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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

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  • $\begingroup$ Who's "he"? Where is this from? $\endgroup$
    – KReiser
    Aug 2 '20 at 7:50
  • $\begingroup$ @KReiser fixed, thank you. $\endgroup$
    – Johnny T.
    Aug 2 '20 at 7:51
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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.

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  • $\begingroup$ Thank you for this. Which assumption implies that the tangent space vanish? $\endgroup$
    – Johnny T.
    Aug 3 '20 at 7:37
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    $\begingroup$ We note you've left out "finite type" as an adjective of varieties. The fiber is finite type over a field (by base change) and thus noetherian. The fiber is zero-dimensional, as any scheme which is finite type over a field and positive-dimensional is has infinitely many points. Since 0-dimensional noetherian schemes are discrete, this means that every point is the spectrum of an 0-dimensional noetherian ring, which is equivalent to the ring being Artinian. Artinian local domains are fields. So it's maximal ideal is 0, and the cotangent space is 0, and thus the tangent space is zero. $\endgroup$
    – KReiser
    Aug 3 '20 at 8:03
  • $\begingroup$ Should $\Omega_{X/Y, x}$ be $\Omega_{X/k, x}$ in the answer? $\endgroup$
    – Johnny T.
    Aug 31 '20 at 11:07
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    $\begingroup$ 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 $\Omega_{f^{-1}(y)/k,x}$. $\endgroup$
    – KReiser
    Aug 31 '20 at 17:47

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