Zariski cotangent space of cusp at origin and at generic point

Question

I want to explicitly compute the Zariski cotangent spaces of the cusp $X=Z(x^3-y^2)\subset \mathbb{C}^2$. I can work with the definitions, but I have no idea how to actually compute this (this is a problem I tend to have...). So by definition we consider the ring $$A=\mathbb{C}[x,y]/\langle x^3-y^2\rangle$$ And then some maximal ideal $\mathfrak{n}=\langle \overline{x-1},\overline{y-1}\rangle\subset A$. Then we compute $$A_{\mathfrak{n}}=\left( \mathbb{C}[x,y]/\langle x^3-y^2\rangle \right)_{\langle \overline{x-1},\overline{y-1}\rangle }$$ and consider its maximal ideal $\mathfrak{m}=\mathfrak{n}A_{\mathfrak{n}}$, and finally we compute $\mathfrak{m}/\mathfrak{m}^2$. Frankly I have no idea how to go about this. I know that $$A\cong \mathbb{C}[t^2,t^3]=A'$$ and it seems that under this isomorphism $\mathfrak{n}$ would be send to $\mathfrak{n}'=\langle t^2-1,t^3-1\rangle$. Now I still don't see how to compute $A'_{\mathfrak{n}'}$, but this might just be because I am bad at this...

Any help would be appreciated.

Answer 1

At a smooth point on a curve, the maximal ideal of the local ring is principal. (See here for more equivalent conditions.) Here is a quick and dirty way of seeing this for your example. \begin{align*} y^2 = x^3 \implies y^2 - 1 = x^3 - 1 \implies (y-1)(y+1) = (x-1)(x^2+x+1) \end{align*} Since $y+1$ and $x^2+x+1$ are both units in $A_\mathfrak{n}$ (they do not vanish at the point $(1,1)$), then $x-1$ and $y-1$ are associate. Then $$ (x-1,y-1) = (x-1) = (y-1) $$ in $A_\mathfrak{n}$, so $$ \frac{\mathfrak{m}}{\mathfrak{m}^2} \cong \frac{(x-1)}{((x-1)^2, (x-1)(y-1), (y-1)^2)} \cong \mathbb{C} $$ is a $1$-dimensional vector space with basis $x-1$ (or $y-1$). This corresponds to our intuition: at a smooth point on the curve there is a unique tangent line, so the tangent and cotangent spaces should be $1$-dimensional.

A more general method is the following. Consider the Taylor expansion of $f = y^2 - x^3$ at the point $(a,b)$: $$ f = -3a^2(x-a) + 2b (y-b) - 3a (x-a)^2 + (y-b)^2 - (x-a)^3 \, . $$ Since $f = 0$ in the quotient, this allows us to relate $x-a$ and $y-b$ for any point $(a,b) \neq (0,0)$. You could try this for your example $(a,b) = (1,1)$, or see the linked post for more detail.

You can probably also see why this sort of argument breaks down for $(a,b) = (0,0)$. There we have a double tangent line $y = 0$, and correspondingly, $f_x(a,b) = -3a^2$ and $f_y(a,b) = 2b$ both vanish. Then \begin{align*} \frac{\mathfrak{m}}{\mathfrak{m}^2} \cong \frac{(x,y)}{(x^2, xy, y^2)} \end{align*} which is $2$-dimensional, since $x,y$ is a basis: no linear combination $c_1 x + c_2 y$ lies in $(x^2, xy, y^2)$.

(As an aside, beware of using the term "generic" in its intuitive sense. The generic point of $\operatorname{Spec}(A)$ is the zero ideal $(0)$.)