MODERN ALGEBRA (MATH 210) THANKSGIVING LEFTOVERS: PROOF OF TUESDAY’S THEOREM

RAVI VAKIL

Here is the proof of the theorem that I was in the process of proving at the end of class.

Theorem. If σ_{1}, …, σ_{n} is a group of automorphisms of E, and if F is the fixed field, then [E : F] = n. In particular, E/F is Galois.

Proof. We use linear algebra again. Suppose [E : F] ≥ n + 1. Say σ_{1} is the identity. Then let α_{1}, …, α_{n+1} be elements of E that are linearly independent with respect to F. By linear algebra, there is a nontrivial solution (x_{1}, …, x_{n}) ≠ (0, …, 0) in E to the system of equations

x_{1}σ_{1}(α_{1}) + x_{2}σ_{1}(α_{2}) + ⋯ + x_{n+1}σ_{1}(α_{n+1}) = 0
x_{1}σ_{2}(α_{1}) + x_{2}σ_{2}(α_{2}) + ⋯ + x_{n+1}σ_{2}(α_{n+1}) = 0
⋮
x_{1}σ_{n}(α_{1}) + x_{2}σ_{n}(α_{2}) + ⋯ + x_{n+1}σ_{n}(α_{n+1}) = 0.

The solution can’t lie in F (i.e. all x_{i} can’t lie in F) or else the first equation would be a dependence over F.

Among all nontrivial solutions (x_{1}, …, x_{n}), we choose one with the least number of nonzero elements. We’ll get a contradiction out of this. We may suppose this solution is (a_{1}, …, a_{r}, 0, …, 0), where the first r are non-zero. Also, dividing by a_{r}, we may assume a_{r} = 1. So we have:

(1)  a_{1}σ_{1}(α_{1}) + a_{2}σ_{1}(α_{2}) + ⋯ + σ_{1}(α_{r}) = 0
a_{1}σ_{2}(α_{1}) + a_{2}σ_{2}(α_{2}) + ⋯ + σ_{1}(α_{r}) = 0
⋮
a_{1}σ_{n}(α_{1}) + a_{2}σ_{n}(α_{2}) + ⋯ + σ_{1}(α_{r}) = 0.

Note that r ≠ 1. We may also assume a_{1} is in E, but not in F (as not all a_{i} can lie in F, as observed above). Thus there is an automorphism σ_{k} for which σ_{k}(a_{1}) ≠ a_{1}. Apply σ_{k} to (1)to get

Date: Tuesday, November 26, 2002. Revised Dec. 5, patching an error caught by Oren.

σ_{k}(a_{1})σ_{k}σ_{1}(α_{1}) + σ_{k}(a_{2})σ_{k}σ_{1}(α_{2}) + ⋯ + σ_{k}σ_{1}(α_{r}) = 0
σ_{k}(a_{1})σ_{k}σ_{2}(α_{1}) + σ_{k}(a_{2})σ_{k}σ_{2}(α_{2}) + ⋯ + σ_{k}σ_{1}(α_{r}) = 0
⋮
σ_{k}(a_{1})σ_{k}σ_{n}(α_{1}) + σ_{k}(a_{2})σ_{k}σ_{n}(α_{2}) + ⋯ + σ_{k}σ_{1}(α_{r}) = 0.

Using the fact that σ_{1}, …, σ_{n} form a group (this is where we use the hypothesis!), we see that σ_{k} · σ_{1}, …, σ_{k} · σ_{2}, …, σ_{k} · σ_{n} is a permutation of σ_{1}, …, σ_{k}. Thus we can rearrange the rows to get

σ_{k}(a_{1})σ_{1}(α_{1}) + σ_{k}(a_{2})σ_{1}(α_{2}) + ⋯ + σ_{1}(α_{r}) = 0
σ_{k}(a_{1})σ_{2}(α_{1}) + σ_{k}(a_{2})σ_{2}(α_{2}) + ⋯ + σ_{1}(α_{r}) = 0
⋮
σ_{k}(a_{1})σ_{n}(α_{1}) + σ_{k}(a_{2})σ_{n}(α_{2}) + ⋯ + σ_{1}(α_{r}) = 0.

Subtracting these from (1), we get

(a_{1} − σ_{k}(a_{1}))σ_{1}(α_{1}) + (a_{2} − σ_{k}(a_{2}))σ_{1}(α_{2}) + ⋯ + 0σ_{1}(α_{r}) = 0
(a_{1} − σ_{k}(a_{1}))σ_{2}(α_{1}) + (a_{2} − σ_{k}(a_{2}))σ_{2}(α_{2}) + ⋯ + 0σ_{1}(α_{r}) = 0
⋮
(a_{1} − σ_{k}(a_{1}))σ_{n}(α_{1}) + (a_{2} − σ_{k}(a_{2}))σ_{n}(α_{2}) + ⋯ + 0σ_{1}(α_{r}) = 0.

Thus we have a solution with a fewer number of nonzero entries. Contradiction!

  E-mail address: vakil@math.stanford.edu