Proposition 3

— Let A be an etale algebra over K. There exist only a finite number of subalgebras and ideals of A. Moreover, every extension of K which diagonalizes A also diagonalizes every subalgebra and every quotient algebra of A; in particular these algebras are etale.

It suffices to show that an algebra $K^n$ has only a finite number of subalgebras and ideals, and that the subalgebras and quotient algebras of $K^n$ are diagonalizable. We denote by $(\varepsilon_1, \ldots, \varepsilon_n)$ the canonical basis of $K^n$.

Let A be a subalgebra of $K^n$ and let $v_1, \ldots, v_n$ be the restrictions to A of the n projections $K^n \to K$. Since the intersection of the kernels of the $v_i$ is clearly 0, the $v_i$ generate the vector K-space dual to A (II, p. 302, Cor. 1); hence the K-algebra A is diagonalizable (V, p. 29, Prop. 1).

For every subset I of $\{1, 2, \ldots, n\}$ put $\varepsilon_I = \sum_{i \in I} \varepsilon_i$. It is clear that the elements $\varepsilon_I$ are idempotents of $K^n$; we have $\varepsilon_I = 0$ if and only if I is empty, and $\varepsilon_I \varepsilon_J = \varepsilon_{I \cap J}$. By what has been said, every subalgebra A of $K^n$ is diagonalizable; by condition b) of Prop. 1 every subalgebra A of $K^n$ therefore admits a basis $(\varepsilon_{I_1}, \ldots, \varepsilon_{I_p})$, where $(I_1, \ldots, I_p)$ is a partition of $\{1, 2, \ldots, n\}$, and there are only a finite number of such subalgebras.

For every subset I of $(1, 2, \ldots, n)$ let $a_i$ be the vector subspace of $K^n$ having as basis the idempotents $\varepsilon_i$ for $i \in I$; then it is clear that $a_i$ is an ideal of $K^n$; moreover if $J = (1, 2, \ldots, n) - I$, then the residue classes $\overline{\varepsilon}_j$ of $\varepsilon_j$ mod $a_i$, for $j \in J$ form a basis of $K^n / a_i$. We have $\overline{\varepsilon}_j^2 = \overline{\varepsilon}_j$ and $\overline{\varepsilon}_j \overline{\varepsilon}_k = 0$ if $j \neq k$, hence the algebra $K^n / a_I$ is diagonalizable, by Prop. 1 of V, p. 29.

It remains to show that every ideal of $K^n$ is of the form $a_I$. Let I be the set of integers $i$ such that $1 \leq i \leq n$ and $\varepsilon_i \in a$, then $a_i \subset a$. Let $x = x_1 \varepsilon_1 + \cdots + x_n \varepsilon_n$ be an element of $a$ (with $x_1, \ldots, x_n$ in $K$) and let $i$ be in $(1, 2, \ldots, n) - I$. We have $x_i \varepsilon_i = x \varepsilon_i \in a_i$, and $\varepsilon_i \notin a$, whence $x_i = 0$. Thus $x = \sum_{i \in I} x_i \varepsilon_i$ and this shows that $x \in a_i$. We have now shown $a \subset a_I$, whence finally $a = a_i$.