The setup I have is as follows: Let $f: X \to Y$ be a morphism of non-singular $n$-dimensional varieties (separated reduced irreducible scheme of finite type over $k$) over $k$ an algebraically closed field. For all closed points $y \in Y$, the fibre $f^{-1}(y)$ is a finite set of reduced points. Assume $Y = \operatorname{Spec}B$
Let $y = f(x)$ for some closed point $y \in Y$. In the proof I'm going through we have deduced that $f^{-1}(y)$ is locally $\operatorname{Spec}k[X_1, ,,,, X_N]/( \bar{f}_1,..., \bar{f}_N )$. Up to here I understand, but I am struggling to see the next two lines and I'd appreciate explanations. He states (This is from Mumford's red book Theorem 4 III.5): By assumption, this fibre has no tangent space at all at $x$. Therefore, the differentials $d \bar{f}_i$ must be independent at $x$.
Thank you!
edit. I forgot to mention fintie type over $k$ and this has been added
On the one hand, the cotangent space is the dual of the cotangent space. On the other, it's $\Omega_{X/Y,x}\otimes k(x)$, and $\Omega_{X/Y,x}$ has a presentation (near $x$) as a free module on $dX_i$ with relations coming from $df_j$. The assumptions imply that the tangent space vanishes, so it's dual must vanish as well, which is equivalent to the $df_j$ being independent at $x$ - if they weren't, the module wouldn't vanish, and $\Omega_{X/Y,x}\otimes k(x)$ would give a nonzero module in contradiction to our assumptions.