Fix some algebraically closed field $k$. Let $F$ be an irreducible affine-plane curve, let $P=(0,0)$ and $m>1$ be the multiplicity of $P$ in $F$. Let $\Gamma(F)=k[X,Y]/(F)$ (i.e. the coordinate ring of $F$). Let $$\mathcal{O}_P(F)=\{\frac{f}{g}\mid f,g\in \Gamma(F)\text{ such that }g(P)\neq 0\},$$
(i.e. rational functions defined at $P$). Let $\mathfrak{m}$ be the ring of non-units in $\mathcal{O}_P(F)$. Notice that $\mathfrak{m}=(x,y)_{\mathcal{O}_P(F)}$ (i.e. the ideal generated by $x$ and $y$ in $\mathcal{O}_P(F)$).
For $G\in k[X,Y]$, denote by $g$ its residue in $\Gamma(F)$ (so upper case letter $\rightarrow$ lowercase letter);
and for $g\in\mathfrak{m}$ denote its residue in $\frac{\mathfrak{m}}{\mathfrak{m}^2}$ by $\overline{g}$.
Show that the mapping
$$
\Psi:\{\text{forms of degree }1\}\longrightarrow \frac{\mathfrak{m}}{\mathfrak{m}^2}
$$
given by
$$
aX+bY\longmapsto \overline{ax+by}
$$
is an isomorphism of vector spaces.
This is exercise 3.24.a in Fulton's book.
I have the following proposed solution (we will assume a couple of facts):
We know that (by 3.13 and 2.35):
$$
\text{dim}_k\frac{\mathfrak{m}}{\mathfrak{m}^2}=2=\text{dim}_k\{\text{forms of degree 1}\}.
$$
Therefore, it is enough to show that the mapping is surjective to conclude it's an isomorphism.
Recall $\mathfrak{m}=(x,y)_{\mathcal{O}_P(F)}$ and $\mathfrak{m}^2=(x,y)_{\mathcal{O}_P(F)}^2$. (The subindex here is denoting where the ideal is being calculated).
Therefore,
$$
\mathfrak{m}=(x)_{\mathcal{O}_P(F)}+(y)_{\mathcal{O}_P(F)}+\mathfrak{m}^2
$$
and thus
$$
\tag{$\ast$}\frac{\mathfrak{m}}{\mathfrak{m}^2}=(\overline{x})_{O_P(F)}+(\overline{y})_{O_P(F)}.
$$
As $\{\text{forms of degree 1}\}=(X)+(Y)$, it follows that $\Psi$ is surjective.
However, I now think that the conclusion does not follow from $(\ast)$. I need to show that $\frac{\mathfrak{m}}{\mathfrak{m}^2}=\text{span}_k(\overline{x},\overline{y})$ and I think $(\ast)$ just shows the weaker fact $\frac{\mathfrak{m}}{\mathfrak{m}^2}=\text{span}_{\mathcal{O}_P(F)}(\overline{x},\overline{y})$. I'm struggling to fix this.
I'd very much appreciate any help, either filling this hole or with a new solution. Thanks!
EDIT: Please beware: "Fulton uses the word "form" to refer to any homogeneous polynomial, whereas most geometric contexts reserve the word "form" for differential forms. As it happens, in this case I'm not sure the difference is really substantive". (As helpfully commented below by Will R).
Best Answer
What you call $\mathcal{O}_P$-span and $k$-span coincide. I'll try to explain the general case. Let $A$ be a commutative ring with unity, $M$ an $A$-module and $\mathfrak{a}\subset A$ an ideal. Then $M/\mathfrak{a}M$ inherits as structure of $A$-module by the rule $x\cdot (m + \mathfrak{a}M) = xm + \mathfrak{a}M$ where $x\in A$ and $m \in M$. This is just its structure as $A/\mathfrak{a}$ module, since $(x + \mathfrak{a})\cdot(m + \mathfrak{a}M) = xm + \mathfrak{a}M$; that is, the action of $x$ on $M/\mathfrak{a}M$ is completely determined by the action of $x + \mathfrak{a}$ on $M/\mathfrak{a}M$. Particularly, $A$-span and $A/\mathfrak{a}$-span coincide for $M/\mathfrak{a}M$. Apply this to $A = \mathcal{O}_P$, $\mathfrak{a} = \mathfrak{m}$, $M = \mathfrak{m}$, $A/\mathfrak{a} \cong k$ and $M/\mathfrak{a}M = \mathfrak{m}/\mathfrak{m}^2$.