# calculating the result modulo 3, as follows: p(0) = = 1 0,

Save this PDF as:

Size: px
Start display at page:

Download "calculating the result modulo 3, as follows: p(0) = 0 3 + 0 + 1 = 1 0,"

## Transcription

1 Homework #02, due 1/27/10 = 9.4.1, 9.4.2, 9.4.5, 9.4.6, Additional problems recommended for study: (9.4.3), 9.4.4, 9.4.9, , , (9.4.14), 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] By Prop the quadratic polynomial p(x) = x 2 +x+1 is irreducible in F 2 [x] iff p(x) has no root in F 2. This we check by substituting all the elements of the field F 2 (there are only 0 and 1) into p(x) and calculating the result modulo 2, as follows: p(0) = = 1 0 and p(1) = = = 1 0. Since p(x) has no roots in F 2, p(x) is irreducible (b) x 3 + x + 1 in F 3 [x]. By Prop the cubic polynomial q(x) = x 3 + x + 1 is irreducible in F 3 [x] iff p(x) has no root in F 3. This we check by substituting all the elements of the field F 3 (there are only 0, 1, and 2) into p(x) and calculating the result modulo 3, as follows: p(0) = = 1 0, p(1) = = = 0, and p(2) = = = 11 = 2 0. Since p(x) has root 1 in F 2, p(x) has linear factor x 1 and factorization p(x) = (x 1)(x 2 + x 1) = (x + 2)(x 2 + x + 2) (note 1 = 2 mod 3) (c) x in F 5 [x]. Let p(x) = x By Prop we can show that p(x) has no linear factor by substituting the elements 0, 1, 2, 3, 4 of F 5 in p(x) and calculating modulo 5: p(0) = 1 0, p(1) = = 2 0, p(2) = = 17 = 2 0, p(3) = = 82 = 2 0, and p(4) = = = 257 = 2 0. It remains to consider factorizations into monic quadratic polynomials. Suppose p(x) = (x 2 + ax + b)(x 2 + cx + d) where a, b, c, d F 5. Then x = p(x) = x 4 + (a + c)x 3 + (b + ac + d)x 2 + (ad + bc)x + bd

2 2 so 0 = a + c, 0 = b + ac + d, 0 = ad + bc, and 1 = bd. From 0 = a + c we have c = a, so 0 = b + a( a) + d and 0 = ad + b( a), so a 2 = b + d and ad = ab. If a 0 then b = d by cancellation from ad = ab, so a 2 = 2b and b 2 = 1 from a 2 = b + d and 1 = bd. Now 1 2 = 1, 2 2 = 4, 3 2 = 9 = 4, 4 2 = 1, so b is 1 or 4. But then either a 2 = 2 or a 2 = 2(4) = 8 = 3, and neither of these equations has a solution in F 5 (since 1 2 = 1, 2 2 = 4, 3 2 = 4, 4 2 = 1). On the other hand, if a = 0 then 0 2 = 0 = b + d, so b = d, so b 2 = bd = 1 = 4, so b = 2, d = 2, a = 0, c = 0, giving a factorization p(x) = x = (x 2 + 2)(x 2 2) = (x 2 + 2)(x 2 + 3). Thus p(x) is a reducible polynomial in F 5 [x] (d) x x in Z[x]. This polynomial has no real-valued roots, since x 4 0 and x 2 0 for all x R, hence x x for all x R. This shows, by Prop. 9.10, that x 4 +10x 2 +1 has no linear factors in Z[x]. We consider factoring x x into quadratic polynomials. If a, b, c, d Z and x x = (x 2 + ax + b)(x 2 + cx + d) = x 4 + (a + c)x 3 + (b + ac + d)x 2 + (ad + bc)x + bd then 0 = a+c, b+ac+d = 10, ad+bc = 0, and bd = 1. From 0 = a+c we get c = a, so, by ad + bc = 0 we have ad ab = 0, hence a = 0 or d = b. If a = 0 then c = 0 and the factorization becomes x x = (x 2 + b)(x 2 + d) = x 4 + (b + d)x 2 + bd so b + d = 10 and bd = 1. These equations have no solution in Z. Indeed, we have either b = d = 1 or b = d = 1 by bd = 1, and neither choice satisfies the equation b + d = 10. If a 0 then b = d and the factorization becomes x x = (x 2 + ax + b)(x 2 ax + b) = x 4 + (2b a 2 )x 2 + b 2

3 where 2b = a and b 2 = 1. From 2b = a we get b 5, but from b 2 = 1 we get b = ±1, a contradiction. Thus there is no factorization into quadratic factors, and x x is irreducible over Z[x] Prove that the following polynomials are irreducible in Z[x] (a) x 4 4x This polynomial is irreducible by Eisenstein s Criterion, since the prime 2 does not divide the leading coefficient, does divide all the coefficients of the low order terms, namely 4, 0, 0, and 6, but the square of 2 does not divide the constant (b) x x 5 15x 3 + 6x 120 The coefficients of the low order terms are 30, 0, 15, 0, 6, and 120. These numbers are divisible by the prime 3, but 3 2 = 9 does not divide 120 = , so the polynomial is irreducible by Eisenstein s Criterion (c) x 4 + 6x 3 + 4x 2 + 2x + 1 [Substitute x 1 for x.] Let p(x) = x 4 + 6x 3 + 4x 2 + 2x + 1. Then to calculate p(x 1) we first note that (x 1) 4 = x 4 4x 3 + 6x 2 4x + 1 6(x 1) 3 = 6x 3 18x x 6 4(x 1) 2 = 4x 2 8x + 4 2(x 1) = 2x 2 1 = 1 Then we add these up and get p(x 1) = (x 1) 4 + 6(x 1) 3 + 4(x 1) 2 + 2(x 1) + 1 = x 4 + 2x 3 8x 2 + 8x 2 Now x 4 + 2x 3 8x 2 + 8x 2 is irreducible by Eisenstein s Criterion, since the prime 2 divides all the lower order coefficients, but 2 2 = 3

4 4 4 does not divide the constant 2. Any factorization of p(x) would provide a factorization of p(x 1) as well, since if p(x) = a(x)b(x) then x 4 + 2x 3 8x 2 + 8x 2 = p(x 1) = a(x 1)b(x 1), contradicting the irreducibility of p(x 1). Therefore p(x) is itself irreducible in Z[x] (d) (x+2)p 2 p x, where p is an odd prime, in Z[x]. To express (x+2)p 2 p x as a polynomial we expand (x + 2) p according to the binomial theorem. The final constant 2 p cancels with 2 p, so every remaining term has x as a factor, which we cancel, leaving ( ( ) ( ) x p 1 p + 2 x 1) p p x 2 p p 1 p p 1 Every lower order coefficient in this polynomial has the form ( ) 2 k p x k p k 1 = 2 k p (p 1) (p k 1) with 0 < k < p. Each lower order coefficient has p as a factor but does not have p 2 as a factor, so the polynomial is irreducible by Eisenstein s Criterion Find all the monic irreducible polynomials of degree 3 in F 2 [x], and the same in F 3 [x]. All the monic linear (degree 1) polynomials x a are irreducible. For F 2 these irreducible linear polynomials are x and x 1(= x + 1). For F 3 these irreducible linear polynomials are x, x 1(= x + 2), and x 2(= x + 1). In F 2 [x] the only (monic) quadratic polynomials are x 2, x 2 +1, x 2 +x, and x 2 + x + 1. Obviously, x 2 and x 2 + x are reducible since x 2 = xx and x 2 + x = x(x + 1). Less obviously, x has 1 as a root, which leads to the factorization x = (x + 1)(x + 1). Finally, x 2 + x + 1 has no root in F 2 since = 1 0 and = 1 0, so, because it is quadratic, it is irreducible. In F 3 [x] the nine monic quadratic polynomials are x 2, x 2 + 1, x 2 + 2, x 2 + x, x 2 + 2x, x 2 + x + 1, x 2 + 2x + 1, x 2 + x + 2, and x 2 + 2x + 2. Obviously, x 2, x 2 + x, and x 2 + 2x are reducible. Since the remaining

5 polynomials x 2 + 1, x 2 + 2, x 2 + x + 1, x 2 + 2x + 1, x 2 + x + 2, and x 2 + 2x + 2 are quadratic, they are reducible iff they have a root in F 3, so we check each of them for roots: x = = = 2 x = = = 0 x 2 + x = = = 1 x 2 + 2x = = = 0 x 2 + x = = = 2 x 2 + 2x = = = 1 Since x has 1 and 2 as roots, we get the factorization x = (x 1)(x 2) = (x + 2)(x + 1). Since x 2 + x + 1 has 1 as a (multiple) root, we get a factorization x 2 + x + 1 = (x 1)(x 1) = (x + 2)(x + 2). Finally, x + 2x + 1 = (x + 1)(x + 1). The remaining polynomials, which are the only irreducible quadratic polynomials in F 3 [x], are x 2 +1 (used below), x 2 + x + 2, and x 2 + 2x + 2. There are quite a few monic cubic polynomials, so I thought it best to write some code in GAP to do the computations. The following GAP code computes, counts, and prints the irreducible cubic polynomials over the field F p, for p ranging over the first 12 primes. irrcubic:=function(p) local f,c,t,roots,irr; f:=function(c,p) return (c[4]^3+c[1]*c[4]^2+c[2]*c[4]+c[3]) mod p; end; t:=tuples([0..p-1],4); roots:=set(filtered(t,c->f(c,p)=0),c->c{[1..3]}); irr:=difference(tuples([0..p-1],3),roots); Print("The number of irreducible monic cubic "); Print("polynomials over F_",p," is\t",size(irr)); # Print("The irreducible monic cubic "); # Print("polynomials over F_",p," are\n"); # for c in irr do # Print("x^3"); 5

6 6 # if c[1]=1 then Print(" + x^2");fi; # if c[1] in [2..p-1] then Print(" + ",c[1],"x^2");fi; # if c[2]=1 then Print(" + x");fi; # if c[2] in [2..p-1] then Print(" + ",c[2],"x");fi; # if c[3]=1 then Print(" + 1");fi; # if c[3] in [2..p-1] then Print(" + ",c[3]);fi; # Print("\n"); # od; Print("\n"); return irr; end; LogTo("roots.output"); for p in Primes{[1..12]} do irrcubic(p); od; Running the GAP code without printing the polynomials produces the data in the following table, which shows the number n p of irreducible monic cubic polynomials over F p for the first 12 primes. p n p By uncommenting the Print commands we get lists of irreducible monic cubic polynomials. The 2 irreducible monic cubic polynomials in F 2 [x] are x 3 + x + 1 x 3 + x The 8 irreducible monic cubic polynomials over F 3 [x] are x 3 + 2x + 1 x 3 + x 2 + 2x + 1 x 3 + 2x + 2 x 3 + 2x x 3 + x x 3 + 2x 2 + x + 1 x 3 + x 2 + x + 2 x 3 + 2x 2 + 2x + 2

7 7 The 40 irreducible monic cubic polynomials over F 5 are x 3 + x + 1 x 3 + 2x 2 + 2x + 2 x 3 + x + 4 x 3 + 2x 2 + 2x + 3 x 3 + 2x + 1 x 3 + 2x 2 + 4x + 2 x 3 + 2x + 4 x 3 + 2x 2 + 4x + 4 x 3 + 3x + 2 x 3 + 3x x 3 + 3x + 3 x 3 + 3x x 3 + 4x + 2 x 3 + 3x 2 + x + 1 x 3 + 4x + 3 x 3 + 3x 2 + x + 2 x 3 + x x 3 + 3x 2 + 2x + 2 x 3 + x x 3 + 3x 2 + 2x + 3 x 3 + x 2 + x + 3 x 3 + 3x 2 + 4x + 1 x 3 + x 2 + x + 4 x 3 + 3x 2 + 4x + 3 x 3 + x 2 + 3x + 1 x 3 + 4x x 3 + x 2 + 3x + 4 x 3 + 4x x 3 + x 2 + 4x + 1 x 3 + 4x 2 + x + 1 x 3 + x 2 + 4x + 3 x 3 + 4x 2 + x + 2 x 3 + 2x x 3 + 4x 2 + 3x + 1 x 3 + 2x x 3 + 4x 2 + 3x + 4 x 3 + 2x 2 + x + 3 x 3 + 4x 2 + 4x + 2 x 3 + 2x 2 + x + 4 x 3 + 4x 2 + 4x Construct fields of each of the following orders: (a) 9, (b) 49, (c) 8, (d) 81. (you may exhibit these as F [x]/(f(x)) for some F and f). [Use Exercises 2 and 3 in Section 2.]

8 8 By Exercise 9.2.3, F [x]/(f(x)) is a field when f F [x] is irreducible, and by Exercise 9.2.2, F [x]/(f(x)) has q n elements if F = q and deg(f) = n. (a) To get a field of order 9, we choose F = F 3 so that q = F 3 = 3, and we choose an irreducible quadratic, so that n = 2 and the order of the field F [x]/(f(x)) is q n = 3 2 = 9, as desired. For f we may use x 2 + 1, which is an irreducible quadratic polynomial in F 3 [x] by Exercise 9.4.5, solved above. (b) To get a field of order 49, we choose F = F 7 so that q = F 7 = 7, and we let f be any irreducible quadratic in F 7 [x]. Then n = 2 and the order of the field F [x]/(f(x)) is q n = 7 2 = 49. One choice that works is f(x) = x By the way, the quadratic irreducible polynomials in F 7 [x] are x x x x 2 + x + 3 x 2 + x + 4 x 2 + x + 6 x 2 + 2x + 2 x 2 + 2x + 3 x 2 + 2x + 5 x 2 + 3x + 1 x 2 + 3x + 5 x 2 + 3x + 6 x 2 + 4x + 1 x 2 + 4x + 5 x 2 + 4x + 6 x 2 + 5x + 2 x 2 + 5x + 3 x 2 + 5x + 5 x 2 + 6x + 3 x 2 + 6x + 4 x 2 + 6x + 6 (c) To get a field of order 8, we choose F = F 2 so that q = F 2 = 2, and we choose an irreducible cubic for f, so that n = 3 and the order of the field F [x]/(f(x)) is q n = 2 3 = 8. For f we may use x 3 + x + 1 or x 3 + x (the two irreducible monic cubics in F 2 [x], as shown above). (d) To get a field of order 81, we choose F = F 3 so that q = F 3 = 3, and we choose an irreducible quartic polynomial f in F 3 [x], so that n = 4 and the order of the field F [x]/(f(x)) is q n = 3 4 = 81. Let f(x) = x 4 + x + 2. Then f has no linear factors because it has no roots in F 3, since = 2, = 1, and = 2.

9 Suppose f(x) = (x 2 + ax + b)(x 2 + cx + d) where a, b, c, d F 3. Then x 4 + x + 2 = x 4 + (a + c)x 3 + (b + ac + d)x 2 + (ad + bc)x + bd, which implies that 0 = a + c, 0 = b + ac + d, 1 = ad + bc, and 2 = bd. From 0 = a + c we have c = a. Substituting a for c in the equations 0 = b + ac + d and 1 = ad + bc (and simplifying) yields a 2 = b + d and 1 = a(d b). By the latter equation, a = a 2 (d b), so by a 2 = b + d we have a = (d b)(b + d) = d 2 b 2. From bd = 1 we know b 0 and d 0. The only nonzero element that is a square in F 3 is 1, since 1 2 = 1 and 2 2 = 1 in F 3. Therefore a = b 2 d 2 = 1 1 = 0, but then 1 = a(d b) = 0(d b) = 0, a contradiction. This shows that f is actually irreducible, so F 3 [x]/(f) is a field with 81 elements in it Show that R[x]/(x 2 + 1) is isomorphic to the field of complex numbers. Let p(x) = x and let P = (x 2 + 1) be the principal ideal of R[x] generated by p(x). Obviously p(x) has no real roots since, for every r R, r Since p(x) is quadratic and has no real roots, p(x) is irreducible in R[x] by Prop Therefore R[x]/P is a field by Exercise (or the upcoming Prop. 9.15). The element x = x + P R[x]/P is a solution to x 2 = 1 and plays the same role as the imaginary i = 1 in C, because x 2 = (x + P ) 2 = x 2 + P = x 2 (x 2 + 1) + P = 1 + P. The elements of R[x]/P are cosets of the form a + bx + P, a, b, R. Define a map ϕ : R[x]/P C by ϕ(a + bx + P ) = a + b 1. To show ϕ is well-defined, let us assume that a + bx + P = c + dx + P for some a, b, c, d R. Then a c+(b d)x P, hence a c+(b d)x = p(x)q(x) for some q(x) R[x]. If b d 0 then 1 = deg(a c + (b d)x) = deg(p(x)q(x)) = deg(p(x)) + deg(q(x)) = 2 + deg(q(x)) 2, a contradiction, so b = d. But then a c P, hence a c = p(x)q (x) for some q (x) R[x]. If a c 0 then 0 = deg(a c) = deg(p(x)q (x)) = deg(p(x)) + deg(q (x)) = 2 + deg(q (x)) 2, a contradiction, so a = c. 9

10 10 Thus, each element of R[x]/P has the form a + bx + P for uniquely determined reals a, b R. Next are calculations that show ϕ is a ring homomorphism because it preserves differences and products. ϕ((a + bx + P ) (c + dx + P )) = ϕ((a c) + (b d)x + P ) = (a c) + (b d) 1 = (a + b 1) (c + d 1) = ϕ(a + bx + P ) ϕ(c + dx + P ) ϕ((a + bx + P )(c + dx + P )) = ϕ((a + bx)(c + dx) + P ) = ϕ(ac + (ad + bc)x + bdx 2 + P ) = ϕ(ac + (ad + bc)x + bdx 2 + ( bd)(x 2 + 1) + P ) = ϕ(ac + (ad + bc)x + bdx 2 bdx 2 bd + P ) = ϕ(ac bd + (ad + bc)x + P ) = ac bd + (ad + bc) 1 = ac + (ad + bc) 1 + bd( 1) 2 = (a + b 1)(c + d 1) = ϕ(a + bx + P )ϕ(c + dx + P ) Finally, to see that ϕ is injective, we note that kernel of ϕ is trivial since if ϕ(a + bx + P ) = 0 then a + b 1 = 0, hence a = b = 0, so a + bx + P = P = 0 R[x]/P.

### Quotient Rings and Field Extensions

Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.

### it is easy to see that α = a

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 UF. Therefore

### 3 1. Note that all cubes solve it; therefore, there are no more

Math 13 Problem set 5 Artin 11.4.7 Factor the following polynomials into irreducible factors in Q[x]: (a) x 3 3x (b) x 3 3x + (c) x 9 6x 6 + 9x 3 3 Solution: The first two polynomials are cubics, so if

### H/wk 13, Solutions to selected problems

H/wk 13, Solutions to selected problems Ch. 4.1, Problem 5 (a) Find the number of roots of x x in Z 4, Z Z, any integral domain, Z 6. (b) Find a commutative ring in which x x has infinitely many roots.

### Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize

### 7. Some irreducible polynomials

7. Some irreducible polynomials 7.1 Irreducibles over a finite field 7.2 Worked examples Linear factors x α of a polynomial P (x) with coefficients in a field k correspond precisely to roots α k [1] of

### (a) Write each of p and q as a polynomial in x with coefficients in Z[y, z]. deg(p) = 7 deg(q) = 9

Homework #01, due 1/20/10 = 9.1.2, 9.1.4, 9.1.6, 9.1.8, 9.2.3 Additional problems for study: 9.1.1, 9.1.3, 9.1.5, 9.1.13, 9.2.1, 9.2.2, 9.2.4, 9.2.5, 9.2.6, 9.3.2, 9.3.3 9.1.1 (This problem was not assigned

### a 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4)

ROOTS OF POLYNOMIAL EQUATIONS In this unit we discuss polynomial equations. A polynomial in x of degree n, where n 0 is an integer, is an expression of the form P n (x) =a n x n + a n 1 x n 1 + + a 1 x

### Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2)

Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2) Kevin Broughan University of Waikato, Hamilton, New Zealand May 13, 2010 Remainder and Factor Theorem 15 Definition of factor If f (x)

### Math 345-60 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field

Math 345-60 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field 1. Throughout this section, F is a field and F [x] is the ring of polynomials with coefficients in F. We will

### minimal polyonomial Example

Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We

### CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY

January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics.

### Winter Camp 2011 Polynomials Alexander Remorov. Polynomials. Alexander Remorov alexanderrem@gmail.com

Polynomials Alexander Remorov alexanderrem@gmail.com Warm-up Problem 1: Let f(x) be a quadratic polynomial. Prove that there exist quadratic polynomials g(x) and h(x) such that f(x)f(x + 1) = g(h(x)).

### The Division Algorithm for Polynomials Handout Monday March 5, 2012

The Division Algorithm for Polynomials Handout Monday March 5, 0 Let F be a field (such as R, Q, C, or F p for some prime p. This will allow us to divide by any nonzero scalar. (For some of the following,

### 1 = (a 0 + b 0 α) 2 + + (a m 1 + b m 1 α) 2. for certain elements a 0,..., a m 1, b 0,..., b m 1 of F. Multiplying out, we obtain

Notes on real-closed fields These notes develop the algebraic background needed to understand the model theory of real-closed fields. To understand these notes, a standard graduate course in algebra is

### Introduction to Finite Fields (cont.)

Chapter 6 Introduction to Finite Fields (cont.) 6.1 Recall Theorem. Z m is a field m is a prime number. Theorem (Subfield Isomorphic to Z p ). Every finite field has the order of a power of a prime number

### PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.

PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include

### 3 Factorisation into irreducibles

3 Factorisation into irreducibles Consider the factorisation of a non-zero, non-invertible integer n as a product of primes: n = p 1 p t. If you insist that primes should be positive then, since n could

### SOLVING POLYNOMIAL EQUATIONS

C SOLVING POLYNOMIAL EQUATIONS We will assume in this appendix that you know how to divide polynomials using long division and synthetic division. If you need to review those techniques, refer to an algebra

### ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS

ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS DANIEL RABAYEV AND JACK SONN Abstract. Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Q p for

### POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS

POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS RUSS WOODROOFE 1. Unique Factorization Domains Throughout the following, we think of R as sitting inside R[x] as the constant polynomials (of degree 0).

### Unique Factorization

Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon

### 1 Homework 1. [p 0 q i+j +... + p i 1 q j+1 ] + [p i q j ] + [p i+1 q j 1 +... + p i+j q 0 ]

1 Homework 1 (1) Prove the ideal (3,x) is a maximal ideal in Z[x]. SOLUTION: Suppose we expand this ideal by including another generator polynomial, P / (3, x). Write P = n + x Q with n an integer not

### CM2202: Scientific Computing and Multimedia Applications General Maths: 2. Algebra - Factorisation

CM2202: Scientific Computing and Multimedia Applications General Maths: 2. Algebra - Factorisation Prof. David Marshall School of Computer Science & Informatics Factorisation Factorisation is a way of

### MOP 2007 Black Group Integer Polynomials Yufei Zhao. Integer Polynomials. June 29, 2007 Yufei Zhao yufeiz@mit.edu

Integer Polynomials June 9, 007 Yufei Zhao yufeiz@mit.edu We will use Z[x] to denote the ring of polynomials with integer coefficients. We begin by summarizing some of the common approaches used in dealing

### 9. POLYNOMIALS. Example 1: The expression a(x) = x 3 4x 2 + 7x 11 is a polynomial in x. The coefficients of a(x) are the numbers 1, 4, 7, 11.

9. POLYNOMIALS 9.1. Definition of a Polynomial A polynomial is an expression of the form: a(x) = a n x n + a n-1 x n-1 +... + a 1 x + a 0. The symbol x is called an indeterminate and simply plays the role

### JUST THE MATHS UNIT NUMBER 1.8. ALGEBRA 8 (Polynomials) A.J.Hobson

JUST THE MATHS UNIT NUMBER 1.8 ALGEBRA 8 (Polynomials) by A.J.Hobson 1.8.1 The factor theorem 1.8.2 Application to quadratic and cubic expressions 1.8.3 Cubic equations 1.8.4 Long division of polynomials

### 1.7. Partial Fractions. 1.7.1. Rational Functions and Partial Fractions. A rational function is a quotient of two polynomials: R(x) = P (x) Q(x).

.7. PRTIL FRCTIONS 3.7. Partial Fractions.7.. Rational Functions and Partial Fractions. rational function is a quotient of two polynomials: R(x) = P (x) Q(x). Here we discuss how to integrate rational

### RESULTANT AND DISCRIMINANT OF POLYNOMIALS

RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. All results

### x 2 x 2 cos 1 x x2, lim 1. If x > 0, multiply all three parts by x > 0, we get: x x cos 1 x x, lim lim x cos 1 lim = 5 lim sin 5x

Homework 4 3.4,. Show that x x cos x x holds for x 0. Solution: Since cos x, multiply all three parts by x > 0, we get: x x cos x x, and since x 0 x x 0 ( x ) = 0, then by Sandwich theorem, we get: x 0

### 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 of Gallian s Contemporary Abstract Algebra. Most of the important

### Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any.

Algebra 2 - Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers - {1,2,3,4,...}

### Applications of Fermat s Little Theorem and Congruences

Applications of Fermat s Little Theorem and Congruences Definition: Let m be a positive integer. Then integers a and b are congruent modulo m, denoted by a b mod m, if m (a b). Example: 3 1 mod 2, 6 4

### Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given.

Polynomials (Ch.1) Study Guide by BS, JL, AZ, CC, SH, HL Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given. Sasha s method

### THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS

THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS KEITH CONRAD 1. Introduction The Fundamental Theorem of Algebra says every nonconstant polynomial with complex coefficients can be factored into linear

### 0.4 FACTORING POLYNOMIALS

36_.qxd /3/5 :9 AM Page -9 SECTION. Factoring Polynomials -9. FACTORING POLYNOMIALS Use special products and factorization techniques to factor polynomials. Find the domains of radical expressions. Use

### Factoring of Prime Ideals in Extensions

Chapter 4 Factoring of Prime Ideals in Extensions 4. Lifting of Prime Ideals Recall the basic AKLB setup: A is a Dedekind domain with fraction field K, L is a finite, separable extension of K of degree

### FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z

FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization

### Factorization Algorithms for Polynomials over Finite Fields

Degree Project Factorization Algorithms for Polynomials over Finite Fields Sajid Hanif, Muhammad Imran 2011-05-03 Subject: Mathematics Level: Master Course code: 4MA11E Abstract Integer factorization is

### FACTORING IN QUADRATIC FIELDS. 1. Introduction. This is called a quadratic field and it has degree 2 over Q. Similarly, set

FACTORING IN QUADRATIC FIELDS KEITH CONRAD For a squarefree integer d other than 1, let 1. Introduction K = Q[ d] = {x + y d : x, y Q}. This is called a quadratic field and it has degree 2 over Q. Similarly,

### Factoring Polynomials

Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent

### FACTORING AFTER DEDEKIND

FACTORING AFTER DEDEKIND KEITH CONRAD Let K be a number field and p be a prime number. When we factor (p) = po K into prime ideals, say (p) = p e 1 1 peg g, we refer to the data of the e i s, the exponents

### Chapter 13: Basic ring theory

Chapter 3: Basic ring theory Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 3: Basic ring

### Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013

Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents A Cyclotomic Polynomials 79 A.1 Minimum Polynomials of Roots of

### SMT 2014 Algebra Test Solutions February 15, 2014

1. Alice and Bob are painting a house. If Alice and Bob do not take any breaks, they will finish painting the house in 20 hours. If, however, Bob stops painting once the house is half-finished, then the

### Factoring polynomials over finite fields

Factoring polynomials over finite fields Summary and et questions 12 octobre 2011 1 Finite fields Let p an odd prime and let F p = Z/pZ the (unique up to automorphism) field with p-elements. We want to

### Partial Fractions Examples

Partial Fractions Examples Partial fractions is the name given to a technique of integration that may be used to integrate any ratio of polynomials. A ratio of polynomials is called a rational function.

### 3.3. Solving Polynomial Equations. Introduction. Prerequisites. Learning Outcomes

Solving Polynomial Equations 3.3 Introduction Linear and quadratic equations, dealt within Sections 3.1 and 3.2, are members of a class of equations, called polynomial equations. These have the general

### Factoring Polynomials and Solving Quadratic Equations

Factoring Polynomials and Solving Quadratic Equations Math Tutorial Lab Special Topic Factoring Factoring Binomials Remember that a binomial is just a polynomial with two terms. Some examples include 2x+3

### PROBLEM SET 6: POLYNOMIALS

PROBLEM SET 6: POLYNOMIALS 1. introduction In this problem set we will consider polynomials with coefficients in K, where K is the real numbers R, the complex numbers C, the rational numbers Q or any other

### Integer roots of quadratic and cubic polynomials with integer coefficients

Integer roots of quadratic and cubic polynomials with integer coefficients Konstantine Zelator Mathematics, Computer Science and Statistics 212 Ben Franklin Hall Bloomsburg University 400 East Second Street

### A number field is a field of finite degree over Q. By the Primitive Element Theorem, any number

Number Fields Introduction A number field is a field of finite degree over Q. By the Primitive Element Theorem, any number field K = Q(α) for some α K. The minimal polynomial Let K be a number field and

### 15. Symmetric polynomials

15. Symmetric polynomials 15.1 The theorem 15.2 First examples 15.3 A variant: discriminants 1. The theorem Let S n be the group of permutations of {1,, n}, also called the symmetric group on n things.

### Polynomial Factoring. Ramesh Hariharan

Polynomial Factoring Ramesh Hariharan The Problem Factoring Polynomials overs Integers Factorization is unique (why?) (x^2 + 5x +6) (x+2)(x+3) Time: Polynomial in degree A Related Problem Factoring Integers

### College Algebra - MAT 161 Page: 1 Copyright 2009 Killoran

College Algebra - MAT 6 Page: Copyright 2009 Killoran Zeros and Roots of Polynomial Functions Finding a Root (zero or x-intercept) of a polynomial is identical to the process of factoring a polynomial.

### p e i 1 [p e i i ) = i=1

Homework 1 Solutions - Sri Raga Velagapudi Algebra Section 1. Show that if n Z then for every integer a with gcd(a, n) = 1, there exists a unique x mod n such that ax = 1 mod n. By the definition of gcd,

### Die ganzen zahlen hat Gott gemacht

Die ganzen zahlen hat Gott gemacht Polynomials with integer values B.Sury A quote attributed to the famous mathematician L.Kronecker is Die Ganzen Zahlen hat Gott gemacht, alles andere ist Menschenwerk.

### expression is written horizontally. The Last terms ((2)( 4)) because they are the last terms of the two polynomials. This is called the FOIL method.

A polynomial of degree n (in one variable, with real coefficients) is an expression of the form: a n x n + a n 1 x n 1 + a n 2 x n 2 + + a 2 x 2 + a 1 x + a 0 where a n, a n 1, a n 2, a 2, a 1, a 0 are

### 6 EXTENDING ALGEBRA. 6.0 Introduction. 6.1 The cubic equation. Objectives

6 EXTENDING ALGEBRA Chapter 6 Extending Algebra Objectives After studying this chapter you should understand techniques whereby equations of cubic degree and higher can be solved; be able to factorise

### Tim Kerins. Leaving Certificate Honours Maths - Algebra. Tim Kerins. the date

Leaving Certificate Honours Maths - Algebra the date Chapter 1 Algebra This is an important portion of the course. As well as generally accounting for 2 3 questions in examination it is the basis for many

### 1 Lecture: Integration of rational functions by decomposition

Lecture: Integration of rational functions by decomposition into partial fractions Recognize and integrate basic rational functions, except when the denominator is a power of an irreducible quadratic.

### The Dirichlet Unit Theorem

Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if

### Integrals of Rational Functions

Integrals of Rational Functions Scott R. Fulton Overview A rational function has the form where p and q are polynomials. For example, r(x) = p(x) q(x) f(x) = x2 3 x 4 + 3, g(t) = t6 + 4t 2 3, 7t 5 + 3t

### ALGEBRA HW 5 CLAY SHONKWILER

ALGEBRA HW 5 CLAY SHONKWILER 510.5 Let F = Q(i). Prove that x 3 and x 3 3 are irreducible over F. Proof. If x 3 is reducible over F then, since it is a polynomial of degree 3, it must reduce into a product

### 5. Factoring by the QF method

5. Factoring by the QF method 5.0 Preliminaries 5.1 The QF view of factorability 5.2 Illustration of the QF view of factorability 5.3 The QF approach to factorization 5.4 Alternative factorization by the

### Galois Theory. Richard Koch

Galois Theory Richard Koch April 2, 2015 Contents 1 Preliminaries 4 1.1 The Extension Problem; Simple Groups.................... 4 1.2 An Isomorphism Lemma............................. 5 1.3 Jordan Holder...................................

### Algebra Practice Problems for Precalculus and Calculus

Algebra Practice Problems for Precalculus and Calculus Solve the following equations for the unknown x: 1. 5 = 7x 16 2. 2x 3 = 5 x 3. 4. 1 2 (x 3) + x = 17 + 3(4 x) 5 x = 2 x 3 Multiply the indicated polynomials

### EMBEDDING DEGREE OF HYPERELLIPTIC CURVES WITH COMPLEX MULTIPLICATION

EMBEDDING DEGREE OF HYPERELLIPTIC CURVES WITH COMPLEX MULTIPLICATION CHRISTIAN ROBENHAGEN RAVNSHØJ Abstract. Consider the Jacobian of a genus two curve defined over a finite field and with complex multiplication.

### The Method of Partial Fractions Math 121 Calculus II Spring 2015

Rational functions. as The Method of Partial Fractions Math 11 Calculus II Spring 015 Recall that a rational function is a quotient of two polynomials such f(x) g(x) = 3x5 + x 3 + 16x x 60. The method

### I. GROUPS: BASIC DEFINITIONS AND EXAMPLES

I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called

### Name Intro to Algebra 2. Unit 1: Polynomials and Factoring

Name Intro to Algebra 2 Unit 1: Polynomials and Factoring Date Page Topic Homework 9/3 2 Polynomial Vocabulary No Homework 9/4 x In Class assignment None 9/5 3 Adding and Subtracting Polynomials Pg. 332

### SOLVING POLYNOMIAL EQUATIONS BY RADICALS

SOLVING POLYNOMIAL EQUATIONS BY RADICALS Lee Si Ying 1 and Zhang De-Qi 2 1 Raffles Girls School (Secondary), 20 Anderson Road, Singapore 259978 2 Department of Mathematics, National University of Singapore,

### 1.3 Algebraic Expressions

1.3 Algebraic Expressions A polynomial is an expression of the form: a n x n + a n 1 x n 1 +... + a 2 x 2 + a 1 x + a 0 The numbers a 1, a 2,..., a n are called coefficients. Each of the separate parts,

### Solving Cubic Polynomials

Solving Cubic Polynomials 1.1 The general solution to the quadratic equation There are four steps to finding the zeroes of a quadratic polynomial. 1. First divide by the leading term, making the polynomial

### Factoring Cubic Polynomials

Factoring Cubic Polynomials Robert G. Underwood 1. Introduction There are at least two ways in which using the famous Cardano formulas (1545) to factor cubic polynomials present more difficulties than

### 10 Splitting Fields. 2. The splitting field for x 3 2 over Q is Q( 3 2,ω), where ω is a primitive third root of 1 in C. Thus, since ω = 1+ 3

10 Splitting Fields We have seen how to construct a field K F such that K contains a root α of a given (irreducible) polynomial p(x) F [x], namely K = F [x]/(p(x)). We can extendthe procedure to build

### 2. Let H and K be subgroups of a group G. Show that H K G if and only if H K or K H.

Math 307 Abstract Algebra Sample final examination questions with solutions 1. Suppose that H is a proper subgroup of Z under addition and H contains 18, 30 and 40, Determine H. Solution. Since gcd(18,

### Equations, Inequalities & Partial Fractions

Contents Equations, Inequalities & Partial Fractions.1 Solving Linear Equations 2.2 Solving Quadratic Equations 1. Solving Polynomial Equations 1.4 Solving Simultaneous Linear Equations 42.5 Solving Inequalities

### Partial Fractions. (x 1)(x 2 + 1)

Partial Fractions Adding rational functions involves finding a common denominator, rewriting each fraction so that it has that denominator, then adding. For example, 3x x 1 3x(x 1) (x + 1)(x 1) + 1(x +

### Decomposing Rational Functions into Partial Fractions:

Prof. Keely's Math Online Lessons University of Phoenix Online & Clark College, Vancouver WA Copyright 2003 Sally J. Keely. All Rights Reserved. COLLEGE ALGEBRA Hi! Today's topic is highly structured and

### Galois theory for dummies

Galois theory for dummies Ruben Spaans May 21, 2009 1 Notes on notation To help avoid vertical figures, I use the notation E/F if E is an extension to the field F. This is the same notation as Wikipedia

### Putnam Notes Polynomials and palindromes

Putnam Notes Polynomials and palindromes Polynomials show up one way or another in just about every area of math. You will hardly ever see any math competition without at least one problem explicitly concerning

### The cyclotomic polynomials

The cyclotomic polynomials Notes by G.J.O. Jameson 1. The definition and general results We use the notation e(t) = e 2πit. Note that e(n) = 1 for integers n, e(s + t) = e(s)e(t) for all s, t. e( 1 ) =

### 1.3 Polynomials and Factoring

1.3 Polynomials and Factoring Polynomials Constant: a number, such as 5 or 27 Variable: a letter or symbol that represents a value. Term: a constant, variable, or the product or a constant and variable.

### SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS

(Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.1 SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS LEARNING OBJECTIVES Be able to identify polynomial, rational, and algebraic

### 3.6 The Real Zeros of a Polynomial Function

SECTION 3.6 The Real Zeros of a Polynomial Function 219 3.6 The Real Zeros of a Polynomial Function PREPARING FOR THIS SECTION Before getting started, review the following: Classification of Numbers (Appendix,

### Factoring Polynomials

Factoring Polynomials Any Any Any natural number that that that greater greater than than than 1 1can can 1 be can be be factored into into into a a a product of of of prime prime numbers. For For For

### Cyclotomic Extensions

Chapter 7 Cyclotomic Extensions A cyclotomic extension Q(ζ n ) of the rationals is formed by adjoining a primitive n th root of unity ζ n. In this chapter, we will find an integral basis and calculate

### 3.2 The Factor Theorem and The Remainder Theorem

3. The Factor Theorem and The Remainder Theorem 57 3. The Factor Theorem and The Remainder Theorem Suppose we wish to find the zeros of f(x) = x 3 + 4x 5x 4. Setting f(x) = 0 results in the polynomial

### The Notebook Series. The solution of cubic and quartic equations. R.S. Johnson. Professor of Applied Mathematics

The Notebook Series The solution of cubic and quartic equations by R.S. Johnson Professor of Applied Mathematics School of Mathematics & Statistics University of Newcastle upon Tyne R.S.Johnson 006 CONTENTS

### UNCORRECTED PAGE PROOFS

number and and algebra TopIC 17 Polynomials 17.1 Overview Why learn this? Just as number is learned in stages, so too are graphs. You have been building your knowledge of graphs and functions over time.

### Zeros of a Polynomial Function

Zeros of a Polynomial Function An important consequence of the Factor Theorem is that finding the zeros of a polynomial is really the same thing as factoring it into linear factors. In this section we

### 7.1 Graphs of Quadratic Functions in Vertex Form

7.1 Graphs of Quadratic Functions in Vertex Form Quadratic Function in Vertex Form A quadratic function in vertex form is a function that can be written in the form f (x) = a(x! h) 2 + k where a is called

### Linear Equations in One Variable

Linear Equations in One Variable MATH 101 College Algebra J. Robert Buchanan Department of Mathematics Summer 2012 Objectives In this section we will learn how to: Recognize and combine like terms. Solve

### Factoring Polynomials

Factoring Polynomials Factoring Factoring is the process of writing a polynomial as the product of two or more polynomials. The factors of 6x 2 x 2 are 2x + 1 and 3x 2. In this section, we will be factoring

### Introduction to Algebraic Geometry. Bézout s Theorem and Inflection Points

Introduction to Algebraic Geometry Bézout s Theorem and Inflection Points 1. The resultant. Let K be a field. Then the polynomial ring K[x] is a unique factorisation domain (UFD). Another example of a

### Algebra 3: algorithms in algebra

Algebra 3: algorithms in algebra Hans Sterk 2003-2004 ii Contents 1 Polynomials, Gröbner bases and Buchberger s algorithm 1 1.1 Introduction............................ 1 1.2 Polynomial rings and systems

### MA107 Precalculus Algebra Exam 2 Review Solutions

MA107 Precalculus Algebra Exam 2 Review Solutions February 24, 2008 1. The following demand equation models the number of units sold, x, of a product as a function of price, p. x = 4p + 200 a. Please write