How do I translate the intersection of two affine curves in a plane into a statement about ideals in $k[X, Y]$?

Question

Let $E : y^2 = x^3 + Ax + B$ be an elliptic curve over a field $k$ with characteristic not $2$ or $3$ (not necessarily algebraically closed), and let $x = x_0$ be a line that intersects $E$ in $(x_0, \pm y_0) \in E(k)$.

According to Lemma 10 of this paper, this is expressed algebraically as the following equation of ideals over $k[X, Y]$: $$ (X - x_0, Y - y_0) (X - x_0, Y + y_0) \equiv (X - x_0) \pmod{(Y^2 - (X^3 + AX + B))}, $$ or more directly $$ (X - x_0, Y - y_0) (X - x_0, Y + y_0) = (X - x_0, Y^2 - (X^3 + AX + B)). $$

The paper gives a purely algebraic proof of this statement, although I can't help but think there must be a way to derive this algebro-geometrically.

I'd imagine it'd go something like: $$ V((X - x_0, Y - y_0) (X - x_0, Y + y_0)) = \\{ (x_0, \pm y_0) \\} = V(X - x_0, Y^2 - (X^3 + AX + B)) $$ We'd like to conclude that $$ (X - x_0, Y - y_0) (X - x_0, Y + y_0) = (X - x_0, Y^2 - (X^3 + AX + B)). $$

I know that $V(I) = V(J)$ doesn't imply $I = J$ in general; the most common condition is that $I$ and $J$ are radical ideals, and $k$ has to be algebraically closed, and then we can use the Nullstellensatz directly.

It seems doable to show that $(X - x_0, Y - y_0) (X - x_0, Y + y_0)$ is radical, via $$ \sqrt{I J} = \sqrt{I \cap J} = \sqrt{I} \cap \sqrt{J} $$ but it seems less obvious that $(X - x_0, Y^2 - (X^3 + AX + B))$ is radical. (Edit: It now seems doable by showing that $k[X, Y]/(X - x_0, Y^2 - (X^3 + AX + B))$ is reduced.)

Also, even if we assume that $k$ is algebraically closed and use the Nullstellensatz, maybe we can extend this to non-algebraically closed subfields of $k$ given some conditions, e.g. if the points of intersection are in the subfield. (Edit: this answer seems to be what I need.)

Am I on the right track?

Answer 1

Yes, all of this is fine. If two ideals have the same zero set, then their radicals are equal, so if both ideals are radical, they are the same. It's not difficult to see that both of your ideals are radical by showing that the quotients by them are reduced:

• For the case of $I=(x-x_0,y-y_0)(x-x_0,y+y_0)$, the ideals $(x-x_0,y-y_0)$ and $(x-x_0,y+y_0)$ are comaximal, as $(y+y_0)-(y-y_0)=2y_0$, which is a unit as $2\neq 0$ and $y_0\neq 0$ by the assumption that the line intersects the elliptic curve in two points. Therefore $k[x,y]/I\cong k[x,y]/(x-x_0,y-y_0) \times k[x,y]/(x-x_0,y+y_0)\cong k\times k$ which is reduced.
• For the case of $J=(x-x_0,y^2-(x^3+Ax+B))$, we have $k[x,y]/J\cong k[y]/(y^2-(x_0^3+Ax_0+B))$, and $x_0^3+Ax_0+B$ is a nonzero square by the assumption that the line intersects the elliptic curve in two $k$-rational points. So $k[x,y]/J\cong k[y]/(y^2-c^2)$ for some $c\neq 0$, and as we're not in characteristic two, $y^2-c^2=(y-c)(y+c)$, and again $k[y]/(y-c)(y+c)\cong k\times k$ by the Chinese remainder theorem.

This will work correctly when $k$ is not algebraically closed assuming you can fix the nullstellensatz to remove the "algebraically closed" assumption. The right way of doing this is by using schemes, which add enough extra info that you can make that upgrade. (Another way to do this is that you see sometimes in print is to work with things defined over $k$ but look at their zero sets over $\overline{k}$.)

Answer 2

Comment: "@FredAkalin - Let us assume $y_0≠0$ and let $I:=(x−x_0,y−y_0),J:=(x−x_0,y+y_0)$ and $N:=(x−x_0)$. Let $A:=k[x,y]/(y^2−x^3−Ax−B)$. You are trying to prove there is an equality of ideals $IJ=N$ in $A$. The ring $A/N≅k[y]/(y^2−f(x_0))$, and this ring is a field if $f(x_0)$ is not a square in $k$. The ideals $I,J$ are coprime maximal ideals,hence the ring $A/IJ≅A/I⊕A/J≅k⊕k$ which is never a field. Hence it seems the claimed equality is not correct.

Response: If $(x_0,y_0)∈E(k)$ doesn't it follow that $f(x_0)$ is a square in $k$, i.e. $y^2_0=f(x_0)$? I suppose I wasn't clear about that -- edited to clarify.

Answer: You must be careful when working with maximal ideals in the ring $A$.

Let $I:=(x−x_0,y−y_0),J:=(x−x_0,y+y_0)$ and $N:=(x−x_0)$ with $(x_0,y_0) \in E(k)$ be $k$-rational points. Since $I$ and $J$ are coprime maximal ideals (if $char(k) \neq 2$) it follows $A/IJ \cong k \oplus k$ and this ring has no nilpotents. Let us consider $A/N$. We get the following

$$A/(x-x_0) \cong k[y]/(y^2-f(x_0))$$

and we have the following equality of ideals in $k[y]$:

$$(y-y_0)(y+y_0)=(y^2-y_0^2)=(y^2-f(x_0))$$

hence

$$A/(x-x_0) \cong k[y]/(y^2-f(x_0)) \cong k[y]/(y-y_0)(y+y_0) \cong k \oplus k$$

if $y_0 \neq 0$. This is because $(x_0,y_0)\in E(k)$ and hence $y_0^2=f(x_0)$.

You get an exact sequence

$$0 \rightarrow N \rightarrow A \stackrel{\rho}\rightarrow k \oplus k \rightarrow 0$$

And the map $\rho$ sends the ideal $IJ$ to the following ideal:

$$\rho(IJ)= (x_0-x_0, y-y_0)(x_0-x_0, y+y_0) =(y-y_0)(y+y_0)=0 $$

Hence there is an iduced map $A/IJ \cong k \oplus k \rightarrow A/N \cong k \oplus k$ which is an isomorphism.

If $y_0=0$ it follows $IJ:=(x-x_0, y)^2 \neq (x-x_0):=N,$

hence the ideals are not equal. Hence the following holds in general (over any field of $char(k) \neq 2$).

Summing up: 1. If $(x_0,y_0) \in E(k)$ and if $y_0^2 \neq 0$ it follows there is an equality of ideals $IJ =N$ in $A$.

2. If $y_0=0, f(x_0)=0$ it follows $IJ \neq N$.

Note: There is a "maximal spectrum" defined for an affine (or quasi projective ) algebraic variety over a non algebraically closed field. This may have some interest:

Grothendieck point of view of algebraic geometry

Im unsure what is your background, but the notion "algebraic variety" as defined in Hartshornes book "Algebraic geometry" chapter I, needs the base field to be algebraically closed. Over a non algebraically closed field $k$ there is no 1-1 correspondence between maximal ideals in $A$ and points $(x_0,y_0) \in k^2$ with $y^2_0=f(x_0)$.

The above link defines a "max spectrum" for your ring $A$ when $k$ is non algebraically closed. Using this the construction in the link you may speak of lines, tangents etc.