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

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1 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 + 2) + ( x 2 1) to find that the gcd is x x 3 + x 2 + x + 1 = ( x 2 1)( x 1) For two polynomials f(x), g(x) Fx, set J f,g = {a(x)f(x) + b(x)g(x) a(x), b(x) Fx} to be the set of polynomial combinations of f(x) and g(x). (a) Prove that J f,g is an ideal in Fx. First we will show that J f,g is a vector subspace. For α F and a i, b i Fx, we have α(a 1 f + b 1 g) = α(a 1 f) + α(b 1 g) = (αa 1 )f + (αb 1 )g, (a 1 f + b 1 g) + (a 2 f + b 2 g) = (a 1 f + a 2 f) + (b 1 g + b 2 g) = (a 1 + a 2 )f + (b 1 + b 2 )g are both in J f,g, so J f,g is a vector subspace. Note a 2 (a 1 f + b 1 g) = a 2 (a 1 f) + a 2 (b 1 g) = (a 2 a 1 )f + (a 2 b 1 )g is in J f,g. Furthermore, since multiplication is commutative, (a 1 f + b 1 g)a 2 is also in J f,g. Thus J f,g is an ideal in Fx. (b) Prove that if f(x) and g(x) are both non-zero, then J f,g is the ideal I d(x) of all multiples of d(x), where d(x) = gcd(f(x), g(x)). (Note that I d(x) is also denoted d(x).) By the Euclidean algorithm, there are a(x), b(x) Fx such that a(x)f(x) + b(x)g(x) = d(x), so d(x) is in J f,g. Thus I d(x) J f,g. On the other hand, since d(x) = gcd(f(x), g(x)), every element of J f,g is a multiple of d(x), so J f,g I d(x), concluding our proof. 3. Let F be a field. Prove that a degree 3 polynomial p(x) Fx is reducible if and only if it has a root. Suppose p is reducible. Then there are non-constant polynomials f, g Fx such that p = fg. In particular we have 3 = deg p = deg f + deg g. Since f and g are non-constants, deg f and deg g are both at least 1. However the only way to have 3 be the sum of two positive integers is for deg f = 1 and deg g = 2. Since deg f = 1, we can write f(x) = c(x a) for some a, c F. Then p(a) = f(a)g(a) = 0, and p has a root. Now suppose p has a root a. Then p(a) = 0, so by Corollary 3.4, x a divides p. Hence p is reducible.

2 Math 4310 (Fall 2016) Solution Identify all the monic irreducible polynomials of degree at most 3 in Fx for F = Z/3Z. The monic irreducible polynomials of degree 1 are x, x + 1, and x + 2. Using the same argument as in problem 3, a polynomial of degree 2 is irreducible if and only if it has no roots. Thus the monic irreducible polynomials of degree 2 are x 2 +1, x 2 +x+2, and x 2 +2x+2. Finally we can use problem 3 and check to see which polynomials of degree 3 have no roots. We find that the monic irreducible polynomials of degree 3 are x 3 + 2x + 1, x 3 + 2x + 2, x 3 + x 2 + 2, x 3 + x 2 + 2x + 1, x 3 + 2x 2 + 1, x 3 + 2x 2 + x + 1, and x 3 + 2x 2 + 2x For this problem, let F = Z/2Z. Define F 4 to be the set of all 2 2 matrices { a b F 4 = a, b F}. b a + b (a) Prove that with the operations matrix addition and matrix multiplication, F 4 is a field having exactly 4 elements. (HINT: Use the properties of matrix addition and multiplication to deduce most of the axioms!!!) Note that matrix addition is commutative and associative, matrix multiplication is associative, and matrix multiplication distributes over matrix addition. Furthermore = 0 and = 1 are the additive and multiplicative identities, both are in F 4, and additive inverses exist for matrices. Therefore all that remains to show is that multiplication is commutative, additive and multiplicative inverses exist, and that F 4 is closed under addition and multiplication. Since F 4 is so small, we will demonstrate these by laying out the addition and multiplication tables

3 Math 4310 (Fall 2016) Solution 7 3 Looking the tables, we can see multiplication is commutative, the set is closed under addition and multiplication, and additive and multiplicative inverses exist. (b) Prove that F 4 is isomorphic to Fx/ x 2 + x + 1. (Since these both have four elements, it may be easiest to find the isomorphism by comparing the addition and multiplication tables.) Define the function f from Fx/ x 2 + x + 1 to F 4 to send 0 to 0, 1 to 1, x to x + 1 to. The addition and multiplication tables for Fx/ x 2 + x + 1 are: + x x x x x + 1 x x x x + 1 x + 1 x + 1 x x x x x + 1 x 0 x x + x + x + x Comparing the operation tables, we can see f is an isomorphism., and 6. (Lagrange Interpolation) The goal of this exercise is to prove that for a field F and n + 1 pairs (x 0, y 0 ),..., (x n, y n ) with x i, y j F and x i x j for i j, there is a unique degree n polynomial p(x) Fx so that p(x i ) = y i. As stated, this is false. For instance, if you take three points which are collinear, there is no degree 2 polynomial which passes through those points. You have to weaken it to say there is a unique polynomial of degree at most n. (a) Prove that l j (x) = k j ( ) x xk is a well-defined polynomial of degree n in Fx. Since j k and the x i are distinct, every denominator is nonzero. Furthermore, the product is over n terms, each of which has degree 1. Thus l j (x) is a well-defined polynomial of degree n. n (b) Prove that p(x) = y j l j (x) is a polynomial of the desired form. j=0 Note that for each i j, l j (x i ) = k j ( ) = k i,j ( ) xi x i = 0, x j x i and l i (x i ) = k i ( ) = 1 = 1. x i x k k i

4 Math 4310 (Fall 2016) Solution 7 4 Thus for each i, p(x i ) = n y j l j (x i ) = y i. j=0 Furthermore since p is the sum of degree n polynomials, p has degree at most n. This is where the proof breaks down, when asserting that p has degree n. The leading coefficients of y j l j (x) may cancel each other out. (c) Prove that p(x) is the unique polynomial of degree at most n of this form. (HINT: Suppose there is another, and think about roots of the difference.) Suppose q(x) is a polynomial of degree at most n such that q(x i ) = y i for each i. Then (p q)(x i ) = 0 for each i, which implies p q has n + 1 roots. Since deg p n and deg q n, we have deg(p q) n. Because p q has more roots than its degree, p q must equal 0, and p = q. 7. For polynomials in multiple variables, there are not necessarily interpolating polynomials as in the above example. Let R (k) x, y denote the polynomials with R coefficients in two variables of degree at most k. For a set of points {(a i, b i ) R 2 i = 1,..., n}, there is a linear transformation given by evaluating at these points. E : R (k) x, y R n p(a 1, b 1 ) p(x, y).. p(a n, b n ) (You do not need to prove this is linear, but should convince yourself of this fact.) There will exist an interpolating polynomial taking on certain values at each point (a i, b i ) if and only if the list of values is in the image of E. (Again, don t prove, but convince yourself of this.) Prove that for any list of 10 points on the unit circle in R 2, there is a list of values so that there is not an interpolating polynomial of degree at most 3 taking on those values at the 10 prescribed points. Let k = 3 and n = 10. First note that R (3) x, y has basis { 1, x, y, x 2, xy, y 2, x 3, x 2 y, xy 2, y 3}, so dim(r (3) x, y) = 10. Suppose {(a i, b i ) : 1 i 10} is an arbitrary set of 10 points on the unit circle in R 2. Let p(x, y) = x 2 + y 2 1. Then p(a i, b i ) = 0 for all i, so p is a nonzero element of ker E. This implies dim(ker E) 1. By the first isomorphism theorem, R (3) x, y/ ker E = Im E, so dim(ker E) + dim(im E) = dim(r (3) x, y). Since dim(r (3) x, y) = 10 and dim(ker E) 1, we have dim(im E) < 10. From this we can conclude E is not surjective. Thus there is a vector (c 1,..., c 10 ) in R 10 which is not in Im E, and therefore there is no interpolating polynomial of degree at most 3 which equals c i at (a i, b i ) for each i.

5 Math 4310 (Fall 2016) Solution 7 5 Extended Glossary. You have two choices this week (one inspired by a question in class). 1. Give the definition of a Euclidean domain. Then give an example of a Euclidean domain that we have not yet seen in class, an example which is not a Euclidean domain, and state and prove a theorem about Euclidean domains. (The natural choice is that in a Euclidean domain, there is a Division Algorithm!) 2. Give the definition of a group. Then give an example of a group that we have already seen in class but not called it such, an example from class which is not a group, and state and prove a theorem about groups.

MATH 431 PART 2: POLYNOMIAL RINGS AND FACTORIZATION

MATH 431 PART 2: POLYNOMIAL RINGS AND FACTORIZATION 1. Polynomial rings (review) Definition 1. A polynomial f(x) with coefficients in a ring R is n f(x) = a i x i = a 0 + a 1 x + a 2 x 2 + + a n x n i=0