Let $R=k[x,y]/(y^2-x^2-x^3)$ and $I=(x,y)\cdot R \subset R$. I would like to show that $$ \bigoplus_{i=0}^{\infty} I^i\,/\,I^{i+1} \cong \,k[x,y]\,/\,(x^2-y^2). $$ Could you please help me?
Remark: Geometrically speaking X=$spec(R)$ is the simple nodal curve with node at the origin and $I$ is the maximal ideal of the origin in $X$. Spec of the above ring is by definition the tangent cone of $X$ at the origin, which should geometrically be represented by the lines $x=y$ and $x=-y$ in the plane.
EDIT:
Possible proof that seems plausible, but could probably be formally more correct:
Set $S=\oplus_iI^i/I^{i+1}$ and $J=(y^2-x^2-x^3)$. Consider the map $\varphi:k[x,y]\to S$ defined by $$ x \mapsto (0,[x]_2,0,\dots) $$ $$ x^2 \mapsto (0,0,[x^2]_3,0,\dots) $$ $$ \vdots $$ $$ x^n \mapsto (0,\dots,0,[x^n]_{n+1},0,\dots) $$
where $$[a]_n = (a \mod J) \mod I^n$$ and extend the definition in the same way for $y$. I'm not happy with this definition (how to make it more rigorous?), but I want it to be a ring homomorphism, i.e. the behavior on a generic element should be $$ x^5 +y^5 + y^4 + xy^2 \mapsto (0,0,0,[xy^2]_4, [y^4]_5, [x^5 + y^5]_6, 0, \dots). $$ It's then easy to see that $\varphi$ is surjective and $\ker(\varphi)=(x^2-y^2)$. What do you think? Is this correct? Can it be made more elegant?
You only have to say that you define $\varphi$ by mapping $x$ and $y$ to their respective residue classes in $I/I^2$. This is possible because both $x$ and $y$ are in $I$. The universal property of the polynomial ring then takes care of the rest (a map from the polynomial ring to any $k$-algebra exists and is uniquely determined by the images of the indeterminantes): Note that $[x]_2^k=[x^k]_{k+1}$ in the ring $\bigoplus_d I^d/I^{d+1}$.