Unramified morphism of schemes: why is "finite" put in parentheses in the statement of this proposition Here is a proposition in the book Neron Models giving various equivalent definitions of an unramified morphism of schemes.

In condition (e), why is finite put in parentheses?  Normally when you state a theorem and write a property in parentheses, that is done to remind the reader that this property is always the case.  But here, I see no reason why $\kappa(x)$ should be a finite extension of $\kappa(s)$.
In the affine case, if $\phi: A \rightarrow B$ is an algebra map of finite presentation (so $B$ is the quotient of a polynomial ring $A[t_1, ... , t_n]$ by a finite generated ideal), $\mathfrak P$ is a prime of $B$, and $\mathfrak p = \phi^{-1}\mathfrak P$, there is no reason that the induced map
$$A_{\mathfrak p}/\mathfrak p A_{\mathfrak p} \rightarrow B_{\mathfrak P}/\mathfrak P B_{\mathfrak P}$$
be a finite extension of fields, or even a finitely generated extension of fields.
 A: We'll show that any of the conditions that are not (e) imply that the given field extension is finite. We assume that the equivalence of (a)-(d) has already been proven.
Let $k=\kappa(s)$. By (d) and the fact that locally of finite presentation is stable under base change, we may assume $S=\operatorname{Spec} k$. By (c) and the fact that for a $k$-scheme $X$ the local dimension of $X\times_kX$ at $(x,x)$ is twice the local dimension of $X$ at $x$, we know that the local dimension of $X$ at $x$ must be zero, or in other words that $x$ is an isolated point of $X$. By $X\to S$ locally of finite presentation, we can pick an affine open neighborhood of $x$ of the form $\operatorname{Spec} k[x_1,\cdots,x_n]/(f_1,\cdots,f_m)$, and since $x$ is an isolated point of $X$, we know that in fact $x$ is the spectrum of a field extension of $k$ of the form $k[y_1,\cdots,y_p]/(g_1,\cdots,g_q)$. In particular, this implies that $\kappa(x)$ is a finite extension of $k=\kappa(s)$.
On the other hand, one does need the finite extension property when proving the implication that (e) implies the other conditions - if we do not require it, the extension of fields $k\subset k(x)$ clearly has $\Omega_{k(x)/k}\neq 0$.
