Vector space structure on $\mathcal{O}/\mathfrak{m^n}$ Let $k$ be a field and $F\in k[X,Y]$ irreducible such that $F(0,0)=0$. Let $\mathcal{O}$ the local ring of the plane curve $F$ at $P=(0,0)$ and suppose that $P$ is a simple point of $F$. Suppose that the tangent line of $F$ at $P$ is $Y$ and that the maximal ideal of $\mathcal{O}$ is $\mathfrak{m}=(X)$.
Note that for $n\ge 1$, $\mathcal{O}/\mathfrak{m^n}$ has two $\mathcal{O}/\mathfrak{m}$-vector space structures.


*

*Induced from the inclusion of $\mathcal{O}/\mathfrak{m}$: we have an inclusion $$\mathcal{O}/\mathfrak{m}\to
    \mathcal{O}/\mathfrak{m^n}$$ given by $[f]\mapsto [fX^{n-1}]$, so
$\mathcal{O}/\mathfrak{m^n}$ is a $\mathcal{O}/\mathfrak{m}$-vector
space.

*Induced from the $K$-vector space structure of $k[X,Y]$: The quotient ring $\mathcal{O}/\mathfrak{m}$ is isomorphic to $k$
through the map $[f]\mapsto f(0,0)$. In this point of view, we can also consider the natural $\mathcal{O}/\mathfrak{m}$-vector space structure of $\mathcal{O}/\mathfrak{m^n}$ given by
$$\mathcal{O}/\mathfrak{m}\times \mathcal{O}/\mathfrak{m^n}\to \mathcal{O}/\mathfrak{m^n}$$
$$([f],[g])\mapsto [f(0,0)g].$$


Are those two structure the same (or is there any mistake in the above reasoning) ? 
I remarked that since $F=Y+X(\dots)$, we have that
$$[f(X,Y)X^{n-1}g]=[f(0,Y)X^{n-1}g]=[(f(0,0)+Y(\dots))X^{n-1}g]=[f(0,0)X^{n-1}g]$$
in $\mathcal{O}/\mathfrak{m^n}$, but this isn't exactly the second action.
 A: Your first construction does not in fact determine a vector space structure on $\mathcal O/\mathfrak m.$ It simply defines an isomorphism between $\mathcal O/\mathfrak m = k$ and the subspace generated by $X^{n-1}$ in $\mathcal O/\mathfrak m^n,$ which coincides with scalar multiplication on the vector $X^{n-1}$ under the vector space structure on $\mathcal O/\mathfrak m^n$ defined in the second construction.
Let's consider an example. If $F=Y-X^2,$ then $$\mathcal O=(k[X,Y]/F)_{(X,Y)}=(k[X,Y]/(Y-X^2))_{(X,Y)}=k[X]_{(X)},$$ which consists of elements that are quotients $f/g$ with $f,g\in k[X]$ and $g(0)\neq 0.$  So $\mathfrak m=(X)\subseteq \mathcal O.$
It is not hard to see that $\mathcal O/\mathfrak m^n=\operatorname{span}_k\{1, X,X^2,X^3,\ldots,X^{n-1}\}.$ Then the map from the first construction takes an element $f/g\in\mathcal O/\mathfrak m$ to the element $(f(0)/g(0))X^{n-1}.$ Clearly we have said nothing about scalar multiplication on the elements $1,X,X^2,\ldots, X^{n-2}$ here however.
But, if we take the collection of these isomorphisms from $1,\ldots,n-1$, then they together give the scalar multiplication we're after on all of $\mathcal O/\mathfrak m^n$.
