# Bad reduction at prime numbers for the elliptic curve $y^2+y=x^3-x^2+2x-2$

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

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 $$p$$):

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.

• 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. – dyf Nov 19 '18 at 14:59

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))


$${}$$

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.

• 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 $(x,y)=(2,2)$. – Joe Nov 20 '18 at 0:20
• Opps, forgot to write the actual question. :) My question is, how did you get $y=3$? :) – Joe Nov 20 '18 at 0:21
• @Joe Ah, my mistake: that should be $y = -3 = 2$. Thanks for the correction. – André 3000 Nov 20 '18 at 1:11
• 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 $y\mapsto y-3$? – Joe Nov 20 '18 at 12:56
• @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. – André 3000 Nov 21 '18 at 4:15