Bourbaki

Theorem 1

Tag 06JB, permanent. Cite it as [Bourbaki, Tag 06JB].

Algebra, chapter V, § 6. Etale algebras

— Let L be an extension of a field K and A an algebra over K; let H\mathscr{H} be the set of all K-algebra homomorphisms of A into L. Then H\mathscr{H} is a free subset of the vector space HomK(A,L)\mathrm{Hom}_K(A, L) over L.

Let us show by induction on the integer n≥0n \geq 0 that every sequence (u1,…,un)(u_1, \ldots, u_n) of distinct elements of H\mathscr{H} is free. The case n=0n = 0 being trivial, we may henceforth suppose that n≥1n \geq 1; let a1,…,ana_1, \ldots, a_n be elements of L such that

∑i=1nαiui=0.\sum_{i=1}^n \alpha_i u_i = 0.
For x, yx,\ y in A we have
∑i=1n−1αi[ui(x)−un(x)]⋅ui(y)=∑i=1nαiui(xy)−un(x)∑i=1nαiui(y)=0, \sum_{i=1}^{n-1} \alpha_i [u_i(x) - u_n(x)] \cdot u_i(y) = \sum_{i=1}^n \alpha_i u_i(xy) - u_n(x) \sum_{i=1}^n \alpha_i u_i(y) = 0,
whence
∑i=1n−1αi[ui(x)−un(x)]⋅ui=0.\sum_{i=1}^{n-1} \alpha_i [u_i(x) - u_n(x)] \cdot u_i = 0.
By the induction hypothesis, the elements u1,…,un−1u_1, \ldots, u_{n-1} of H\mathscr{H} are linearly independent, whence αi[ui(x)−un(x)]=0\alpha_i [u_i(x) - u_n(x)] = 0 for 1≤i≤n−11 \leq i \leq n-1 and for all xx in A. Since the uiu_i are distinct, this implies that αi=0\alpha_i = 0 for i≠ni \neq n, hence αnun=0\alpha_n u_n = 0 and so αn=αnun(1)=0\alpha_n = \alpha_n u_n(1) = 0 (on denoting by 1 the unit element of A). We have thus shown that a1,…,an−1,αna_1, \ldots, a_{n-1}, \alpha_n are zero, and this proves the theorem.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Cites

Nothing.

Transcribed from the printed edition, 2003, Springer. Not translated.