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

Size: px
Start display at page:

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

Transcription

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

### CHAPTER I. Rings. Definition A ring R is a set with two binary operations, addition + and

CHAPTER I Rings 1.1 Definitions and Examples Definition 1.1.1. A ring R is a set with two binary operations, addition + and multiplication satisfying the following conditions for all a, b, c in R : (i)

### RINGS: SUMMARY OF MATERIAL

RINGS: SUMMARY OF MATERIAL BRIAN OSSERMAN This is a summary of terms used and main results proved in the subject of rings, from Chapters 11-13 of Artin. Definitions not included here may be considered

### Math 547, Exam 2 Information.

Math 547, Exam 2 Information. 3/19/10, LC 303B, 10:10-11:00. Exam 2 will be based on: Homework and textbook sections covered by lectures 2/3-3/5. (see http://www.math.sc.edu/ boylan/sccourses/547sp10/547.html)

### Part IX. Factorization

IX.45. Unique Factorization Domains 1 Part IX. Factorization Section IX.45. Unique Factorization Domains Note. In this section we return to integral domains and concern ourselves with factoring (with respect

### Computations/Applications

Computations/Applications 1. Find the inverse of x + 1 in the ring F 5 [x]/(x 3 1). Solution: We use the Euclidean Algorithm: x 3 1 (x + 1)(x + 4x + 1) + 3 (x + 1) 3(x + ) + 0. Thus 3 (x 3 1) + (x + 1)(4x

### Honors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35

Honors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35 1. Let R 0 be a commutative ring with 1 and let S R be the subset of nonzero elements which are not zero divisors. (a)

### 1. Group Theory Permutations.

1.1. Permutations. 1. Group Theory Problem 1.1. Let G be a subgroup of S n of index 2. Show that G = A n. Problem 1.2. Find two elements of S 7 that have the same order but are not conjugate. Let π S 7

### Math 120 HW 9 Solutions

Math 120 HW 9 Solutions June 8, 2018 Question 1 Write down a ring homomorphism (no proof required) f from R = Z[ 11] = {a + b 11 a, b Z} to S = Z/35Z. The main difficulty is to find an element x Z/35Z

### 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

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

### ALGEBRA AND NUMBER THEORY II: Solutions 3 (Michaelmas term 2008)

ALGEBRA AND NUMBER THEORY II: Solutions 3 Michaelmas term 28 A A C B B D 61 i If ϕ : R R is the indicated map, then ϕf + g = f + ga = fa + ga = ϕf + ϕg, and ϕfg = f ga = faga = ϕfϕg. ii Clearly g lies

### Algebraic structures I

MTH5100 Assignment 1-10 Algebraic structures I For handing in on various dates January March 2011 1 FUNCTIONS. Say which of the following rules successfully define functions, giving reasons. For each one

### Polynomial Rings. i=0. i=0. n+m. i=0. k=0

Polynomial Rings 1. Definitions and Basic Properties For convenience, the ring will always be a commutative ring with identity. Basic Properties The polynomial ring R[x] in the indeterminate x with coefficients

### Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35

Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35 1. Let R be a commutative ring with 1 0. (a) Prove that the nilradical of R is equal to the intersection of the prime

### Chapter 3. Rings. The basic commutative rings in mathematics are the integers Z, the. Examples

Chapter 3 Rings Rings are additive abelian groups with a second operation called multiplication. The connection between the two operations is provided by the distributive law. Assuming the results of Chapter

### Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman

Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 17, 2006 TALK SLOWLY AND WRITE NEATLY!! 1 0.1 Factorization 0.1.1 Factorization of Integers and Polynomials Now we are going

### Contents. 4 Arithmetic and Unique Factorization in Integral Domains. 4.1 Euclidean Domains and Principal Ideal Domains

Ring Theory (part 4): Arithmetic and Unique Factorization in Integral Domains (by Evan Dummit, 018, v. 1.00) Contents 4 Arithmetic and Unique Factorization in Integral Domains 1 4.1 Euclidean Domains and

### Homework 6 Solution. Math 113 Summer 2016.

Homework 6 Solution. Math 113 Summer 2016. 1. For each of the following ideals, say whether they are prime, maximal (hence also prime), or neither (a) (x 4 + 2x 2 + 1) C[x] (b) (x 5 + 24x 3 54x 2 + 6x

### Factorization in Polynomial Rings

Factorization in Polynomial Rings Throughout these notes, F denotes a field. 1 Long division with remainder We begin with some basic definitions. Definition 1.1. Let f, g F [x]. We say that f divides g,

### CHAPTER 14. Ideals and Factor Rings

CHAPTER 14 Ideals and Factor Rings Ideals Definition (Ideal). A subring A of a ring R is called a (two-sided) ideal of R if for every r 2 R and every a 2 A, ra 2 A and ar 2 A. Note. (1) A absorbs elements

### Quizzes for Math 401

Quizzes for Math 401 QUIZ 1. a) Let a,b be integers such that λa+µb = 1 for some inetegrs λ,µ. Prove that gcd(a,b) = 1. b) Use Euclid s algorithm to compute gcd(803, 154) and find integers λ,µ such that

### g(x) = 1 1 x = 1 + x + x2 + x 3 + is not a polynomial, since it doesn t have finite degree. g(x) is an example of a power series.

6 Polynomial Rings We introduce a class of rings called the polynomial rings, describing computation, factorization and divisibility in such rings For the case where the coefficients come from an integral

### 1. Factorization Divisibility in Z.

8 J. E. CREMONA 1.1. Divisibility in Z. 1. Factorization Definition 1.1.1. Let a, b Z. Then we say that a divides b and write a b if b = ac for some c Z: a b c Z : b = ac. Alternatively, we may say that

### Rings. Chapter 1. Definition 1.2. A commutative ring R is a ring in which multiplication is commutative. That is, ab = ba for all a, b R.

Chapter 1 Rings We have spent the term studying groups. A group is a set with a binary operation that satisfies certain properties. But many algebraic structures such as R, Z, and Z n come with two binary

### be any ring homomorphism and let s S be any element of S. Then there is a unique ring homomorphism

21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UFD. Therefore

### Discrete valuation rings. Suppose F is a field. A discrete valuation on F is a function v : F {0} Z such that:

Discrete valuation rings Suppose F is a field. A discrete valuation on F is a function v : F {0} Z such that: 1. v is surjective. 2. v(ab) = v(a) + v(b). 3. v(a + b) min(v(a), v(b)) if a + b 0. Proposition:

### MATH 326: RINGS AND MODULES STEFAN GILLE

MATH 326: RINGS AND MODULES STEFAN GILLE 1 2 STEFAN GILLE 1. Rings We recall first the definition of a group. 1.1. Definition. Let G be a non empty set. The set G is called a group if there is a map called

### 4.4 Noetherian Rings

4.4 Noetherian Rings Recall that a ring A is Noetherian if it satisfies the following three equivalent conditions: (1) Every nonempty set of ideals of A has a maximal element (the maximal condition); (2)

### 6]. (10) (i) Determine the units in the rings Z[i] and Z[ 10]. If n is a squarefree

Quadratic extensions Definition: Let R, S be commutative rings, R S. An extension of rings R S is said to be quadratic there is α S \R and monic polynomial f(x) R[x] of degree such that f(α) = 0 and S

### SUMMARY ALGEBRA I LOUIS-PHILIPPE THIBAULT

SUMMARY ALGEBRA I LOUIS-PHILIPPE THIBAULT Contents 1. Group Theory 1 1.1. Basic Notions 1 1.2. Isomorphism Theorems 2 1.3. Jordan- Holder Theorem 2 1.4. Symmetric Group 3 1.5. Group action on Sets 3 1.6.

### Math Introduction to Modern Algebra

Math 343 - Introduction to Modern Algebra Notes Rings and Special Kinds of Rings Let R be a (nonempty) set. R is a ring if there are two binary operations + and such that (A) (R, +) is an abelian group.

### 32 Divisibility Theory in Integral Domains

3 Divisibility Theory in Integral Domains As we have already mentioned, the ring of integers is the prototype of integral domains. There is a divisibility relation on * : an integer b is said to be divisible

### Eighth Homework Solutions

Math 4124 Wednesday, April 20 Eighth Homework Solutions 1. Exercise 5.2.1(e). Determine the number of nonisomorphic abelian groups of order 2704. First we write 2704 as a product of prime powers, namely

### Algebra Homework, Edition 2 9 September 2010

Algebra Homework, Edition 2 9 September 2010 Problem 6. (1) Let I and J be ideals of a commutative ring R with I + J = R. Prove that IJ = I J. (2) Let I, J, and K be ideals of a principal ideal domain.

### Mathematical Olympiad Training Polynomials

Mathematical Olympiad Training Polynomials Definition A polynomial over a ring R(Z, Q, R, C) in x is an expression of the form p(x) = a n x n + a n 1 x n 1 + + a 1 x + a 0, a i R, for 0 i n. If a n 0,

### Rings. Chapter Homomorphisms and ideals

Chapter 2 Rings This chapter should be at least in part a review of stuff you ve seen before. Roughly it is covered in Rotman chapter 3 and sections 6.1 and 6.2. You should *know* well all the material

### Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra

Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra D. R. Wilkins Contents 3 Topics in Commutative Algebra 2 3.1 Rings and Fields......................... 2 3.2 Ideals...............................

### 36 Rings of fractions

36 Rings of fractions Recall. If R is a PID then R is a UFD. In particular Z is a UFD if F is a field then F[x] is a UFD. Goal. If R is a UFD then so is R[x]. Idea of proof. 1) Find an embedding R F where

### 12 16 = (12)(16) = 0.

Homework Assignment 5 Homework 5. Due day: 11/6/06 (5A) Do each of the following. (i) Compute the multiplication: (12)(16) in Z 24. (ii) Determine the set of units in Z 5. Can we extend our conclusion

### MATH 3030, Abstract Algebra FALL 2012 Toby Kenney Midyear Examination Friday 7th December: 7:00-10:00 PM

MATH 3030, Abstract Algebra FALL 2012 Toby Kenney Midyear Examination Friday 7th December: 7:00-10:00 PM Basic Questions 1. Compute the factor group Z 3 Z 9 / (1, 6). The subgroup generated by (1, 6) is

### Factorization in Integral Domains II

Factorization in Integral Domains II 1 Statement of the main theorem Throughout these notes, unless otherwise specified, R is a UFD with field of quotients F. The main examples will be R = Z, F = Q, and

### MTH310 EXAM 2 REVIEW

MTH310 EXAM 2 REVIEW SA LI 4.1 Polynomial Arithmetic and the Division Algorithm A. Polynomial Arithmetic *Polynomial Rings If R is a ring, then there exists a ring T containing an element x that is not

### (Rgs) Rings Math 683L (Summer 2003)

(Rgs) Rings Math 683L (Summer 2003) We will first summarise the general results that we will need from the theory of rings. A unital ring, R, is a set equipped with two binary operations + and such that

### Abstract Algebra: Chapters 16 and 17

Study polynomials, their factorization, and the construction of fields. Chapter 16 Polynomial Rings Notation Let R be a commutative ring. The ring of polynomials over R in the indeterminate x is the set

### Section III.6. Factorization in Polynomial Rings

III.6. Factorization in Polynomial Rings 1 Section III.6. Factorization in Polynomial Rings Note. We push several of the results in Section III.3 (such as divisibility, irreducibility, and unique factorization)

### NOTES ON FINITE FIELDS

NOTES ON FINITE FIELDS AARON LANDESMAN CONTENTS 1. Introduction to finite fields 2 2. Definition and constructions of fields 3 2.1. The definition of a field 3 2.2. Constructing field extensions by adjoining

### Finite Fields. Sophie Huczynska. Semester 2, Academic Year

Finite Fields Sophie Huczynska Semester 2, Academic Year 2005-06 2 Chapter 1. Introduction Finite fields is a branch of mathematics which has come to the fore in the last 50 years due to its numerous applications,

### where c R and the content of f is one. 1

9. Gauss Lemma Obviously it would be nice to have some more general methods of proving that a given polynomial is irreducible. The first is rather beautiful and due to Gauss. The basic idea is as follows.

### Homework 10 M 373K by Mark Lindberg (mal4549)

Homework 10 M 373K by Mark Lindberg (mal4549) 1. Artin, Chapter 11, Exercise 1.1. Prove that 7 + 3 2 and 3 + 5 are algebraic numbers. To do this, we must provide a polynomial with integer coefficients

### Math 2070BC Term 2 Weeks 1 13 Lecture Notes

Math 2070BC 2017 18 Term 2 Weeks 1 13 Lecture Notes Keywords: group operation multiplication associative identity element inverse commutative abelian group Special Linear Group order infinite order cyclic

### THE JOHNS HOPKINS UNIVERSITY Faculty of Arts and Sciences FINAL EXAM - FALL SESSION ADVANCED ALGEBRA I.

THE JOHNS HOPKINS UNIVERSITY Faculty of Arts and Sciences FINAL EXAM - FALL SESSION 2006 110.401 - ADVANCED ALGEBRA I. Examiner: Professor C. Consani Duration: take home final. No calculators allowed.

### Commutative Algebra and Algebraic Geometry. Robert Friedman

Commutative Algebra and Algebraic Geometry Robert Friedman August 1, 2006 2 Disclaimer: These are rough notes for a course on commutative algebra and algebraic geometry. I would appreciate all suggestions

### 1 2 3 style total. Circle the correct answer; no explanation is required. Each problem in this section counts 5 points.

1 2 3 style total Math 415 Examination 3 Please print your name: Answer Key 1 True/false Circle the correct answer; no explanation is required. Each problem in this section counts 5 points. 1. The rings

### Polynomial Rings. (Last Updated: December 8, 2017)

Polynomial Rings (Last Updated: December 8, 2017) These notes are derived primarily from Abstract Algebra, Theory and Applications by Thomas Judson (16ed). Most of this material is drawn from Chapters

### Example: This theorem is the easiest way to test an ideal (or an element) is prime. Z[x] (x)

Math 4010/5530 Factorization Theory January 2016 Let R be an integral domain. Recall that s, t R are called associates if they differ by a unit (i.e. there is some c R such that s = ct). Let R be a commutative

### Informal Notes on Algebra

Informal Notes on Algebra R. Boyer Contents 1 Rings 2 1.1 Examples and Definitions................................. 2 1.2 Integral Domains...................................... 3 1.3 Fields............................................

### Mathematics for Cryptography

Mathematics for Cryptography Douglas R. Stinson David R. Cheriton School of Computer Science University of Waterloo Waterloo, Ontario, N2L 3G1, Canada March 15, 2016 1 Groups and Modular Arithmetic 1.1

### Factorization in Polynomial Rings

Factorization in Polynomial Rings These notes are a summary of some of the important points on divisibility in polynomial rings from 17 and 18. PIDs Definition 1 A principal ideal domain (PID) is an integral

### Total 100

Math 542 Midterm Exam, Spring 2016 Prof: Paul Terwilliger Your Name (please print) SOLUTIONS NO CALCULATORS/ELECTRONIC DEVICES ALLOWED. MAKE SURE YOUR CELL PHONE IS OFF. Problem Value 1 10 2 10 3 10 4

### 2a 2 4ac), provided there is an element r in our

MTH 310002 Test II Review Spring 2012 Absractions versus examples The purpose of abstraction is to reduce ideas to their essentials, uncluttered by the details of a specific situation Our lectures built

### Homework problems from Chapters IV-VI: answers and solutions

Homework problems from Chapters IV-VI: answers and solutions IV.21.1. In this problem we have to describe the field F of quotients of the domain D. Note that by definition, F is the set of equivalence

### Solutions to odd-numbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 2

Solutions to odd-numbered exercises Peter J Cameron, Introduction to Algebra, Chapter 1 The answers are a No; b No; c Yes; d Yes; e No; f Yes; g Yes; h No; i Yes; j No a No: The inverse law for addition

### Factorization in Domains

Last Time Chain Conditions and s Uniqueness of s The Division Algorithm Revisited in Domains Ryan C. Trinity University Modern Algebra II Last Time Chain Conditions and s Uniqueness of s The Division Algorithm

### COMMUTATIVE RINGS. Definition 3: A domain is a commutative ring R that satisfies the cancellation law for multiplication:

COMMUTATIVE RINGS Definition 1: A commutative ring R is a set with two operations, addition and multiplication, such that: (i) R is an abelian group under addition; (ii) ab = ba for all a, b R (commutative

### ALGEBRA HANDOUT 2.3: FACTORIZATION IN INTEGRAL DOMAINS. In this handout we wish to describe some aspects of the theory of factorization.

ALGEBRA HANDOUT 2.3: FACTORIZATION IN INTEGRAL DOMAINS PETE L. CLARK In this handout we wish to describe some aspects of the theory of factorization. The first goal is to state what it means for an arbitrary

### Ph.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018

Ph.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018 Do 6 problems with at least 2 in each section. Group theory problems: (1) Suppose G is a group. The

### Finite Fields. Sophie Huczynska (with changes by Max Neunhöffer) Semester 2, Academic Year 2012/13

Finite Fields Sophie Huczynska (with changes by Max Neunhöffer) Semester 2, Academic Year 2012/13 Contents 1 Introduction 3 1 Group theory: a brief summary............................ 3 2 Rings and fields....................................

### School of Mathematics and Statistics. MT5836 Galois Theory. Handout 0: Course Information

MRQ 2017 School of Mathematics and Statistics MT5836 Galois Theory Handout 0: Course Information Lecturer: Martyn Quick, Room 326. Prerequisite: MT3505 (or MT4517) Rings & Fields Lectures: Tutorials: Mon

### Algebra Review. Instructor: Laszlo Babai Notes by Vincent Lucarelli and the instructor. June 15, 2001

Algebra Review Instructor: Laszlo Babai Notes by Vincent Lucarelli and the instructor June 15, 2001 1 Groups Definition 1.1 A semigroup (G, ) is a set G with a binary operation such that: Axiom 1 ( a,

### Name: Solutions Final Exam

Instructions. Answer each of the questions on your own paper. Be sure to show your work so that partial credit can be adequately assessed. Put your name on each page of your paper. 1. [10 Points] All of

### Part IX ( 45-47) Factorization

Part IX ( 45-47) Factorization Satya Mandal University of Kansas, Lawrence KS 66045 USA January 22 45 Unique Factorization Domain (UFD) Abstract We prove evey PID is an UFD. We also prove if D is a UFD,

### Chapter 9, Additional topics for integral domains

Chapter 9, Additional topics for integral domains Many times we have mentioned that theorems we proved could be done much more generally they only required some special property like unique factorization,

### φ(xy) = (xy) n = x n y n = φ(x)φ(y)

Groups 1. (Algebra Comp S03) Let A, B and C be normal subgroups of a group G with A B. If A C = B C and AC = BC then prove that A = B. Let b B. Since b = b1 BC = AC, there are a A and c C such that b =

### M381 Number Theory 2004 Page 1

M81 Number Theory 2004 Page 1 [[ Comments are written like this. Please send me (dave@wildd.freeserve.co.uk) details of any errors you find or suggestions for improvements. ]] Question 1 20 = 2 * 10 +

### (1) A frac = b : a, b A, b 0. We can define addition and multiplication of fractions as we normally would. a b + c d

The Algebraic Method 0.1. Integral Domains. Emmy Noether and others quickly realized that the classical algebraic number theory of Dedekind could be abstracted completely. In particular, rings of integers

### Prime Rational Functions and Integral Polynomials. Jesse Larone, Bachelor of Science. Mathematics and Statistics

Prime Rational Functions and Integral Polynomials Jesse Larone, Bachelor of Science Mathematics and Statistics Submitted in partial fulfillment of the requirements for the degree of Master of Science Faculty

### THE GROUP OF UNITS OF SOME FINITE LOCAL RINGS I

J Korean Math Soc 46 (009), No, pp 95 311 THE GROUP OF UNITS OF SOME FINITE LOCAL RINGS I Sung Sik Woo Abstract The purpose of this paper is to identify the group of units of finite local rings of the

### Algebra Exam Fall Alexander J. Wertheim Last Updated: October 26, Groups Problem Problem Problem 3...

Algebra Exam Fall 2006 Alexander J. Wertheim Last Updated: October 26, 2017 Contents 1 Groups 2 1.1 Problem 1..................................... 2 1.2 Problem 2..................................... 2

### Polynomials, Ideals, and Gröbner Bases

Polynomials, Ideals, and Gröbner Bases Notes by Bernd Sturmfels for the lecture on April 10, 2018, in the IMPRS Ringvorlesung Introduction to Nonlinear Algebra We fix a field K. Some examples of fields

### ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ.

ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ. ANDREW SALCH 1. Hilbert s Nullstellensatz. The last lecture left off with the claim that, if J k[x 1,..., x n ] is an ideal, then

### IUPUI Qualifying Exam Abstract Algebra

IUPUI Qualifying Exam Abstract Algebra January 2017 Daniel Ramras (1) a) Prove that if G is a group of order 2 2 5 2 11, then G contains either a normal subgroup of order 11, or a normal subgroup of order

### Math 547, Exam 1 Information.

Math 547, Exam 1 Information. 2/10/10, LC 303B, 10:10-11:00. Exam 1 will be based on: Sections 5.1, 5.2, 5.3, 9.1; The corresponding assigned homework problems (see http://www.math.sc.edu/ boylan/sccourses/547sp10/547.html)

### Practice problems for first midterm, Spring 98

Practice problems for first midterm, Spring 98 midterm to be held Wednesday, February 25, 1998, in class Dave Bayer, Modern Algebra All rings are assumed to be commutative with identity, as in our text.

### Polynomial Rings : Linear Algebra Notes

Polynomial Rings : Linear Algebra Notes Satya Mandal September 27, 2005 1 Section 1: Basics Definition 1.1 A nonempty set R is said to be a ring if the following are satisfied: 1. R has two binary operations,

### ADVANCED COMMUTATIVE ALGEBRA: PROBLEM SETS

ADVANCED COMMUTATIVE ALGEBRA: PROBLEM SETS UZI VISHNE The 11 problem sets below were composed by Michael Schein, according to his course. Take into account that we are covering slightly different material.

### 1.5 The Nil and Jacobson Radicals

1.5 The Nil and Jacobson Radicals The idea of a radical of a ring A is an ideal I comprising some nasty piece of A such that A/I is well-behaved or tractable. Two types considered here are the nil and

### 1 First Theme: Sums of Squares

I will try to organize the work of this semester around several classical questions. The first is, When is a prime p the sum of two squares? The question was raised by Fermat who gave the correct answer

### Math Introduction to Modern Algebra

Math 343 - Introduction to Modern Algebra Notes Field Theory Basics Let R be a ring. M is called a maximal ideal of R if M is a proper ideal of R and there is no proper ideal of R that properly contains

### PRACTICE FINAL MATH , MIT, SPRING 13. You have three hours. This test is closed book, closed notes, no calculators.

PRACTICE FINAL MATH 18.703, MIT, SPRING 13 You have three hours. This test is closed book, closed notes, no calculators. There are 11 problems, and the total number of points is 180. Show all your work.

### 2. THE EUCLIDEAN ALGORITHM More ring essentials

2. THE EUCLIDEAN ALGORITHM More ring essentials In this chapter: rings R commutative with 1. An element b R divides a R, or b is a divisor of a, or a is divisible by b, or a is a multiple of b, if there

### MT5836 Galois Theory MRQ

MT5836 Galois Theory MRQ May 3, 2017 Contents Introduction 3 Structure of the lecture course............................... 4 Recommended texts..................................... 4 1 Rings, Fields and

### Ideals: Definitions & Examples

Ideals: Definitions & Examples Defn: An ideal I of a commutative ring R is a subset of R such that for a, b I and r R we have a + b, a b, ra I Examples: All ideals of Z have form nz = (n) = {..., n, 0,

### Groups, Rings, and Finite Fields. Andreas Klappenecker. September 12, 2002

Background on Groups, Rings, and Finite Fields Andreas Klappenecker September 12, 2002 A thorough understanding of the Agrawal, Kayal, and Saxena primality test requires some tools from algebra and elementary

### Lecture 7.5: Euclidean domains and algebraic integers

Lecture 7.5: Euclidean domains and algebraic integers Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Modern Algebra M. Macauley

### 5.1 Commutative rings; Integral Domains

5.1 J.A.Beachy 1 5.1 Commutative rings; Integral Domains from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 23. Let R be a commutative ring. Prove the following

### CSIR - Algebra Problems

CSIR - Algebra Problems N. Annamalai DST - INSPIRE Fellow (SRF) Department of Mathematics Bharathidasan University Tiruchirappalli -620024 E-mail: algebra.annamalai@gmail.com Website: https://annamalaimaths.wordpress.com

### 4.5 Hilbert s Nullstellensatz (Zeros Theorem)

4.5 Hilbert s Nullstellensatz (Zeros Theorem) We develop a deep result of Hilbert s, relating solutions of polynomial equations to ideals of polynomial rings in many variables. Notation: Put A = F[x 1,...,x

### EXERCISES. = {1, 4}, and. The zero coset is J. Thus, by (***), to say that J 4- a iu not zero, is to

19 CHAPTER NINETEEN Whenever J is a prime ideal of a commutative ring with unity A, the quotient ring A/J is an integral domain. (The details are left as an exercise.) An ideal of a ring is called proper

### 38 Irreducibility criteria in rings of polynomials

38 Irreducibility criteria in rings of polynomials 38.1 Theorem. Let p(x), q(x) R[x] be polynomials such that p(x) = a 0 + a 1 x +... + a n x n, q(x) = b 0 + b 1 x +... + b m x m and a n, b m 0. If b m