The University of Sydney MATH3002 Rings and Fields

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1 The University of Sydney MATH3002 Rings and Fields Semester 1 Tutorial for Week 13 (Magma) 2001 The topics of this tutorial are irreducible polynomials and finite fields. Some facts about Magma that may be helpful: (a) The function FiniteField(q) or GF(q) returns the finite field with q elements, where q is a prime or a power of a prime. (b) To construct the polynomial ring P in the indeterminate X over the field K, assign K and then type: P<X> := PolynomialRing(K); (c) The function Factorization(a) returns the factorization of the polynomial a, as a sequence of tuples <r i, n i > such that a divided by its leading coefficient equals i rn i i. (d) The function IsIrreducible(f) returns true if f is irreducible. (e) The function Order(a) returns the order of a. 1. (i) How do you know that there is a field with 81 elements? Define it by typing K<g> := FiniteField(81); (ii) The element g K has a special property, to do with its order. Find the order of g, and hence say what kind of element g is in K. (iii) Define the polynomial ring P in the indeterminate X over K, andthepolynomial f = X (iv) Factorize f using Magma. (v) In terms of g, what are the zeros of f? Are there any elements of K which are not zeros of f? (vi) Think of a polynomial whose zeros are every element of K. Test your answer using Magma. 81 = 3 4, which is a power of a prime. For every prime power, there is a field with that number of elements. > K<g> := FiniteField(81); > Order(g); 80 The order of g is 80, which is the number of non-zero elements of K. Therefore g is a generator of the non-zero elements of K. > P<X> := PolynomialRing(K); > f := X^80-1; > Factorization(f); <X + 1, 1>, <X + g, 1>, <X + g^2, 1>, <X + g^3, 1>, <X + g^4, 1>,

2 2.. <X + g^78, 1>, <X + g^79, 1> This shows that the zeros of f are 1, g, g 2, g 3,..., g 79. Since g 40 = 1 we have g i = g i+40 and so the zeros of f can also be described as the powers g j, where 0 j 79. Only 0 K is not a zero of f. The zeros of the polynomial X f(x) =X 81 X are all the zeros of f together with Prove the following statements, on paper and/or using Magma: (i) x is irreducible over Z/7Z. (ii) x 3 9 is irreducible over Z/31Z. (iii) x 3 9 is reducible over Z/11Z. (i) (ii) If x is reducible over Z/7Z, it must be the product of two linear factors, say (x a)(x b) forsomea, b Z/7Z, sincex has degree 2. But then a = 0. If you substitute every element a of Z/7Z in this equation, you will see that there are no solutions. Therefore x is irreducible over Z/7Z. If x 3 9 is reducible over Z/31Z, it must be the product of three linear factors or a linear factor and an irreducible quadratic factor, since x 3 9 has degree 3. Either way, (x a) divides x 3 9forsomea Z/31Z. Butthena 3 9=0. If you substitute every element a of Z/7Z in this equation, you will see that there are no solutions. Therefore x 3 9 is irreducible over Z/31Z. (iii) Same idea as above, but observe that 4 3 9=0overZ/11Z. Therefore (x 4) = (x + 7) is a proper factor of x 3 9overZ/11Z, andsox 3 9is reducible over Z/11Z. Magma solution: > P<x> := PolynomialRing(Integers(7)); > IsIrreducible(x^2 + 1); true > P<x> := PolynomialRing(Integers(31)); > IsIrreducible(x^3-9); true > P<x> := PolynomialRing(Integers(11)); > IsIrreducible(x^3-9); false > Factorization(x^3-9); <x + 7, 1>, <x^2 + 4*x + 5, 1> 3. (i) Explain how to construct a field with: (a) 7 elements (b) 49 elements (c) 31 elements (d) 31 3 elements Hint: Use the results from the previous exercise.

3 (ii) How many zero divisors does each of these fields have? (iii) How many units does each of these fields have? 3 (i) (a) Use Z/7Z, since 7 is prime. (b) Extend the finite field with 7 elements using x 2 + 1, which is irreducible over Z/7Z. The polynomial has degree 2, so the resulting field has 7 2 =49elements. (c) Use Z/31Z, since 31 is prime. (d) Extend the finite field with 31 elements using x 3 9, which is irreducible over Z/31Z. The polynomial has degree 3, so the resulting field has 31 3 elements. (ii) No field has any zero divisors, so the answer in each case is zero. (iii) Every non-zero element of a field is a unit, so the answer in each case is the size of the field minus Let F and K be two fields, such that K is an extension of F. Suppose f(x) and g(x) are relatively prime in F x (i.e. gcd(f(x),g(x)) = 1). Show that they are also relatively prime in Kx. Since f(x) andg(x) are relatively prime in F x, there exist polynomials a(x) and b(x) inf x such that a(x)f(x) + b(x)g(x) = 1. Now let d(x) be the greatest common divisor of f(x) andg(x), considered over Kx, and consider this equation over Kx instead of F x. The polynomial d(x) divides the left side of the equation, so it must also divide the right side, which is the element 1 in K. Hence d(x) divides 1, and since d(x) is monic by definition, d(x) = 1. Therefore f(x) andg(x) are relatively prime in Kx. 5. Let E = Q 2 and K = E 5. (i) Show that K E. (ii) Calculate the degree K : Q. (iii) Observe that t = 2+ 5 K. Can the numbers 1, t, t 2, t 3, t 4 be linearly independent over Q? (iv) Prove that 2+ 5 is algebraic over Q by finding a non-zero polynomial f(x) QX such that f( 2+ 5) = 0. (i) If K = E, then 5 E and we could write 5=x + y 2forsomex, y Q. Squaring this equation gives 5 = x 2 +2y 2 +2xy 2. If xy 0,thiswould show that 2=(5 x 2 2y 2 )/2xy is rational, a contradiction. If x =0, then we would have 5/ 2=y Q, a contradiction. And if y = 0 we would have 5=x Q, yet another contradiction. Thus it cannot be the case that K = E. (ii) The minimal polynomial of 2overQ is X 2 2, since 2 is certainly not the zero of a polynomial of degree 1 in QX. Thus Q 2 : Q =deg(x 2 2) = 2. It follows from (i) that the polynomial X 5 1 is irreducible in EX. Thus K : E = 2 and therefore K : Q =K : E E : Q =4.

4 4 (iii) No. The dimension of K over Q is 4 and so the five numbers 1, t, t 2, t 3, t 4 are not linearly independent. (iv) We have t 2 = and therefore (t 2 7) 2 = 40. That is, t 4 14t 2 +9 = 0 andsowemaytakef(x) =X 4 14X In this exercise, you will construct a generator for the unit group of a finite field (i.e. the group of its non-zero elements). The method could be generalized into a proof that the unit group of every finite field is cyclic. (i) Define F to be the finite field with 625 elements. Use w in the angle brackets (see exercise 2) so that the elements will print nicely. Then define the polynomial ring R in x over F. (ii) Show that the factorization of 624 = 625 1is (iii) We will now look for elements of F of orders 2 4, 3 and 13, and multiply them together to get an element of order 624 this will be a generator. We start with 13. Factorize x 13 1, and hence find its zeros. (iv) The order of each of these zeros must divide 13. What are the positive divisors of 13? Choose an element with order 13 and assign it to a. (v) Factorize x 3 1 and hence find an element with order 3. Assign it to b. (vi) Factorize x 16 1, and hence find its zeros. These zeros have orders dividing 16. Also factorize x 8 1 and find its zeros. These zeros have orders dividing 8. (vii) The elements with order exactly 16 are those in the first list but not the second list. Choose one of them and assign it to c. (viii)calculate d = abc. You should have found a generator for the non-zero elements of F. What do you expect the order of d to be? Test your answer using the Order function. (i) (ii) (iii) > F<w> := FiniteField(625); > R<x> := PolynomialRing(F); > Factorization(x^13-1); <x + w^24, 1>, <x + w^72, 1>, <x + w^120, 1>, <x + w^168, 1>, <x + w^216, 1>, <x + w^264, 1>, <x + w^360, 1>, <x + w^408, 1>, <x + w^456, 1>, <x + w^504, 1>,

5 5 <x + w^552, 1>, <x + w^600, 1> The zeros of x 13 1are w 24, w 72, w 120, w 168, w 216, w 264, 4, w 360, w 408, w 456, w 504, w 552, w 600. (iv) The positive divisors of 13 are 1 and = 1 has order 1, and the other zeros have order 13. Choose any of them to be a. For example: > a := -w^24; (v) > Factorization(x^3-1); <x + w^104, 1>, <x + w^520, 1> The zeros are w 104, 4, w 520. Since 3 is prime, like 13, all of these except 4 = 1 have order 3. Choose one of them: > b := -w^104; (vi) > Factorization(x^16-1); <x + 1, 1>, <x + w^39, 1>, <x + w^78, 1>, <x + w^117, 1>, <x + 2, 1>, <x + w^195, 1>, <x + w^234, 1>, <x + w^273, 1>, <x + w^351, 1>, <x + w^390, 1>, <x + w^429, 1>, <x + 3, 1>, <x + w^507, 1>, <x + w^546, 1>, <x + w^585, 1> > Factorization(x^8-1); <x + 1, 1>, <x + w^78, 1>, <x + 2, 1>, <x + w^234, 1>, <x + w^390, 1>, <x + 3, 1>, <x + w^546, 1>

6 6 Do the zeros as before. Notice that all the zeros for x 8 1 are also in the x 16 1 list, because if the order of an element divides 8 then it also divides 16. (vii) Choosing: > c := -w^39; (viii) The order of d is = 624, which is the number of non-zero elements of F. > d := a*b*c; > Order(d); 624

Math 4310 Solutions to homework 7 Due 10/27/16

Math 4310 Solutions to homework 7 Due 10/27/16 1. Find the gcd of x 3 + x 2 + x + 1 and x 5 + 2x 3 + x 2 + x + 1 in Rx. Use the Euclidean algorithm: x 5 + 2x 3 + x 2 + x + 1 = (x 3 + x 2 + x + 1)(x 2 x