# differentials and tangent space of a fibre

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

• Who's "he"? Where is this from? Aug 2 '20 at 7:50
• @KReiser fixed, thank you. Aug 2 '20 at 7:51

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.
• Should $\Omega_{X/Y, x}$ be $\Omega_{X/k, x}$ in the answer? Aug 31 '20 at 11:07
• No. Think about the case $id:\Bbb A^1_k\to\Bbb A^1_k$. It might be helpful for you to remember what happened in your other question about $\Omega_{X/Y,x}\otimes k(x)$ being $\Omega_{f^{-1}(y)/k,x}$. Aug 31 '20 at 17:47