Dedekind independence lemma

Let $K \subseteq L$ be a field extension , and let $\Omega$ be a field containing (identified copies of) the images of $L$ under the embeddings below. If

are distinct field embeddings that fix $K$, then they are linearly independent over $\Omega$ when viewed as $\Omega$-valued functions on $L$. Concretely:

If $a_1,\dots,a_n \in \Omega$ satisfy

$$> a_1\sigma_1(x)+\cdots+a_n\sigma_n(x)=0 \quad \text{for all } x\in L, >$$

then $a_1=\cdots=a_n=0$.

Equivalently, the $\Omega$-vector space $\Omega^L$ of all functions $L\to\Omega$ contains $\{\sigma_1,\dots,\sigma_n\}$ as a linearly independent set.

This lemma is frequently applied with $\Omega=L$ and $\sigma_i$ ranging over a subgroup of the Galois group (or more generally the field automorphism group) of $L$.

Examples

1. Two embeddings of a quadratic extension.
   In $L=\mathbb{Q}(\sqrt2)\subset \mathbb{C}$, there are two $\mathbb{Q}$-embeddings into $\mathbb{C}$: the identity $\mathrm{id}$ and conjugation $\tau(\sqrt2)=-\sqrt2$. If $a\,\mathrm{id}+b\,\tau=0$ as functions, then evaluating at $1$ gives $a+b=0$, and at $\sqrt2$ gives $a\sqrt2-b\sqrt2=0$, hence $a=b=0$.

2. Cyclotomic embeddings.
   For $L=\mathbb{Q}(\zeta_n)$ with $\zeta_n$ a primitive root of unity , the distinct embeddings $\sigma_k(\zeta_n)=\zeta_n^k$ (for $\gcd(k,n)=1$) are linearly independent as $\mathbb{C}$-valued functions on $L$.

3. Finite-field Frobenius powers.
   In a finite field $L=\mathbb{F}_{p^m}$, the maps $x\mapsto x^{p^i}$ are distinct automorphisms for $0\le i<m$ (see cyclic Galois group of a finite field ). Dedekind independence implies no nontrivial $\mathbb{F}_{p^m}$-linear combination of these maps vanishes identically.