# Checking smoothness of curves and finding multiplicites.

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.

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.