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Hartshorne book Proposition (IV.3. 8) is that

Let $X$ be a curve in $\mathbb{P}^3$, which is not contained in any plane. where, curve means a complete, nonsingular curve over algebraically closed field $k$. Suppose either

(a) every secant of $X$ is a multisecant. or

(b) for any two points $P,Q$ in $X$, the tangent lines $L_P.L_Q$ are coplanar.

Then there is a point $A$ in $\mathbb{P}^3$, which lies on every tangent line of $X$.

In proof, fix a point $R$ in $X$, consider the projection from $R$ , $\phi:X-R \rightarrow \mathbb{P}^2$.

My question is that

(1) If $\phi$ is inseparable, why the tangent line $L_P$ at $X$ passes through $R$ for any point $P$ in $X$?

(2) If $\phi$ is separable, does there exist point $T$ is a nonsingular point of $\phi(X)$ over which $\phi$ is not ramified?

Manolis C. Tsakiris
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Sang Cheol Lee
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2 Answers2

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Here is the proof of (1). If $P =R$, it is trivial. If $P \ne R$, since $\phi$ is inseparable, $\phi$ must be ramified at $P$, this implies the line $\overline {PR}$ must be $L_P$. Check Figure 13 on page 299 of Hartshorne's book.

user41541
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    Inseparability of $\phi$ implies ramification everywhere because it implies ramification at the generic point (see general definition III Ex. 10.3) and the locus of non ramification (=smooth locus) is open. – Gabriel Soranzo Aug 23 '22 at 13:29
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For (2): In the separable case, the union of the singular locus and the ramification locus is a finite set of points.

Manolis C. Tsakiris
  • 161