1 Please find a reducible polynomial over Z2 that has no roo

1. Please find a reducible polynomial over Z₂ that has no roots in Z₂

2. Find the polynomials a(x) and b(x) in Z₅[x] such that a(x)(3x³x+3)+b(x)(x²+x2) = x² 1 (all the coefficients are in Z₅, but, as is customary, we write 3 instead of [3]₅).

3. Let p(x) = x² x + 2 be a polynomial over Z₅.

(a) Please check that p(x) is irreducible.

(b) It follows from the general theory (Kronecker’s theorem) that the congruence class [x]_p(x) is a root of the polynomial p(x) in the field Z₅[x]/<p(x)>. Please find the other root of the polynomial p(x) in that larger field.

4. We have noticed on a number of occasions that the set of polynomials over a field has similar properties to the set of integers and that irreducible polynomials are somewhat similar to prime numbers. We proved in class (as a part of the proof of Kronecker’s theorem) that if p(x) is irreducible, then F[x]/<p(x)> is a field. Please adapt that argument to show that, if p is a prime number, then Zp is a field.

Solution

1/ Since Z2 is a field, no constant polynomial is irreducible. Hence we need to find the irreducible polynomials of degrees 1, 2, and 3. Degree 1: By Theorem 1, every polynomial in Z2[x] of degree 1 is irreducible over Z2. Since the only coefficients are 0 and 1, the polynomials of degree 1 are x and x + 1.

2/

A review of the principal ideal domain F[x] of all polynomials f(x) in the indeterminate x over the field F: degree f(x), evaluation mappings, the polynomial division law, the Euclidean algorithm for gcds. Irreducible polynomials p(x) over F, the extension field F(c) where p(c)=0.

Equivalence of s×t matrices A(x) over F[x], elementary row and column operations over F[x]. Reduction of A(x) to its Smith normal form S(A(x)), the invariant factors d l (x) of A(x). Calculation of invertible matrices P(x) and Q(x) over F[x] satisfying P(x)A(x)=Q(x)S(A(x)).

3/

is an algebraic integer, an 1=,2,…,n1=,2,…,n are its conjugates. The question is about the coefficients of the polynomial

fk(x)=i(xki)=xn+ak,n1xk1++ak,0.fk(x)=i(xik)=xn+ak,n1xk1++ak,0.

The coefficients of the polynomial are symmetric polynomials in the unknowns ii with integer coefficients. By basic results on symmetric polynomials they are thus values of some polynomials g(s1,s2,…,sn)g(s1,s2,…,sn) with integer coefficients evaluated at the elementary symmetric polynomials sisi of 1,2,…,n1,2,…,n. Because is an algebraic integer, the elementary symmetric polynomials are all rational integers (they are up to sign the coefficients of the minimal polynomial of ). The claim follows from this.

Note that fk(x)fk(x) is not necessarily the minimal polynomial of kk. Indeed, when is a root of unity (the case in the text), there will be some repetitions among the roots k1,k2,…,kn1k,2k,…,nk of fk(x)fk(x). At least for some exponents kk.

On second thought. Even though fk(x)fk(x) is not necessarily the minimal polynomial of kk in the problem scenario outlined in the previous paragraph, it is always a power of the minimal polynomial of kk. This is because by basic Galois theory each distinct conjugate of kk appears in the list k1,k2,…,kn1k,2k,…,nk the same number of times. The fact that fk(x)[x]fk(x)Z[x] follows also from this.

4/For p an integer greater than or equal to 2 define Zp to be the set {0, 1, 2, ..., p 1} equipped with addition and multiplication modulo p. Show that Zp is a field if and only if p is a prime number. Note that since this is an “if and only if” proof, we need to show both directions: “if” (a.k.a. “”): Let p be prime. We need to prove that Zp is a field by showing it satisfies the field axioms: Addition: (A1) Closure under addition is inherited from the closure of the integers under addition (A2) Commutativity of addition is also inherited from the integers (A3) Associativity of addition is inherited from the integers (A4) Zp contains an additive identity, 0 which acts in the same way it does in the integers (A5) Each element has an additive inverse: for each x Zp take x (mod p). Then by the properties of modular arithmetic: x + (x) 0 mod p Multiplication: (M1) We get closure under multiplication from the integers since for x, y Zp xy mod p is in Zp (M2) Multiplication is commutative since its commutative in the integers. (M3) Multiplication is associative since it is in the integers. (M4) We have a multiplicative identity, 1 which acts the same as it does in the integers. (M5) Ok so this is the only interesting axiom to prove here! We need multiplicative inverses for each element. Let x Zp. Note that we cannot just say that the inverse is 1 x since this does not look like an element of Zp. Since p is prime, it is relatively prime to every number, including x. Therefore by the hint we know that a, b Z such that ax + bp = 1 Modding out by p gives us: ax 1 mod p