Multiplicity of intersection of $y^2=x^3$ and $x^2=y^3$ at the origin

Question

Those curves intersect at the origin with multiplicity 4, if I did everything correctly. In fact, parametrizing by $t \mapsto (t^2,t^3)$ the first curve and plugging into the second, yields $t^4=t^9$, i.e. $t^4(t^5-1)=0$ and one can see the multiplicity of $t=0$ being $4$.

However, if I try the algebraic geometry approach, by calculating locale Rings, I get a different result and I cannot figure out where I'm making a mistake.

First, $$\mathbb{K}[x,y]/(y^2-x^3) \cong \mathbb{K}[t^2,t^3]$$

Then, $$\mathbb{K}[x,y]/(y^2-x^3,x^2-y^3) \cong$$ $$\mathbb{K}[t^2,t^3]/(t^4-t^9) = \mathbb{K}[t^2,t^3]/(t^4(1-t^5)) $$

Now localizing at the origin, $1-x^5$ becomes a unit and thus the final localized ring is: $$\mathbb{K}[x,y]_P/(y^2-x^3,x^2-y^3) \cong$$ $$\mathbb{K}[t^2,t^3]/(t^4) = \{a+bt^2+ct^3\}$$

But this is a vector space over $\mathbb{K}$ of dimension 3, instead of 4. Where is my error?

Answer 1

In your calculation, the error you made is that in the ring $\mathbb{K}[t^2, t^3]$, $t^5$ is not in the ideal generated by $t^4$. Therefore, the correct basis for the vector space $\mathbb{K}[t^2, t^3] / \langle t^4 \rangle$ should in fact be $\{ 1, t^2, t^3, t^5 \}$.

Answer 2

Just localize $\mathbb K[x,y]$ at the origin. Letting $I=(y^2-x^3,x^2-y^3)$, we have $x^3\equiv xy^3\pmod I$ and so $y^2-x^3\equiv y^2(1-xy)\pmod I$ and, similarly, $x^2-y^3\equiv x^2(1-xy)\pmod I$. Since $(1-xy)$ is invertible in the local ring, we have $\mathbb K[x,y]_0/I \cong \mathbb K[x,y]/(x^2,y^2)$ as a vector space, as desired.