Intuition about formal brances of a curve at a point Consider an algebraic surface $X$ and a curve $Y\subset X$. Here $X$ is a $K$-scheme integral of finite type of dimension $2$ and $Y$ is a closed subscheme of dimension $1$. 
Fix a closed  point $x\in Y\subset X$ and let $\mathfrak p\subset\mathcal O_{X,x}$ be the prime ideal corresponding to $Y$; then a formal branch of $Y$ at $x$ is an element of $f^{-1}(\mathfrak p)$ where:
$$f:\text{Spec }\widehat{\mathcal O_{X,x}}\to\text{Spec }\mathcal O_{X,x}$$
is the canonical morphism induced by the completion.
Now suppose that $x\in Y$ is a singular point. I understand that:


*

*if $Y$ has a simple node at $x$, then $Y$ has $2$ formal branches at $x$

*if $Y$ has a triple node at $x$, then $Y$ has $3$ formal branches at $x$

*...etc (it is enough to count the sub-curves of $Y$ passing through $x$ in a small neighbourhood)


But what happens if $x$ is a cusp point? Do we have one single branch or two branches? This seems to be a limiting case.
 A: Just try and calculate. The easiest example of a cusp is $\text{Spec }k[t^2,t^3]\to\text{Spec }k[x,y]$, where $x\mapsto t^2$ and $y\mapsto t^3$. The completion of the plane at the origin is $\text{Spec }k[[x,y]]$, formal power series in two variables. The fiber of the cusp over the completion is $k[[x,y]]\otimes_{k[x,y]}k[t^2,t^3]=k[[t^2,t^3]]$, formal power series in one variable without linear term. If $p(t)$ is a power series without linear term such that $p(0)\neq 0$, it is easy to check directly that the inverse of $p$ in $k[[t]]$ has no linear term, too. This implies that $k[[t^2,t^3]]$ is local with maximal ideal $(t^2,t^3)$, and hence there is only one formal branch by your definition.
Edit: sorry, I realised my answer was misleading. The answer is still 1, but the reason is quite different. You have to look at minimal ideals of $k[[t^2,t^3]]$, not the maximal ones, to get the $f^{-1}(p)$ you are talking about. The idea is that you want to count the "local irreducible components". Taking simply the local ring is not enough, you use the completion to look "closer". Still, $k[[t^2,t^3]]$ is a domain, hence it has only one minimal ideal. The geometric intuition is easy: locally, the cusp has only "one piece", not two like the nodal curve.
