotent elements of R. (a) show that N is an ideal of R. (b) Show that N is contained in the intersection of all the prime ideals in R. solution. (0) Since 0 1 = 0, we have 0 N. If x N, then x n = 0 for some n 0, so ( x) 2n = (x 2 ) n = x 2 n = 0, so ( x) N. Let x,y N. Then there exists m,n 0 such that x m = 0 and y n = 0. Then (x+y) m+n = m+n ) i=0 x i y m+n i. In each term, note that ( m+n i either i m of m+n i n, so either x i = 0 or y m+n i = 0, which forces (x+y) m+n = 0, so (x+y) N. So N is an additive subgroup of R. Finally if r R and x N, then x n = 0 for some n 0. So (rx) n = r n x n = 0 as well. So N is an ideal in R. (b) Let P be a prime ideal of R. If a,b R such that ab P, then a P or b P. Verify by induction on n that if a 1,,a n R such that a 1 a 2 a n P, then a j P for some j. In words our observation says that, if P contains a product of n elements, then P contains one of them. Now let x N \{0}. then x n = 0 for some n > 0. So x n P. By our observation in the previous paragraph we get that x P. So P contains all the nonzero elements of N. Also 4

5 P contains 0. So N P. Thus we have aruged that N is contained in any prime ideal of R. So N is contained in the intersection of all the prime ideals of R. 15. Let Q be the multiplicative group of positive rational numbers. Show that (Q, ) (Z[x], +) as abelian groups. solution. Let p 0 = 2, p 1 = 3, p 2 = 5, be the prime numbers. If m is a positive integer, by unique factorization, one can write m = i=0 pm i i where m i s are non-negative integers that are uniquely determined by m. Also we should remark that only finitely many m i are nonzero, so the product is actually a finite product. For example 63 = If q is a positive rational number then one can write q = m/n where m,n are relatively prime positive integers and then m, n are uniquely determined by q. By unique factorization, we write m = i=0 pm i i and n = i=0 pn i i and so q = i=0 pk i i where k i = m i n i Z. Note that q uniquely determines m and n, and these determine all the m i s and n i s and hence all the k i s. Also note that only finitely many k i is non-zero since k i is zero whenever m i and n i are both zero. Define : Q Z[x] by f(q) = i=0 k ix i where the k i Z are uniquely determined by q as above. Since only finitely many k i is nonzerof(q) is actually a polynomial. Now if q,q are positive rationals and we write q = i=0 pk i i and q = i=0 pk i i, then qq = i=0 pk i+k i i. So f(qq ) = i=0 (k i +k i)x i = i=0 k ix i + i=0 k ix i = f(q)+f(q ). So f is a homomorphism of abelian groups. 16. If F is a field show that there are an infinite number of irreducible polynomials in F[x]. solution. Note that F[x] always has some irreducible polynomial, for example x and (x 1). Suppose there were only finitely many irreducible polynomials, call these p 1,,p k. Each p j is irreducible, so in particular, they are non-units and have degree at least 1. Let us define f = p 1 p k +1. For each j, the polynomial p j divides the product p 1 p k. So if p j divides f, then p j would divide 1 which would mean that p j is a unit, a contradiction. So f is not divisible by p j for j = 1,,k. Since F[x] is an UFD, the polynomial f be written as a finite product of irreducibles, and in particular it has an irreducible factor since each deg(f) deg(p 1 ) 1. Let p be an irreducible factor of f. Then p divides f, so p must be different from p 1,,p k which contradicts that p 1,,p k were the only irreducibles. 17. Write (1+3i) Z[i] as a product of irreducibles. solution. In the first paragraph we do some preparation before doing the computation. If z Z[i], we write N(z) = z 2 = z z and call it the norm of z. Observe that the norm of every element of Z[i] is an integer since N(a+bi) = a 2 +b 2. If N(z) = ±1 for some z Z[i], then writing z = a+bi we find that a 2 +b 2 = 1, so z = ±1 or z = ±i, which implies that z is a unit in Z[i]. So if an element of Z[i] has norm ±1, then it is an unit. Now suppose z Z[i] has a factorization z = wy for some non-units w and y. Then N(z) = N(w)N(y) and both N(w) and N(y) are integers strictly greater than one since w,y are non-units. So N(z) is not a prime number. It follows that if z Z[i] is such that N(z) is a prime number, then z must an irreducible element of Z[i]. Let z = (1+3i). We write N(z) = z 2 = z z and call it the norm of z. So N(z) = 10 = 2.5. This suggests that z might have a factor of norm 2. If (a + bi) Z[i] has norm 2, then a 2 + b 2 = 2, so a and b are ± = 1. So the elements of norm 2 in Z[i] are just the unit multiples of (1 + i). We calculate 1+3i = (1+3i)(1 i) = (1+3i)(1 i) = (2 + i) Z[i]. So 1+i (1+i)(1 i) 2 5

6 (1 + 3i) = (1 + i)(2 + i). Note that N(1 + i) = 2 and N(2 + i) = 5 are both prime, so (1 + i) and (2 + i) are irreducibles in Z[i]. So (1 + 3i) = (1 + i)(2 + i) is the irreudcible factorization. 18. Show that Q[i] = {a+bi: a,b Q} is the fraction field of Z[i]. solution. Note that Z[i] Q[i]. One easily verifies that Q[i] is a ring. If z = (a + bi) is a a nonzero element of Q[i], then one verifies that + ( b) i Q[i] is the inverse of z, so Q[i] a 2 +b 2 a 2 +b 2 is a field. Each element of Q[i] can be written in the form (m+in)/k where m,n,k Z, so (m+in) Z[i] and k Z[i]. So each element of Q[i] is a quotient of two elements of Z[i]. This already shows that Q[i] is the field of fractions of Z[i]. But we can also verify that the quotient of two elements of Z[i] is an element in Q[i]: If m+ni and p+qi are elements of Z[i] with m,n,p,q Z and (p+qi) 0, then m+ni = (m+ni)(p qi) = mp+nq + np mq i Q[i]. p+qi p 2 +q 2 p 2 +q 2 p 2 +q Let R = Z[ 2] = {a+b 2: a,b Z}. (a) Show that R is a domain. (b) Find all the units of R. (c) Determine the field of fraction of R. (d) Show that Z[isqrt2] is an Euclidean domain under the Euclidean valuation ν(a+bi) = a 2 +2b 2. solution. (a) One can see R is a domain simply by noticing that R is a subring of R and R is a field, hence certainly a domain. One can also check R is a domain directly: Suppose α = (a + b 2) and γ = (c + d 2) be two nonzero elements of R. Suppose 0 = αγ = (ca+2bd)+(ad+bc) 2. Since 2 is not a rational number, we must have (ca+2bd) = 0 and ad+bc = 0. It follows that bc 2 = c(bc) = cad = 2bd 2. If b 0, then we get 2c 2 = d 2, which would imply c = d = 0, contradicting (c +d 2) 0, so we get b = 0. But then we get ca = 0 and ad = 0, and since either c or d is nonzero it follows that (b) You can omit this one for now. We shall talk about it later. If you are curious the answer is that the units are ±( 2 1) n where n is any integer. (c) The field of fractions of Z[ 2] is Q[ 2] and this can be verified just like the previous problem. (d)you can omit this one too for now. But the argument is just like the argument we did for Z[i]. 20. Anideal I of a commutative ring R is said to befinitely generated if there exists elements a 1,,a n I such that every element of I can be written in the form r 1 a 1 + +r n a n for some r 1,,r n R. Prove that R satisfies that ACC if and only if evey ideal of R is finitely generated. (Caution: The statement of the problem in the book has rather obvious typos). solution. Assume R satisfies ACC. Suppose I is an ideal in R that is not finitely generated. if b 1,,b m R, the ideal that they generate will be denoted by (b 1,,b m ). Choose a 1 I. Then I (a 1 ) since I is not finitely generated. So I properly contains (a 1 ). Choose a 2 I \ (a 1 ). Now (a 1,a 2 ) I and again these cannot be equal since I is not finitely generated. So we can choose a 3 I \(a 1,a 2 ). Suppose we have inductively chosen a 1,,a m in I such that I (a 1,,a n ) and each a j does not belong to (a 1,,a j 1 ). Since I is not finitely generated, I must properly contain (a 1,,a n ), so we can choose a n+1 I\(a 1,,a n ). Byinduction, verifythatwegetastrictlyincreasinginfiniteascending chain of ideals (a 1 ) (a 1,a 2 ) (a 1,a 2,a 3 ) which violates ACC. So I must be finitely generated. 6

7 Conversely, suppose every ideal in R is finitely generated. Suppose I 1 I 2 I 3 be an increasing sequence of ideals in R. Then one verifies that I = j=1 I j is also an ideal, so I must be finitely generated and we can write I = (a 1,,a n ) for some a 1,,a n R. Now for each r, we have a r j=1 I j, so a r I jr for some j r. Take k = max{j 1,,j n }. Then a r I jr I k for each r, so I = (a 1,,a n ) I k. Since I k I k+1 I k+2 and each of these ideals are contained in I, it now follows that I = I k = I k+1 = I k+2 =. So R satisfies ACC. 7