I am reading Bourbaki's Algebra, Chapter V.
Bourbaki makes the following definition of Etale Algebra (page V.28, definition 1):
Definition. Let $A$ be an algebra over $K$; then $A$ is said to be diagonalizable if there exists an integer $n\geq0$ such that $A$ is isomorphic to the product algebra $K^n$. We say that $A$ is diagonalized by an extension $L$ of $K$ if the algebra $A_{(L)}$ over $L$ derived from $A$ by extension of scalars is diagonalizable. We shall say that $A$ is etale if there exists an extension of $K$ which diagonalizes $A$.
After shortly reminding the definition of product algebra, he writes:
Every etale algebra over $K$ is commutative and of finite degree over $K$.
but I don't see the reason why an etale algebra has to be of finite degree.
So far, I understood:
- $A$ is diagonalizable if $A\cong K^n$. $A$ is of finite degree in this case, of course.
- $A$ is etale if $A_{(L)}\cong L^n$ for some $n$, or more explicitly, $A\otimes_KL\cong L^n$. (I believe my understanding is correct so far - the first equivalent definition of Wikipedia)
So an etale algebra $A$ is commutative, since $L$ (and therefore $L^n$) is commutative.
Also, $A$ is of finite degree over $L$, but why the degree of $A$ over $K$ is finite? What happens if we take, for example, $K=\mathbb{Q}$ and $L=\mathbb{C}$?
It actually does not make sense to say that $A$ has finite degree over $L$, as $A$ is not an $L$-algebra, but a $K$-algebra.
On the other hand, you can say that $A_L\simeq L^n$ has finite degree over $L$, and use the fact that $\dim_K(A) = \dim_L(A_L)$ (scalar extension preserves the dimension).