Local ring of a scheme whose reduction is the affine line.

Question

Let $k$ be a field and $A=k[X,Y]/\langle Y^2,XY\rangle = k[x,y]$ the algebra corresponding to the well-known non primary ideal $I=\langle Y^2,XY\rangle \subset A$ with embedded associated ideal $M=\langle X,Y\rangle$.
If $S=\operatorname {Spec}A$ what is the local ring $\mathcal O_{S,m}=A_m$, where $m=M/I\in S$ is the closed point corresponding to the maximal ideal $M/I\subset A=k[X,Y]/I$?
Is there a reference for this result?

Answer 1

We have $$A_m=k[x]_{\langle x\rangle}\oplus k\cdot y$$ This is not so astonishing once you realize that for example $\frac{y}{P(x,y)}=\frac {1}{P(0,0)}\cdot \frac {y}{1} \in A_M$ for any $P(x,y)\in k[x,y]$ satisfying $P(0,0)\neq 0$.
Edit
I have asked several algebraic geometers the same question and they didn't know the answer offhand, or even gave false answers. [Try asking too for the sake of sociological experiment :-)]
It took me some time to arrive at the rather counter-intuitive answer above since it is not so clear a priori that, say, $\frac{y}{1+x^3-y^2}= \frac {y}{1} \in A_m$ (even if it is trivial to prove once you are shown the result).
The indicated direct sum decomposition (as $k$-vector spaces)shows that the germ at $m$ of a function on $S$ is obtained by restricting a function $P(X,Y)$ on $\mathbb A^2_k$ to the line $\mathbb A^1_k=V(Y)$ AND remembering the value of the partial derivative $\frac{\partial P}{\partial Y}(0,0)$ at $M$.