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A proof from Janos Kollar's Lectures on Resolution of Singularities Kollar (p 37) works as follows:

Theorem 1.58 (M. Noether, 1871). Let $k$ be an algebraically closed field and $C \subset \mathbb{P}^2$ an irreducible plane curve. Then the algorithm (1.57) eventually stops with a curve $C_m \subset \mathbb{P}^2$, which has only ordinary multiple points (1.54).

We give two proofs. The first one assumes that we already know embedded resolution as in (1.52). The second, following Noether’s original approach, gives another proof of embedded resolution.

Proof using resolution. Pick any point $p \in C \subset \mathbb{P}^2$, and let $\pi : C \to \mathbb{P}^1$ be the projection from $p$. In characteristic zero or if the characteristic does not divide $\operatorname{deg} C − \operatorname{mult}_p C$, the projection $\pi$ is separable.

(This fact we can look up on page 15 but also simply assumed as black box: i.e. from now we assume that $\pi$ is separable: field extension $K(C)/K(\overline{\pi(C)}$ is separable.)

Now the part I not understand:

Thus we can take two general lines through $p$ such that they are not contained in the tangent cone of $C$ at $p$ and they have only transverse intersections with $C$ at other points. [...]

Question: Why separability of $\pi$ is neccessary to be able to choose these two lines with described properties and what fails if $\pi$ would be not separable?

We know that seprableness of $\pi$ implies that a general fiber of $\pi$ has $\operatorname{deg} \pi$ points. Possibly, if $\operatorname{deg} \pi >1$ this might be give rise for two "candidates" for the two lines trough $p$ with desired properties, but it can also happen that $\operatorname{deg} \pi =1$ & $\pi$ separable. Thus I unfortunately not see the crucialness of separateness of $\pi$ for the justification of the existence of the the lines.

#Edit: As GTA pointed out at first glance it's not clear where $p$ is mapped by the considered projection. On page 12 I found Kollar's explanation on this issue:

(2)"Assume that $p \in C \subset \mathbb{P}^2$ is a singular point, where $m$ smooth branches of the curve pass through with different tangent directions. Projecting $C$ from $p$ separates these tangent directions, and the singular point $p$ is replaced by $m$ points with only one local branch through each of them."

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    $\begingroup$ Where does p go through this projection from p map? $\endgroup$ – GTA Apr 10 at 15:42
  • $\begingroup$ @GTA:on page 12 Kollar says: ..."Assume that $p \in C \subset \mathbb{P}^2$ is a singular point, where $m$ smooth branches of the curve pass through with different tangent directions. Projecting $C$ from $p$ separates these tangent directions, and the singular point $p$ is replaced by $m$ points with only one local branch through each of them." I think this is what he also implicitly means in the projection map from above. $\endgroup$ – Tim Grosskreutz Apr 10 at 15:53
  • $\begingroup$ You are right, that's an important point. I will add it in the question. $\endgroup$ – Tim Grosskreutz Apr 10 at 15:56
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    $\begingroup$ Separability is needed for the intersections of the line with $C$ to be transverse. A general fibre of the projection will consist of distinct reduced points iff the map is separable. $\endgroup$ – naf Apr 11 at 4:58
  • $\begingroup$ @ulrich: Yes, I think I understand. That's a consequence of Bézout's theorem, right? The general fiber coinsides with intersection of a line through $p$ with $C$. By Bezout & counting at every point of the intersection with such general line must be transversal. And separable implies that we have enough such lines. Thank you! $\endgroup$ – Tim Grosskreutz Apr 11 at 13:26

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