Homework 3

• Real fields
  • Let Q(sqrt(2)) = { a + b*sqrt(2) | a and b are in Q}. Show that this set, with addition and multiplication, is a field.
  • Show that sqrt(3) is not an element of Q(sqrt(2)).
  • Similarly to Q(sqrt(2)), we can define Q(sqrt(3)) = { a + b*sqrt(3) | a and b are in Q} and show that it is a field. Recall that if F₁ and F₂ are fields, a function f : F₁ -> F₂ is a field isomorphism if it is one-to-one and onto, and f(a+b)=f(a)+f(b) and f(ab)=f(a)f(b) for all a and b in F₁. Consider the function f : Q(sqrt(2)) -> Q(sqrt(3)) given by f(a+b*sqrt(2)) = a+b*sqrt(3). Show that f is not a field isomorphism.
  • Show that Q(sqrt(2)) and Q(sqrt(3)) are not isomorphic as fields (i.e. that there is no field isomorphism between these two fields).
    Hint: assume there exists a field isomorphism f : Q(sqrt(2)) -> Q(sqrt(3)), let f(sqrt(2)) = x+y*sqrt(3). Consider f(sqrt(2)*sqrt(2)) and derive a contradiction.
• Compute the following:
  • (5+2i)(3-i)
  • (5+2i) / (3-i)
  • i²⁰⁰⁹ (show your work and/or explain how you got your answer)
  • (i+1)²⁰⁰⁹ (show your work and/or explain how you got your answer)