Let $V = V(Y^2 - X^3 - X) \subset \mathbb{A}^2$ be a curve, how can I show $\dim_{\mathbb{C}} \mathcal{M} / \mathcal{M}^2 = 1$

Question

Let $V = V(Y^2 - X^3 - X) \subset \mathbb{A}^2$ be a curve, and let $P = (0, 0)$. let $\mathcal{M} = \mathcal{M}_P(V) \subset \mathcal{O}_P(V)$ be the maximal ideal of the local ring of $V$ at $P$. Prove that $\dim_{\mathbb{C}} \mathcal{M} / \mathcal{M}^2 = 1$.

I just started to learn algebraic geometry. I tried to read some relevant answers for any hints, like: Dimension of $m/m^2$, where $m$ is the maximal ideal of $\mathcal{O}_{X\times Y,(x,y)}$, Dimension of $\mathfrak{m}^k/\mathfrak{m}^{k+1}$?. But they are a little bit advanced to me.

I am thinking that by having an explicit curve, we might be able to calculate the dimension in a more direct way. However, I have no clue where to start.

Answer 1

This is my attempt after receiving hints from @Jyrki Lahtonen.

First, we recall the definition:

$$\mathcal{O}_P(V) = \left\{ f \in K(V) \mid f \text{ is defined at } P \right\}$$ $$\mathcal{M} = \left\{ f \in K(V) \mid f = \frac{a}{b} \text{ for } a(P) = 0, b(P) \neq 0 \right\}$$

We also know that $\mathcal{M}_P(V) = I(P)\mathcal{O}_P(V)$.

We can see that for all polynomial $f \in \Gamma(W)$, $f(0, 0) = 0 \Leftrightarrow \text{constant term} = 0$.

We can see that for the denominator, any term with $X$ or $Y$ will get evaluated to 0. Hence, the denominator, $b(P) = c$, is the constant term in polynomial $b$.

Collecting all the information above, we can say that $$\mathcal{M} = \langle X, Y\rangle$$ $$\mathcal M^2 = \langle X^2, XY, Y^2 \rangle = \langle X^2, XY, X^3 + X \rangle$$

Since $X = X^3 + X - X \cdot X^2, XY = Y \cdot X$.

We have: $$\mathcal{M}^2 = \langle X \rangle$$

If we consider $\mathcal{M} / \mathcal{M}^2$ as a quotient of vector space:

$$\mathcal{M} / \mathcal{M}^2 = \left\{ [v] \mid v \in \mathcal{M} \right\}$$ with the equivalence class defined as: $$[v] = v + \mathcal{M}^2$$

Since scalar multiplication and addition are defined on this equivalence class as:

• $\alpha[v] = [\alpha v]$ for all $\alpha \in \mathbb{A}$.
• $[u] + [v] = [u + v]$.

Hence, the generator is preserved: \begin{align*} \mathcal{M} / \mathcal{M}^2 & = \langle [X], [Y] \rangle \\ & = \langle X + \mathcal{M}^2, Y + \mathcal{M}^2 \rangle \\ & = \langle X + \langle X \rangle, Y + \langle X \rangle \rangle \\ & = \langle \langle X \rangle, Y + \langle X \rangle \rangle \\ & = \langle [0], [Y] \rangle \\ & = \langle [Y] \rangle \end{align*}

Hence, $\dim_{\mathbb{C}} \mathcal{M} / \mathcal{M}^2 = 1$.

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I still have some questions, if we treat $\mathcal M/\mathcal M^2$ as a quotient space of two vector spaces, what is the field that $\mathcal M$ and $\mathcal M^2$ are over?

And, for concluding that $\dim_{\mathbb{C}} \mathcal{M} / \mathcal{M}^2 = 1$:

If my calculation is correct, that is $\mathcal{M} / \mathcal{M}^2 = \langle [Y] \rangle$, then is it equivalent to $\mathcal{M} / \mathcal{M}^2 = \operatorname{Span}([Y])$?

If so, using my linear algebra knowledge, I could say for sure that $\dim_{\mathbb{C}} \mathcal{M} / \mathcal{M}^2 = 1$.

Answer 2

Question: "But they are a little bit advanced to me. I am thinking that by having an explicit curve, we might be able to calculate the dimension in a more direct way. However, I have no clue where to start."

Response: Here is a method that at first seems "advanced" but it is in fact "elementary". If $F:=y^2-x^3-x$ and $k$ is the complex numbers, let $A:=k[x,y]/(F)$ with $C:=Spec(A)$. If $\mathfrak{m} \subseteq A$ is a maximal ideal it follows there is an isomorphism of $k$-vector spaces

$$F1.\text{ }\phi: Der_k(A, A/\mathfrak{m}) \cong Hom_k(\mathfrak{m}/\mathfrak{m}^2,k)$$

defined as follows: Let $\delta: A \rightarrow k:=A/\mathfrak{m}$ be a $k$-linear derivation. It follows since $\mathfrak{m} \subseteq A$ there is an induced map (we restrict $\delta$ to the sub vector space $\mathfrak{m}$):

$$\delta: \mathfrak{m} \rightarrow k$$

and if $z:=\sum_i u_i v_i \in \mathfrak{m}^2 \subseteq \mathfrak{m}$ it follows

$$\delta(z):= \sum_i \delta(u_i v_i)= \sum_i \delta(v_i)u_i + v_i \delta(u_i) =0$$

since $\delta(u)=0$ for all $u \in \mathfrak{m}$. You may check that the above map is an isomorphism. Hence if $\mathfrak{m}:=(x-a,y-b) \subseteq A$ is a maximal ideal in $A$ corresponding to the point $p$ on your curve it follows the tangent space of $C$ at $p$ may be calulated using the formula $F1$.

Since the curve $C$ is regular/smooth it follows the Jacobian ideal $J:=(F_x,F_y) \subseteq A$ has as zero set the empty set. Hence there is no point $p:=(a,b)\in C$

with $F_x(p)=F_y(p)=0$ where $F_x$ is partial derivative of $F(x,y)$ wrto the $x$-variable, etc. Hence either $F_x(p) \neq 0$ or $F_y(p) \neq 0$.

A derivation $\delta: A \rightarrow k$ at $p$ is well defined iff

$$\delta(0)=\delta(F):=F_x(p)\delta(x)+F_y(p)\delta(y)=0.$$

If $F_y(p)\neq 0$ we get the relation

$$F_y(p)\delta(y)=-F_x(p)\delta(x).$$

Any derivation $\delta: A \rightarrow k$ must satisfy

$$\delta(f(x,y))=f_x(p)\delta(x)+f_y(p)\delta(y)=\delta(x)(f_x(p)-\frac{F_x(p)}{F_y(p)}f_y(p))$$

hence there is an equality

$$\delta(f)=\delta(x)(\partial_x -\frac{F_x(p)}{F_y(p)}\partial_y)(f) $$

and we get

$$\delta =\delta(x)(\partial_x -\frac{F_x(p)}{F_y(p)}\partial_y):=\delta(x)\partial $$

where $\delta(x)\in k$ is any complex number. It follows there is an equality of $k$-vector spaces

$$Der_k(A,k)\cong Hom_k(\mathfrak{m}/\mathfrak{m}^2,k) \cong k\partial.$$

Hence we have shown that the tangent space of $C$ at $p$ is a one dimensional $k$vector space with the derivation $\partial$ as a basis, and since the cotangent space is the dual it follows $\mathfrak{m}/\mathfrak{m}^2$ is also one dimensional at any point $p\in C$ with a basis given by the duale $\partial^*$.

In fact: The only thing used in this argument is that the curve $C$ is regular at any point $p$ (the jacobian ideal defines the empty subscheme of $C$), hence the argument given above proves the claim in general.

There is a similar type question (and answers) here that may be of interest. Here I do a similar calculation for an arbitrary regular curve over an algebraically closed field:

$aX + bY$ is an element of $M^2$ if and only if the line $aX + bY = 0$ is tangent to $W$ at $(0, 0)$