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Let $f:X \to Y$ be a morphism of finite type of irreducible schemes over an algebraically closed field of characteristic $0$. Assume that $Y$ is non-singular. Let $x \in X$ be a closed point and $T_xf:T_xX \to T_{f(x)}Y$ the induced linear map. Then, Proposition III.$10.6$ of Hartshorne's "Algebraic geometry" implies that for a general $x \in X$, $\dim \mathrm{Im}(f) \le \mathrm{rk}(T_xf)$ (with notations same as in Hartshorne).

Suppose now that we consider the same setup except that the algebraically closed field in the beginning is of positive characteristic. My question: Is there any known condition on $f$, other than smoothness, under which the above inequality still holds true? Any reference/hint will be most welcome.

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    $\begingroup$ This is not correct and am sure not what Hartshorne says. As I said earlier, have you tried the map $\mathbb{A}^1$ to itself, given by $x\mapsto x^2$ at the origin? $\endgroup$
    – Mohan
    Oct 4, 2015 at 0:48
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    $\begingroup$ @Mohan : May be I am wrong, but isn't in this case $T_0 f:T_0 \mathbb{A}^1 \to T_0 \mathbb{A}^1$ takes $\frac{\partial}{\partial X}$ to $2X\frac{\partial }{\partial X}$, where $X$ is the local parameter around $0$ of $\mathbb{A}^1$. Now, $X\frac{\partial }{\partial X}$ is non-zero on \textit{any} open neighbourhood of the origin, hence the rank of the linear tranformation $T_0 f$ is at least $1$. Am I missing something? $\endgroup$
    – Kali
    Oct 4, 2015 at 11:30
  • $\begingroup$ Does Hartshorne really use that notation for the stalk of the tangent shead at a point? It is mightily confusing. $\endgroup$ Oct 4, 2015 at 12:16
  • $\begingroup$ @potentiallydense Sorry. You are right. He does not. I agree that the notation is a bit confusing. The current notation is from a text by Le Potier. The result in Hartshorne states that for a 'general' $x \in X$, the above inequality holds (under the notation that $T_x X$ and $T_{f(x)} Y$ are fibers and not just stalks). I will try to edit the question. $\endgroup$
    – Kali
    Oct 4, 2015 at 12:44
  • $\begingroup$ @Mohan : I have edited the question, hope I have got it right this time. $\endgroup$
    – Kali
    Oct 4, 2015 at 13:17

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In characteristic $p$, the Frobenius map $\mathbb A^1\to\mathbb A^1$ given by $x\mapsto x^p$ has zero derivative everywhere, but the map itself is surjective (so image has dimension one).

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  • $\begingroup$ Is there any known criterion on $f$, other than smooth morphism, under which the inequality does hold true? $\endgroup$
    – Kali
    Oct 4, 2015 at 14:09

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