isolating locally the branches of a curve

Question

Let $k$ be an algebraically closed field and $C$ an irreducible curve in $\mathbb{A}^2$ passing through the point $P=(0,0)$, corresponding to the polynomial $F = F_r + F_{r+1} + \dots + F_n$, where each term $F_i$ is homogeneous of degree $i$. After multiplication of $F$ with a nonzero constant (which does not change $V(F)$), we can assume that $F_r = (Y-a_1 X) \cdots (Y-a_r X)$, where the $a_i$ are not necessarily distinct. If $r=1$, we have that $\nabla F|_P \neq 0$. In fact $\nabla F|_P = (1, -a_1)$ and by calculus this is the tangent line of $F$ at $P$. However, if $r>1$, then $\nabla F|_P=0$. Nevertheless, the "tangents" of $F$ at $P$ are still defined to be the lines $Y-a_i X =0$. What i find surprising is that these lines are actually tangents of $F$ at $P$. They are the tangents of each "branch" of $F$ through $P$. Why is this true? How can this be seen rigorously? For example, if we plot the curve $F=Y^2-X^2-X^3$, we do see that it has two tangents at $P$ given by $Y+X=0$ and $Y-X=0$. This could be "deduced" by examining the $Y^2-X^2$ term of $F$, however there is a theorem to be proved, which would say that if we factor the $F_r$ term we get the actual tangents. The issue here is that it does not seem that we have "access" to each of the branches of $F$ separately: if that was the case we could link the factors of the $F_r$ term to these branches. How do we "access" these branches rigorously? For example, starting with the curve $F=Y^2-X^2-X^3$, how can we "isolate" locally its branches at $P$ and then we go and look at their gradients? How about constructing a curve that looks exactly as one of the branches at $P$ and "far" from $P$ behaves arbitrarily?

Answer 1

The local ring at a simple node is analytically isomorphic to the local ring at two intersecting lines. To see this, pass to the completion.

Concretely: In the power series ring $k[[X,Y]]$, $F=Y^2 - X^2 - X^3$ is a unit times $Y^2 - X^2$, so the ring $k[[X,Y]]/(F)$ is isomorphic to $k[[X,Y]]/(XY)$ (with a change of coordinates), and the branches correspond precisely to the minimal prime ideals.

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Another way to think of a "branch" of a node $p\in C$ is to consider the projection $\pi: \tilde{C}\to C$ from the desingularized curve $\tilde{C}$, and an element of the fiber $\pi^{-1}(p)$.

This can be seen algebraically by looking at the integral closure of the coordinate ring. In the case $R = k[X,Y]/(Y^2 - X^2 - X^3)$, there are two maximal ideals of the integral closure of $R$ whose restriction to $R$ is $(X,Y)$, hence two branches.

To fill in some details: $k[X,Y]/(Y^2 - X^2 - X^3)$ is isomorphic to $k[T^2-1,T(T^2-1)]$, and its integral closure is $k[T]$. So the maximal ideals lying above $(X,Y)=(T^2-1,T(T^2-1))$ are exactly $(T-1)$ and $(T+1)$.

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To see that the two ideas are equivalent: A paper of Nagasawa, "Some Remarks on One-Dimensional Local Domains", cites in Nagata the following result:

If $R$ is a $1$-dimensional local domain, there exists a one-to-one correspondence between maximal ideals of $\overline{R}$ and minimal prime ideals of $R^\ast$.

This implies that the points lying above a node in the normalization correspond with the irreducible components of the completed (or "analytified") node. This makes either a good definition of "branch", though it's probably worth comparing both with the analytic topology when $k=\mathbb{C}$.