Math 421 Algebra II - Exam II: Polynomial Irreducibility & Ring Homomorphisms Solutions, Exams of Algebra

Download Math 421 Algebra II - Exam II: Polynomial Irreducibility & Ring Homomorphisms Solutions and more Exams Algebra in PDF only on Docsity! Math 421 Algebra II - Exam II April 10, 2006 Instructions: You have 50 min to complete the exam (6 problems). This is a closed book, closed notes exam. Use of calculators is not permitted. Show all your work for full credit. Write your name and your instructor’s name on the blue book cover. Print your name : Problem Max Points Your Score Problem Max Points Your Score 1 20 4 15 2 20 5 10 3 20 6 15 Extra credit 5 Total 100 1 (1) True or False? If false, give a counterexample. (Rules: Leave it blank if you are unsure. You will get -1 for an incorrect answer! No purchase necessary. ) (a) Reduction modulo p is cool. T. (b) Q is an ideal of R. False. √ 2 · 1 6∈ Q. (c) If a ring R is of prime characteristic, then R is finite. False. Z2(x) is infinite and of characteristic 2. (d) If a ring R is finite then R is of prime characteristic. T. (e) A finite integral domain is a field. T. (f) If a field F contains Z, then it also contains Q. T. (g) If D is an integral domain, then so is D[x]. T (h) A prime ideal is maximal. False. In Z ⊕ Z, Z ⊕ 0 is a prime ideal contained in a maximal ideal Z⊕ 2Z. (i) A maximal ideal is prime. T. (j) Given f(x) ∈ F [x], there is an extension field E of F in which f has a root. T. (k) If A and B are integral domains, then so is A ⊕ B. False. A direct sum is never an integral domain, since (a, 0) · (0, b) = (0, 0). (l) If R is a PID then a nonzero prime ideal of it is maximal. T. (m) If f(x) ∈ Z[x] is irreducible, then its reduction f(x) ∈ Zp[x] is also irreducible. False. x2 + 1 is irreducible in Z[x] but x2 + 1 = (x + 1)2 is reducible in Z2[x]. (2) Determine if each of the following is irreducible: (a) x4 + 10x3 + 4x2 − 3x + 11 over Q. (Hint : Use reduction modulo p.) Answer: Let f(x) denote the given polynomial. Upon reducing modulo 2, we obtain f(x) = x4 + x + 1. Since deg f(x) = deg f(x) and f(x) is irreducible over Z2 (from the Useful data below), f(x) is irreducible over Z. (b) x4 + 1 over Z. Answer: Substitute x = y + 1 and get y4 + 4y3 + 6y2 + 4y + 2 which is irreducible by Eisenstein’s criterion (p = 2). Hence x4 + 1 is irreducible over Z. (c) x3 + 3x2 + 5x + 5 over Z3. Answer: Let f(x) denote the given polynomial. It has no root in Z3 since f(0) = 5 = 2, f(1) = 2 and f(−1) = 2. It follows that f is irreducible over Z3. (d) x3 + 3x2 + 5x + 5 over Z. Answer: Since deg f(x) = deg f(x) and f(x) is irreducible over Z3, it follows from reduction modulo 3 argument that f(x) is irreducible over Z. (3) Find all ring homomorphisms from Z⊕ Z to Z. Prove that they are all. Answer: Let φ : Z⊕ Z→ Z be a ring homomorphism. Let φ((1, 0)) = m and φ((0, 1)) = n. Then φ((a, b)) = φ(a(1, 0) + b(0, 1)) = aφ((1, 0)) + bφ((0, 1)) = am + bn. We have φ((1, 0)(0, 1)) = mn = φ((0, 0)) = 0. Hence m = 0 or n = 0. Also, φ((1, 0)(1, 0)) = φ((1, 0)) yields m2 = m and φ((0, 1)(0, 1)) = φ((0, 1)) yields n2 = n. Thus all possibilities are: (m,n) = (0, 0), (1, 0), (0, 1). These correspond to homomorphisms φ((a, b)) = 0, φ((a, b)) = a and φ((a, b)) = b respectively. (4) (a) Find all ring homomorphisms from Z12 to Z8 and briefly explain why they are all. Answer: Let φ : Z12 → Z8 be a ring homomorphism. It is a group homomorphism and must be of the form φ(x) = mx for some m ∈ Z (this from group theory). (i) φ(12) = φ(0) = 0 yields 8|12m or 2|3m. Hence m = 0, 2, 4, 6, 8, 10. (ii) φ(12) = (φ(1))2 = φ(1) yields 8|m2 −m. From (a) and (b), we conclude that m = 0, 8 and both yields φ(x) = 0 for all x ∈ Z12. (b) Find all ring homomorphisms from Z[x] to Z[x]. 2