Dedekind Independence Lemma

Distinct field homomorphisms are linearly independent as functions.
Dedekind Independence Lemma

Lemma (Dedekind independence).
Let LL be a field and σ1,,σm:LΩ\sigma_1,\dots,\sigma_m: L\to \Omega be distinct field homomorphisms into a commutative ring (or field) Ω\Omega. If

a1σ1(x)++amσm(x)=0for all xL a_1\sigma_1(x)+\cdots+a_m\sigma_m(x)=0 \quad \text{for all }x\in L

with coefficients aiΩa_i\in \Omega, then a1==am=0a_1=\cdots=a_m=0.
Equivalently, {σ1,,σm}\{\sigma_1,\dots,\sigma_m\} is linearly independent over Ω\Omega inside the Ω\Omega-module of functions LΩL\to\Omega.

This is a key input for .

Examples.

  1. Over L=CL=\mathbb{C}, the maps id\mathrm{id} and complex conjugation x\overline{\phantom{x}} are distinct, hence linearly independent as C\mathbb{C}-valued functions on C\mathbb{C}.
  2. If L/KL/K is and G=Gal(L/K)G=\mathrm{Gal}(L/K), then the distinct σG\sigma\in G are linearly independent. In particular, a relation σGaσσ=0\sum_{\sigma\in G}a_\sigma\sigma=0 forces all aσ=0a_\sigma=0.
  3. For L=FpnL=\mathbb{F}_{p^n}, the maps xxpix\mapsto x^{p^i} (0i<n0\le i<n) are distinct automorphisms, hence independent.

Related. , .