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

Question

Let $F(X, Y) = Y^2 - X^3 + X \in \mathbb{C}[X, Y]$, and let $a$ and $b$ be constants (elements of $\mathbb{C}$). Write $W = V(F)$, and let $P$ be the point $(0, 0)$ on $W$.

1. Show that $aX + bY$ is an element of the maximal ideal $M = M_P(W)$ of the local ring $\mathcal{O}_P(W)$ at the point $P = (0, 0)$.
2. Show that $aX + bY$ is an element of $M^2$ if and only if the line $aX + bY = 0$ is tangent to $W$ at $(0, 0)$. ($M$ is the same as in part (a).)

The definition I am using:

$K(W)$ is the fraction field of the coordinate ring $\Gamma(W)$.

The local ring $\mathcal{O}_P(W) = \left\{ f \in K(W) \mid f \text{ is defined at } (0, 0) \right\}$

The maximal ideal $M = M_P(W) = \left\{ f \in K(W) \mid f = \frac{a}{b} \text{ for } a(P) = 0, b(P) \neq 0 \right\}$

My attempt to part (1):

Since $Y^2 - X^3 + X$ is irreducible, $W = V(F)$ is an infinite set, $W = V(F)$ is irreducible with $I(W) = V(F) = \langle F \rangle$.

The coordinate ring is given as: $$\Gamma(W) = \mathbb{C}[X, Y] / I(W) = \mathbb{C}[X, Y] / \langle F \rangle$$

Since the coordinate ring $\Gamma(W)$ is the ring of polynomials from $W$ to $\mathbb{A}^1 = \mathbb{C}$, then we can see that $aX + bY \in \Gamma(W)$ for all $a, b \in \mathbb{C}$. Hence, $aX + bY \in K(W)$. Since $aX + bY$ a polynomial, then $aX + bY$ is defined on all $\mathbb{A}^2$. Naturally, $aX + bY \in M_P(W)$ as we can let $f = k / m$ with $k = aX + bY$ and $m = 1$ such that $k(P) = 0$ and $m(P) = 1 \neq 0$.

I am not sure how to approach part (2).

I know that $M^2$ is the ideal generated by products of an element from $M$ and an element from $M$. $$M^2 = \left\{ \sum_{k = 1}^n m_k m_k \mid m_1, \dots, m_n \in M \right\}$$ but this does not seem to be immediately helpful...

Also, I know that $$M_P(W) = I(P) \mathcal O_P(W)$$ where $I(P) \subseteq \Gamma(W)$ is the ideal of all the functions that vanish at $P$. That is, $M_P(W)$ is just the ideal of $\mathcal O_P(W)$ generated by $I(P)$.

I feel like that I am missing something obvious...

Also, in the book Algebraic Curves by Fulton, there is an exercise 2.18 which states

2.18. Let $\mathcal O_P(V)$ be the local ring of a variety $V$ at a point $P$. Show that there is a natural one-to-one correspondence between the prime ideals in $\mathcal O_P(V)$ and the subvarieties of $V$ that pass through $P$. (Hint: If $I$ is prime in $\mathcal O_P(V)$, $I \cap \Gamma(V)$ is prime in $\Gamma(V)$, and $I$ is generated by $I \cap \Gamma(V)$; use Problem 2.2.)

Answer 1

Comment: "I am not sure how to approach part (2)."

Answer: If $k$ is a field and $A$ is a finitely generated $k$-algebra with $\mathfrak{m} \subseteq A$ a maximal ideal with $A/\mathfrak{m}\cong k$ there is a "canonical" isomorphism of vector spaces

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

If $\partial: A \rightarrow A/\mathfrak{m}$ is a derivation, you get canonically a $k$-linear map

$$\partial: \mathfrak{m} \rightarrow A/\mathfrak{m}$$

and if $uv \in \mathfrak{m}^2$ it follows

$$\partial(uv)=u\partial(v)+\partial(u)v=0$$

and you get an induced map $\partial^*: \mathfrak{m}/\mathfrak{m}^2 \rightarrow A/\mathfrak{m}$.

In explicit examples it is easier to use dervations.

Example: In your case with $k$ the complex numbers, $F:=y^2-x^3+x$, $C':=Spec(k[x,y]/(F))$ and $p:=(0,0)$ you get a similar calculation. The tangent space $T_p(C')$ is the vector space of derivations

$$ \partial:=a \partial_x+b\partial_y$$

with $\partial(F):=f_x(p)\partial(x)+F_y(p)\partial(y)=\partial(x)=0$. Hence the tangent space $T_p(C')$ is spanned by the vector field $\delta_p:=\partial_y$:

$$T_p(C') \cong k\partial_y.$$

The induced map $\delta_p^*: \mathfrak{m}/\mathfrak{m}^2 \rightarrow k$ is the map

$$\delta_p^*(fx+gy):=g(p) \in k.$$

Hence there is an isomorphism of $k$-vector spaces

$$(\mathfrak{m}/\mathfrak{m}^2)^* \cong k \delta_p^*.$$

This approach gives a general metod to calculate a basis for the tangent space. In your case $(2)$ you get the following: There is a pairing

$$<,>: (\mathfrak{m}/\mathfrak{m}^2)^* \oplus \mathfrak{m}/\mathfrak{m}^2 \rightarrow k$$

defined by

$$<\phi,v>:=\phi(v)\in k$$

and your vector $v:=ax+by$ is zero in the cotangent space iff

$$\delta^*_p(ax+by):=\partial_y(ax+by):=b=0,$$

which is iff $v\in \mathfrak{m}^2$. Hence $v$ is tangent to $C'$ at $p$ iff $v\in \mathfrak{m}^2$.

Example: Let $F(x,y) \in k[x,y]$ be an arbitrary polynomial and let $A:=k[x,y]/(F)$ with $C:=Spec(A)$ a regular curve. Let $\mathfrak{m}:=(x-a,y-b) \in A$ be a $k$-rational point, denoted $p$. You may use the above approach to prove that an element

$$l(x,y):=u(x-a)+v(y-b) \in \mathfrak{m}$$

is in $\mathfrak{m}^2$ iff there is an element $s\in k$ with

$$l(x,y)=s(F_x(p)(x-a)+F_y(p)(y-b)).$$

Hence $l(x,y) \in \mathfrak{m}^2$ iff the linear variety $L:=Spec(k[x,y]/(l)$ is tangent to the curve $C$ at the point $p$. Here $F_x(x,y)$ denotes the partial derivative of $F(x,y)$ wrto the $x$-variable, and $F_x(p)$ denotes the "evaulation of $F_x(x,y)$ at the point $p$".

The line $L$ intersects $C$ at $p$ iff $l \in \mathfrak{m}$. The line $L$ intersects $C$ at $p$ with multiplicity $i$ iff $l \in \mathfrak{m}^{i+1}$. This calculation is elementary.

Note: An even more elementary approach is the following: If $F(x,y)\in k[x,y]$ defines a regular curve $C:=Spec(k[x,y]/(F))$, you get for any element $p:=(a,b)\in k^2$ an equality

$$F(x,y)=F(p)+F_x(p)(x-a)+F_y(p)(y-b)+ \sum_{i+j\geq 2}a_{ij}(x-a)^i(y-b)^j $$

$$:=F(p)+l_p(x,y) +\text{ higher order terms}$$

where $l_p(x,y):=F_x(p)(x-a)+F_y(p)(y-b)$.

If $p\in C$ is a point you get

$$F(x,y)=l_p(x,y)+ \text{higher order terms}.$$

If $\mathfrak{m}:=(x-a,y-b)\in A:=k[x,y]/(F)$ it follows

$$ 0=F(x,y)=l_p(x,y) \in \mathfrak{m}/\mathfrak{m}^2.$$

Hence the "tangent line" $l_p(x,y)$ of $C$ at $p$ lives in the ideal $\mathfrak{m}^2$.

Conversely assume $l(x,y):=u(x-a)+v(y-b) \in \mathfrak{m}$ is a line and $l \in \mathfrak{m}^2$. Since $C$ is regular at $p$ it follows $\mathfrak{m}/\mathfrak{m}^2$ is one dimensional and we may use $(x-a)$ as a basis. Since $l_p(x,y)\in \mathfrak{m}^2$ it follows (assume $ F_y(p)\neq 0$)

$$ (y-b)=-\frac{F_x(p)}{F_y(p)}(x-a).$$

We may write

$$l(x,y)=u(x-a)+v(-F_x(p)/F_y(p))(x-a)=(u-v(F_x(p)/F_y(p)))(x-a)=0$$

hence $uF_y(p)=vF_x(p)$. It follows

$$F_y(p)l(x,y)=uF_x(p)(x-a)+vF_y(p)(y-b)=vl_p(x,y)$$

hence if $l(x,y)\in \mathfrak{m}^2$ it follows $l(x,y)$ is a mutiple of $l_p(x,y)$.

Comment:"I feel like that I am missing something obvious..."

Response: Above I have given an elementary proof that an arbitrary line $l(x,y):=u(x-a)-v(y-b)$ is in the ideal $\mathfrak{m}^2$ iff $l(x,y)$ is a mutliple of the tangent line $l_p(x,y)$ at $p$:

$$F_y(p)l(x,y)=vl_p(x,y).$$

Hence a line $l(x,y)\in k[x,y]$ is in $\mathfrak{m}^2$ iff the "line" $L:=Spec(k[x,y]/(l(x,y))$ is tangent to $C$ at $p$.