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I am asked to check whether or not this curve is smooth (and if not provide singular points): $x_2^2x_0 = x_1^3 - x_1x_0^2$.

The way I approached this was by to use the projective Jacobi criterion on the polynomial $ x_1^3 - x_1x_0^2 - x_2^2x_0 = 0$ and see that the matrix with partial entries is $J = \begin{pmatrix} -2x_1x_0 - x_2^2 & -3x_1^2-x_0^2 & -2x_2x_0\\ \end{pmatrix}$. I believe that this has rank $1 = 2 - codim_X{a}$ for when $a \neq (0:0:0)$. Therefore it would be smooth everywhere except the origin. Is this correct?

The second part of the question is how many points in the line $x_1=0$ intersect the curve? Compute the multiplicity of every point of the intersection.

I believe the way to answer this is to show that they only intersect at $(0:0:1)$ and $(1:0:0)$ and then the multiplicity of these points are the degrees of the curves multiplied by each other: $3 \times 1 = 3$.

Are these correct? Sorry, I am leaving out a fair amount of detail because I am in a rush. Would anybody confirm my solutions or point me in the right direction? For reference, I am using Gathmann 2014 edition: smoothness and multiplicity.

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Smoothness When I first saw this post, I recognised the curve as a Weierstrass cubic - an elliptic curve embedded in $\mathbb P^2$. This curve is well known to be smooth.

You're right that the way to prove that the curve is smooth is to show that the Jacobian has rank $1$ everywhere. However, I'll make a minor remark. You only need to check that the Jacobian has full rank at points that lie on the curve - you don't need to check that the Jacobian has full rank at all points in $\mathbb P^2$. You probably know this already, but this didn't come across entirely clearly in your original post.

Also, you mentioned the "origin point" $[0: 0: 0]$. That's not even a point in $\mathbb P^2$ (let alone a point on the curve)! So you don't need to worry about this.

Intersection multiplicities. It's true that the elliptic curve is a degree $3$ curve. But this doesn't mean that each intersection between the elliptic curve and the line has intersection multiplicity $3$! The fact that the elliptic curve is a degree $3$ curve actually tells you that if you compute the intersection multiplicities at each of the intersection points, and then add them up, then this sum is equal to $3$. You've identified two intersection points. Therefore, one of these points is an intersection of multiplicity $1$, while the other point is an intersection of multiplicity $2$.

I'm not sure what definition you're using for the intersection multiplicity. One definition, which is applicable for plane curves, goes as follows. If $C_1$ and $C_2$ are curves in $\mathbb A^2$, defined by the polynomials $f$ and $g$ respectively, and if $p$ is a point where $C_1$ and $C_2$ intersect, then the intersection multiplicity of $C_1$ and $C_2$ at $p$ is $$ \text{mult}_p(C_1 \cap C_2) = \text{dim}_k \frac{\mathcal O_{p}(\mathbb A^2)}{(f, g) }.$$ Here, $\mathcal O_{p}(\mathbb A^2)$ is the coordinate ring for the affine plane $\mathbb A^2$, localised at the point $p$.

If you like, you may study this ring directly and compute its dimension as $k$-vector space. That's one way to solve the problem, and you're welcome to give this a go. However, personally, I tend to find it quite fiddly to work with this ring directly; instead, I prefer to tackle the problem using discrete valuations. This is the approach that I'll describe in my answer.

By simple commutative algebra, one can show that $$ \frac{\mathcal O_{p}(\mathbb A^2)}{(f, g) } \cong \frac{\mathcal O_{p}(C_1)}{(g)}.$$ If $C_1$ is smooth at $p$, then $\mathcal O_p(C_1)$ is a discrete valuation ring, so $$ \text{mult}_p(C_1 \cap C_2) = \text{dim}_k \left(\frac{\mathcal O_{p}(C_1)}{(g)} \right) = v_p(g),$$ where $v_p(g)$ is the valuation of $g$ in $\mathcal O_p(C_1)$.

[Edit: Having quickly skimmed the lecture notes you attached, it seems like the $ \text{mult}_p(C_1 \cap C_2) = \text{dim}_k \left(\frac{\mathcal O_{p}(C_1)}{(g)} \right)$ definition is closer to what you're familiar with than the $\text{mult}_p(C_1 \cap C_2) = \frac{\mathcal O_{p}(\mathbb A^2)}{(f, g) }$ definition. Hopefully this means you'll find this answer easy to follow.]

Example: $p=[1:0:0]$. The point $p = [1: 0: 0]$ lies in the affine patch $$ \mathbb A^2 \cong \{ [1 : x: y] : (x, y ) \in \mathbb A^2 \} \subset \mathbb P^2.$$

In this affine patch:

  • The line (call it $C_1$) is defined by the polynomial $f = x$.
  • The elliptic curve (call it $C_2$) is defined by the polynomial $g = y^2 - x^3 + x$.
  • The point $p$ is the point $(0, 0)$.

$y$ is a local parameter on $C_1$ at $p$. In $\mathcal O_p(C_1)$, we have $ g = y^2$. So the valuation of $g$ at $p$ is $2$. Therefore, the intersection multiplicity of $C_1$ and $C_2$ at $p$ is $2$.

Example: $p=[0:0:1]$. The point $p = [0: 0: 1]$ lies in a different affine patch, namely, $$ \mathbb A^2 \cong \{ [u: v : 1] : (u, v) \in \mathbb A^2 \} \subset \mathbb P^2.$$

In this affine patch:

  • The line (call it $C_1$) is defined by the polynomial $f = v$.
  • The elliptic curve (call it $C_2$) is defined by the polynomial $g = u - v^3 - vu^2$.
  • The point $p$ is the point $(0, 0)$.

$u$ is a local parameter on $C_1$ at $p$. In $\mathcal O_p(C_1)$, we have $ g = u$. So the valuation of $g$ at $p$ is $1$. Therefore, the intersection multiplicity of $C_1$ and $C_2$ at $p$ is $1$.

Finally - lest we get too bogged down in algebra - I should point out that the fact that $[1: 0: 0]$ is an intersection point of multiplicity $2$ whereas $[0:0:1]$ is an intersection point of multiplicity $1$ is a manifestation of the fact that the line is tangent to the elliptic curve at $[1:0:0]$, whereas the line intersects the elliptic curve transversely at $[0:0:1]$. It's always a good idea to interpret things using "high school" intuition as a sanity check.

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