Dedekind independence lemma - Imported Knowlpedia Content

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

\sigma_1,\dots,\sigma_n : L \hookrightarrow \Omega

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 $a_1=\cdots=a_n=0$ .

Equivalent characterizations

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

Remarks

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

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 $a=b=0$ .

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 $L$ .

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.