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Consider the elliptic curve $$E:y^2+y=x^3-x^2+2x-2.$$ My goal is to compute the conductor of the elliptic curve, the example is from

https://planetmath.org/conductorofanellipticcurve.

My problem isn't about the actual definition of the conductor but the definition of the different types of bad reductions that may occur.

Please, correct me if something I have typed is wrong, since I am not totally sure about anything of what I have written here.


Let $K$ be a field. We have that $E/K$ is singular if and only if $\Delta_E=0$ ($\Delta_E$ is the discriminant of the curve $E$).

I start by computing the discriminant

$$\Delta_E=-875=-5^3\cdot 7.$$ If I do reduction modulo $p=5$, we have $$\Delta_E\equiv_p 0.$$ Similarly I do a reduction modulo $q=7$; $$\Delta_E\equiv_q 0.$$ Thus, they are bad primes.

My next task is to classify the reductions modulo $p$ and $q$ respectively. In the link they claim that a reduction modulo $p$ is an additive reduction while modulo $q$ is a multiplicative reduction. Before we try to solve this problem, let me just mention:

(This is taken from The Arithmetic of Elliptic curves written by Silverman). The Weierstrass equation is given by $$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6.$$ We furthermore define,

$$ \begin{cases} b_2=a_1^2+4a_4\\ b_4=2a_4+a_1a_3\\ c_4=b_2^2-24b_4. \end{cases} $$

Case 1 (Reduction Modulo $p$):

In the link

https://planetmath.org/badreduction,

they define the reduction to be additive if the reduction has a cusp which holds if and only if both $\Delta_E$ and $c_4$ are equivalent to $0$.

Notice that $a_1=0$ and $a_4=2$. this gives us $$b_2=8,$$ and $$b_4=4.$$ Thus $$c_4=8^2-24\cdot 4=-32\equiv_p 3.$$ Thus, I would not claim that the reduction is additive, but multiplicative.

The same holds if I reduce it modulo $q$, then it is not $0$ in $\mathbb{F}_q$ and thus I would also claim that this reduction is multiplicative, since the reducion is multiplicative if $\Delta_E$ is equivalent to $0$, while $c_4\not = 0$.


Probably I have misunderstood the definition of bad reduction and the different types that can occur. I would be really happy if someone could explain this concept to me and how I may finish this example.

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    $\begingroup$ Since $p \neq 2,3$, we can write it in a short Weierstrass equation. Over $\mathbb{F}_5$, the equation is $(y+3)^2 = (x-2)^3$, which says it's additive reduction. I didn't read quite carefully enough to determine whether Silverman and I or Planetmath is incorrect. $\endgroup$ – dyf Nov 19 '18 at 14:59
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There's a unfortunate typo in Silverman's formula for $b_2$: it should be $b_2 = a_1^2 + 4 a_2$. See here for a list of errata. Using the corrected formula, I get $c_4 = -80 = -2^4 \cdot 5$, which confirms that the curve has additive reduction at $p = 5$.

But there's no need to trust formulas in a book: we can simply reduce the equation for the curve modulo $p$ and $q$ and check if the singularities are nodes or cusps, as dalbouvet has begun to do in the comments. To do this, we first find the singular points. Letting $$ F = y^2+y-(x^3-x^2+2x-2) \, , $$ these are exactly the solutions to $F = F_x = F_y=0$ where $F_x$ and $F_y$ are the specified partial derivatives. Let $(a,b)$ be the singular point. Making the change of variable $x \mapsto x - a, y \mapsto y - b$ moves the singularity to the origin, and then we can simply read off the tangent lines.

For $p = 5$, I find the singular point $(x,y) = (2,-3) = (2,2)$ and centered Weierstrass equation $y^2 = x^3$, which has a cusp. For $q = 7$, I find the singular point $(x,y) = (-3,-4) = (4,3)$ and centered Weierstrass equation $y^2 = x^3 - 3x^2$, which has a node.

Here are SageMathCells that compute do the above for $p = 5$ and $q = 7$ using the following code.

R.<x,y> = PolynomialRing(GF(5),2,order="lex")
f = x^3 - x^2 + 2*x - 2
F = y^2 + y - f
Fx = F.derivative(x)
Fy = F.derivative(y)
I = ideal([F,Fx,Fy])
show(I)
show(I.groebner_basis())
show(F(x=x+2,y=y-3))

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R.<x,y> = PolynomialRing(GF(7),2,order="lex")
f = x^3 - x^2 + 2*x - 2
F = y^2 + y - f
Fx = F.derivative(x)
Fy = F.derivative(y)
I = ideal([F,Fx,Fy])
show(I)
show(I.groebner_basis())
show(F(x=x-3,y=y-4))

As a note, to compute the conductor of your elliptic curve you may find Tate's algorithm useful.

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    $\begingroup$ Ah, thank you very much for a really clear and insightful answer! I understand in principle everything you wrote. I just have one question, which is quite stupid. You compute $F_y=2y+1=0\iff 2y=4$, which gives us the solution $y=2$. Then we also have $F_x= 3x^2-2x+2=0\iff x^2-4x+4=0$. $F_x=0$ has a double root $x=2$, which gives us the point $(x,y)=(2,2)$. $\endgroup$ – Joe Nov 20 '18 at 0:20
  • $\begingroup$ Opps, forgot to write the actual question. :) My question is, how did you get $y=3$? :) $\endgroup$ – Joe Nov 20 '18 at 0:21
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    $\begingroup$ @Joe Ah, my mistake: that should be $y = -3 = 2$. Thanks for the correction. $\endgroup$ – André 3000 Nov 20 '18 at 1:11
  • $\begingroup$ Hmm, André. I have one more question, just to check my understanding. In the second computation, that is when we work in $\mathbb{F}_7$, shouldn't the substitution be $x\mapsto x-4$ and $y\mapsto y-3$? $\endgroup$ – Joe Nov 20 '18 at 12:56
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    $\begingroup$ @Joe It depends on how you do the change of variable. Since $F(x,y) = y^2+y-(x^3-x^2+2x-2) = -(x+3)^3 + 3 (x+3)^2 + (y+4)^2$, then $F(x-3, y-4) = -x^3 + 3 x^2 + y^2$, so the automorphism of $K[x,y]$ given by $x \mapsto x-3, y \mapsto y-4$ takes $F$ to $-x^3 + 3 x^2 + y^2$. Equivalently, you could let $u = x+3$ and $v= y+4$, and then $F = - u^3 + 3u^2 + v^2$, which I think is what you're thinking of. $\endgroup$ – André 3000 Nov 21 '18 at 4:15

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