I have been reading the notes here. There, a finite locally free morphism of schemes is defined as as a morphism of schemes $f: X \rightarrow Y$ which is finite and for which the sheaf $f_{*} \mathcal{O}_{X}$ is locally free as an $\mathcal{O}_{Y}$-module. A finite etale morphism is then defined as a morphism of schemes $f: X \rightarrow Y$ which is finite locally free, and for which the fiber over any point $q \in Y$ is an etale $\kappa(q)$-algebra.
I am struggling to understand what the part about the fibers being etale algebras adds at all. It seems like every finite locally free morphism would trivially satisfy this.
Take $f: X \rightarrow Y$ to be a finite locally free morphism of schemes. For any $q \in Y$, choose an affine $\operatorname{spec}A$ around $q$ small enough so that $f_{*} \mathcal{O}_{X}$ is free, say of degree $d$. Then since $f$ is finite, it is in particular affine, so the morphism looks locally like $\operatorname{spec}(A^{\oplus d}) \rightarrow \operatorname{spec}A$. Then the fiber over $q$ is just $\operatorname{spec}(\kappa(q)^{\oplus d})$. This seems to trivially be an etale $\kappa(q)$-algebra, since tensoring with some algebraic closure $\Omega$ would give $d$ copies of $\Omega$.
Where is the flaw in my reasoning? Or is it just the case that every finite locally free morphism is etale?
Further to this, the same notes say that for a finite etale morphism, the degree of the morphism at a point $q \in Y$ is the same as the cardinality of the fiber over $q$. But again the above reasoning seems to show this for any finite locally free morphism. Where does etale come into it at all?
Best Answer
Here is an example of a finite locally free morphism which is not etale: take spec of the natural inclusion $\Bbb F_2(t^2)\subset \Bbb F_2(t)$. This fails to be etale because it's a non-separable field extension.
The flaw in your reasoning is that your identification of $f_*\mathcal{O}_X$ as $A^d$ locally is only as a module - it needs to be as a ring to say anything interesting. (You should also note that even in the case of $\Bbb R\subset\Bbb C$, your idea is wrong - this is a point mapping to a point, while your reasoning would have two points mapping to one point.)