What is the Galois group of a polynomial over a finite field?

Abstract Algebra: Galois Theory construction for a polynomial HELP?

• I've basically constructed the automorphisms for f(x) = (x^2 - 2)(x^2 - 3), however I need to find the Galois group. Does anyone know how to do this? I turned in my work to the professor but I wasn't sure if my solution is complete. I have the following mapping: sqrt(2) - > -sqrt(2) sqrt(3) -> sqrt(3) sqrt(2) - > sqrt(2) sqrt(3) -> sqrt(3) sqrt(2) -> -sqrt(2) sqrt(3) -> -sqrt(3) sqrt(2) -> sqrt(2) sqrt(3) -> -sqrt(3) My question is, how do I verify that each of them are bijections? There are 4!=24 possible ways but only 4 of them are automorphic functions. What are all the subfields generated by the Galois group? I'm also have a tough time understanding how a root from a splitting field maps to a splitting field.

• Answer:

  Since both Q(√2) and Q(√3) are subfields of Q(√2, √3), the automorphisms of Q(√2, √3) restricted to either of these subfields must restrict to the automorphisms of the two subfields. Since the automorphisms of Q(√2) and Q(√3) are defined by √2 → ±√2, and √3 → ±√3, respectively (and then extend Q-linearly), this forces the automorphisms of Q(√2, √3) to be one of the following: (i) √2 → √2, √3 → √3 (ii) √2 → -√2, √3 → √3 (iii) √2 → √2, √3 → -√3 (iv) √2 → -√2, √3 → -√3 I'll verify that (iv) yields a bijection (the others are done similarly). First of all, any element in Q(√2, √3) can be written (uniquely) in the form a + b√2 + c√3 + d√6 for some a, b, c, d in Q. So, the automorphism "F" in (iii) is written more explicitly as F(a + b√2 + c√3 + d√6) = a - b√2 - c√3 + d√6. (Note that since √2 → -√2, √3 → -√3, we have √6 = √2 * √3 → -√2 * -√3 = √6.) F is onto: Given w + x√2 + y√3 + z√6 in Q(√2, √3): w - x√2 - y√3 + z√6 is in Q(√2, √3), and F(w - x√2 - y√3 + z√6) = w + x√2 + y√3 + z√6. F is 1-1: ker F = {a + b√2 + c√3 + d√6 in Q(√2, √3): F(a + b√2 + c√3 + d√6) = 0} ........= {a - b√2 - c√3 + d√6 = 0} ........= {0}, since {1, √2, √3, √6} is a basis for Q(√2, √3). ------------------ For the second question, I inadvertently answered much of this already. Check that the 4 automorphisms are each of order 2. Hence, it is isomorphic to the Klein 4-Group, whose subgroups are easy to enumerate. Via fixed fields Fix(subgroup of automorphisms): Fix({(i)} = Q Fix({(i), (ii)} = Q(√2) Fix({(i), (iii)} = Q(√3) Fix({(i), (iv)} = Q(√6) Fix({(i), (ii), (iii), (iv)}) = Q(√2, √3). I hope this helps!