The Galois Correspondence · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Symmetries of fields and the groups that mirror them

Mathematics · Abstract Algebra · ages 22-24
Name ______________________   Date ____________

  1. In K=Q(√2,√3), how many automorphisms fixing Q can send √2 to √3?

    Answer: ______________

  2. The polynomial (x-3)(x-7) splits over Q. How many automorphisms of its splitting field fix Q?

    • 2
    • 4
    • 1
    • 3

  3. Adjoin the square root of a prime to Q. How many automorphisms fix Q?

    • 4
    • 3
    • 2

  4. The nonsolvability of S5 rules out a radical formula that solves every quintic.

    Circle one:   True   False

  5. Let a,b be rational, with a not a rational square and √b outside Q(√a). What is the order of the Galois group of Q(√a,√b) over Q?

    Answer: ______________

  6. For K=Q(√2,√3) over Q, the identity subgroup {e} has fixed field Q.

    Circle one:   True   False

  7. In Q(√2,√3), let st flip both √2 and √3. What is st(2√2+3√3+4√6)?

    • -2√2-3√3+4√6
    • -2√2-3√3-4√6
    • +2√2+3√3+4√6
    • -2√2+3√3-4√6

  8. Which list gives all fields strictly between Q and Q(√2,√5)?

    • Q(√2), Q(√5), Q(√10)
    • Q(√2), Q(√5), Q(√11)
    • Q(√2), Q(√5), Q(√15)

  9. How many automorphisms of the splitting field of x³-5 fix Q?

    • 3
    • 6
    • 9

  10. The splitting field M of x³-7 over Q has Galois group S3. How many fields lie strictly between Q and M, excluding both endpoints?

    Answer: ______________

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Answer key

For grown-ups. Fold this page away before handing over the rest.

Symmetries of fields and the groups that mirror them W1-mt_30kyliQR_L-s1

  1. 0 · Preserving (√2)²=2 requires the image to square to 2. The proposed image √3 squares to 3, so no such automorphism exists.
  2. 1 · Both roots already lie in Q, so the splitting field is Q. Its only automorphism fixing Q is the identity.
  3. 2 · The root can map to itself or its negative. Both choices give an automorphism, so there are 2.
  4. True · Some quintics over Q have Galois group S5. They cannot be solved by radicals, so no radical formula solves all quintics.
  5. 4 · Each adjoining step has degree 2, so the total degree is 4. The field splits (x²-a)(x²-b), making it Galois over Q with group order 4.
  6. False · The identity fixes every element of K. Its fixed field is K; the whole Galois group has fixed field Q.
  7. -2√2-3√3+4√6 · The √2 and √3 terms change sign. Their product √6 stays fixed because the two sign changes cancel.
  8. Q(√2), Q(√5), Q(√10) · For two distinct primes, the three proper intermediate fields adjoin either square root or their product. Here they are Q(√2), Q(√5) and Q(√10).
  9. 6 · Eisenstein at 5 gives degree 3 for the real root. Adjoining a nonreal cube root of unity adds degree 2, so the splitting-field degree and automorphism count are 6.
  10. 4 · Exclude the identity subgroup and S3. The remaining subgroups are the three swap subgroups and the one three-cycle subgroup, giving 4 fields.
Worksheet · LightMySky