5 Question with Solution - Modern Algebra | Quiz 1 | MAT 150B, Quizzes of Algebra

Download 5 Question with Solution - Modern Algebra | Quiz 1 | MAT 150B and more Quizzes Algebra in PDF only on Docsity! Quiz 1 Math 150B March 5, 2009 (1) Let f(x) = x3 + x2 + 3x+ 1 in F5[x]. Answer each of the following questions. (a) If g(x) is any polynomial in F5[x] such that f divides g, what is the relationship between the ideals (f) and (g)? Proof. Since f divides g, we may write g(x) = q(x)f(x) for q(x) some polynomial in F5[x]. So g ∈ (f), and this implies (g) ⊂ (f).  (b) Determine all α ∈ F5 such that f(α) = 0. Proof. Since F5 = {0̄, 1̄, 2̄, 3̄, 4̄} we can compute f(α) for each α. We get f(0̄) = 1̄, f(1̄) = 1̄, f(2̄) = 4̄, f(3̄) = 1̄, f(4̄) = 3̄. So there are no α ∈ F5 such that f(α) = 0. That is, f is irreducible over F5.  (c) Describe all ideals I such that (f) ⊂ I ⊂ F5[x]. Proof. Since f is irreducible, the ideal (f) is maximal. So the only ideals containing (f) are (f) itself and F5[x].  (d) Is F5[x]/(f) a field? Why or why not? Proof. Since (f) is maximal, the quotient F5[x]/(f) is a field.  (e) How many units are in F5[x]/(f)? Explain your answer. Proof. Any element of F5[x]/(f) is of the form a0 + a1x̄+ a2x̄2 where x̄ is the image of x in the quotient. Since ai ∈ F5, there are 125 elements in the field, and since every nonzero element is a unit, there are 124 units.  1