# Tag Info

10

A $\mathbb Q$-point corresponds to a DVR whose residue field equals $\mathbb Q$. The function field can't tell the point at infinity from any other point; the notion of "point a infinity" is not intrinsic, but is defined relative to the embedding of the curve into $\mathbb P^2$ (and a choice of coords. on $\mathbb P^2$, so that we know what the line at ...

6

Let $A,B$ be integers, and let $E_{A,B}: y^2=x^3+Ax^2+Bx$ be an elliptic curve. Then, the rank of the Mordell-Weil group $E_{A,B}(\mathbb{Q})$ is less or equal to $\nu(A^2-4B)+\nu(B)-1$, where $\nu(N)$ is the number of positive prime divisors of $N$. This is shown in Milne's "Elliptic Curves", Proposition 5.6, or here (Proposition 1.1). In particular, ...

4

Intuitively such a thing cannot exist, even with level structure, since the "forget isomorphism class and remember only isogeny class" map from the usual moduli space ought to be algebraic, but an algebraic map of curves has finite fibers but (over $\mathbb{C}$, say) isogeny classes of non-isomorphic curves are infinite. Of course there is the problem that ...

3

When we try to classify elliptic curves up to isogeny, then we use the modular curves $X_0(N)/\mathbb{Q}$, whose non-cuspidal $K$-rational points (for some number field $K$) classify triples $(E/K,E'/K,\phi/K)$ of elliptic curves $E$ and $E'$ defined over $K$, together with an isogeny $\phi:E\to E'$ defined over $K$, with cyclic kernel of size $N$. In other ...

3

Consider the following example. Let $E: y^2+y=x^3-x^2$. Its discriminant is $-11$, and $E(\mathbb{Q})\cong \mathbb{Z}/5\mathbb{Z}$, generated by $(0,0)$. The torsion points are, in fact: $$P=(0,0),\ 2P=(1,-1),\ 3P=(1,0),\ 4P=(0,-1),\ 5P=\infty=[0,1,0].$$ In particular, $p=5$ is a prime of good reduction, and the reduction mod $5$ of each of the points above ...

3

Hints: Prove that the curve has genus 1. Find the primes of bad reduction. If $p$ is a prime of good reduction, and large enough, then the Hasse-Weil bounds tell you that there is an affine point (not just at infinity) over $\mathbb{F}_p$. Now use Hensel's lemma to lift it up to $\mathbb{Q}_p$. If $p$ is a prime of bad reduction, or not large enough, you ...

3

The statement: "The number of elliptic curves of discriminant $D$ is bounded above by number of nontrivial pairs $(a,b)∈ \mathbb{Z}^{2}$ such that $D=−16(4a^{3}+27b^{2})$." is certainly false as stated. For example, the discriminant of the model for the curve $$E:y^2 + y = x^3 - x^2 - 10x - 20$$ is $-11^5$. Any model of the form $y^2=x^3+Ax+B$ with ...

3

Let $E$ be given by a Weierstrass equation $$y^2+a_1xy+a_3y = x^3+a_2x^2+a_4x+a_6.$$ If the characteristic of the field is not $2$, then we can complete the square in $y$ to obtain $$(y+(a_1x+a_3)/2)^2 = x^3+a_2x^2+a_4x+a_6+(a_1x+a_3)^2/4.$$ You can write the last equation as $E':Y^2=f(x)$. In this model, it is clear that if $P'=(x_0,Y_0)$, then $-P'$ (the ...

3

Oh, well. I cannot find this claim in Wikipedia... one case is easy, the other, not. Any primitive Pythagorean triple is $2xy, x^2 - y^2, x^2 + y^2$ with one of $x,y$ odd and the other even, and $\gcd(x,y) =1.$ We cannot have a Pythagorean triple with legs $x^2 - y^2, x^2 + y^2,$ because we would be solving $$2 x^4 + 2 y^4 = z^2$$ with one of $x,y$ odd. ...

2

In general, if a curve is given by $Y^2=X^3+AX^2+BX+C$, a change of variables $Y=y$ and $X=x-A/3$ will provide a model for the curve of the form $y^2=x^3+A'x+B'$. The reason is that $$X^3+AX^2+\cdots = (x-A/3)^3 + A(x-A/3)^2+\cdots$$ and the coefficient in $x^2$ is given by $3(-A/3)+A=0$.

2

Let $E$ and $E'$ be elliptic curves, and let $\phi:E\to E'$ be a $p$-isogeny (i.e., $\phi$ is an isogeny of degree $p$), where $p$ is prime. In particular, $\phi$ is a group homomorphism from $E$ to $E'$ and its kernel $\ker(\phi)$ is a group of size $p$. Prove that every $P$ in $\ker(\phi)$ has order dividing $p$. Prove that the prime-to-$p$ torsion ...

2

First, let me point out that $E:x^3-2y^3=1$, together with the point $[1,0,1]$, is an elliptic curve. Its Mordell-Weil group can be calculated by finding a Weierstrass model, which can be chosen to be $y^2=x^3-27$, and then verify that $E(\mathbb{Q})\cong \mathbb{Z}/2\mathbb{Z}$. Thus, there are only two rational points on $E$, namely $[1,0,1]$ and ...

2

If $f_P(S)$ is a function with divisor $m[P]-m[\mathcal{O}]$, then $f_P$ has a zero at $P$ and a pole at $\mathcal{O}$. Then, for a fixed constant $T$, the function $h_{P,T}(S)=f_P(T+S)$ is simply a translation of $f_P$ by $T$. Thus, $h_{P,T}$ has a zero when $T+S=P$, i.e., at $S=P-T$ and a pole at $T+S=\mathcal{O}$, i.e., at $S=-T$. Moreover, the ...

2

The group structure of the torsion subgroup may be the same, but the group law may look very different! I find the following example to be interesting and related to your question. Let $E:y^2=x^3+1$ with zero at $[0,1,0]$, and consider $E':y^2=x^3+1$ where we now declare zero to be $[2,3,1]$. Then, $E$ and $E'$ are clearly birationally equivalent via the ...

2

You are right. The computation is just too inefficient. The best known attack on ECDLP, the pollard rho attack, would be useless against elliptic curves over the rationals. Consider this, if you were to do the computations over a finite field of say 512-bits, you will only have to deal with 512-bit intermediate values along the way. Considering the same ...

2

Your original model, i.e., $y^2+xy=x^3-x^2-50x+111$, is minimal. Its discriminant is $$\Delta=3901625=5^3\cdot 7^4\cdot 13.$$ Since the coefficients of the model are integral, and the valuation of the discriminant at every prime is less than $12$, it follows that the curve is minimal at every prime number (Silverman's "The Arithmetic of Elliptic Curves", ...

2

Let us begin with the quartic $$m^4-4m^3n-6m^2n^2+4mn^3+n^4=d^2.$$ Dividing through by $n^4$, we obtain $$m'^4-4m'^3-6m'^2+4m'+1=d'^2,$$ with $m'=m/n$ and $d'=d/n^2$. Now we can homogenize this equation to obtain $$E:M^4-4M^3N-6M^2N^2+4MN^3+N^4=D^2N^2,$$ which has a rational point $P=[M,N,D]=[0,0,1]$. The curve $E$ is in fact non-singular, and of genus $1$, ...

1

A sketch of a proof may look like this: As what was said in the comments, it is obvious that $P+Q\in E$. So let's just say that $S=P+Q$. Now, since points on an elliptic curve form a group (abelian group), there is an inverse for $S$. Notice that $S$ must satisfy the equation of $E$ to lie on $E$. So say $E$ is given by the equation $y^2=x^3+ax+b$. Now, ...

1

This is a great observation! Let $E/\mathbb{Q}$ be a curve given by a model $v^3=q(u)$, for some cubic polynomial $q\in\mathbb{Q}[u]$, and assume that the projective closure of this model, i.e., $V^3=W^3q(U/W)$, is smooth, and it has a rational point $P$. Then, $(E,P)$ is an elliptic curve defined over $\mathbb{Q}$. Moreover, $E$ admits an endomorphism ...

1

Theorem: Let #$E(Fq) = 1 - a + q$ Write $X_2 − aX + q = (X − α)(X − β)$. Then # $E(F_{q^n}) = 1 − (α^n + β^n) + q^n$ for all n ≥ 1. Now for the problem: It is easy. Write $x^2 + 2$ = $(x+i \sqrt2)(x-i \sqrt2)$ and so #$E(F_{2^n})=$ $2^n+1 -(x+i \sqrt2)(x-i \sqrt2)$ and from there it is then easy to use a phase argument to duduce the answer. Note: ...

1

If you eliminate $Y$ from both the equations you get $$-2\,X\,Z^2\,\left(Z-X\right)=0$$ Thus there is a pair of roots at $Z=0$ (points at infinity). Set $Z=0$ and you get $X^2+Y^2=0$. Setting $X=1$, you get the two other roots $$(1, i, 0)$$ and $$(1, -i, 0)$$ I had a typo in the earlier answer.

1

Quote from Example section of the Wikipedia entry on Bézout's Theorem (see here): Two circles never intersect in more than two points in the plane, while Bézout's theorem predicts four. The discrepancy comes from the fact that every circle passes through the same two complex points on the line at infinity. Writing the circle $(x-a)^2+(y-b)^2 = r^2$ ...

1

For Beozout's theorem to work we must work in the complex projective plane, as you noted. Mathematica's Reduce transforms the system $$x^2 + y^2 = z^2 \ \ \text{and}\ \ x^2 + y^2 - x z - y z = 0$$ to $$(y\neq 0\land ((x=0\land y=z)\lor (z=0\land (y=-i x\lor y=i x))))\lor (x\neq 0\land ((y=0\land x=z)\lor (z=0\land (y=-i x\lor y=i x))))$$ We can read off ...

1

The missing two points are the http://en.wikipedia.org/wiki/Circular_points_at_infinity From the complex projective point of view, a circle is the same as any other conic, and 5 points are needed specify a conic. The reason 3 points in the plane are enough to define a circle is that circles are the conics that pass through the 2 "circular points at ...

1

Let me explain why I would not expect this to be true. Suppose that $X$ is a $K3$ surface with the following properties: First we ask that the automorphism group $G=\operatorname{Aut}(X)$ is finite. (Remark: by a theorem of Sterk, this is equivalent to requiring that $X$ has fintely many elliptic pencils, though we don't use this.) Note that this gives ...

1

We can use $\mathbf{R}$ in place of $\mathbf{Q}$, if we like. Draw a picture of a curve that has three real $2$-torsion points in Weierstrass form. How many of the $2$-torsion points can be the double of some other point? Recall that every line intersects an elliptic curve in exactly three points, counting multiplicity and the point at infinity.

1

$b$ need not be a square. Instead, the trick is to show that you can actually divide the equation by $b^2$. Let $p$ be a prime that divides $b$ and is not 2. Let $p^n$ be the highest power of $p$ that divides $b$. Since $\gcd(a,b) = 1$, and we know that $2a^2 d^4 \equiv 0 \pmod{p^{2n}}$, thus $d^4 \equiv 0 \pmod{p^{2n}}$ and this implies that $p^n \mid ... 1 As far as I know there is no code to solve this problem distributed with Magma as standard. Your choices are: use the Sage implementation; implement the algorithms in Magma for yourself; or find someone who has appropriate Magma code and persuade them to share it with you. EDIT. This answer is totally wrong, as ccorn's comment above shows. I had apparently ... 1 In positive characteristics, isogenies may ramify. In general, see Silverman's "The Arithmetic of Elliptic Curves", Theorem 4.10 in Chapter III, where he shows that if$\phi:E_1\to E_2$is a non-constant isogeny then (1) For every$Q\in E_2$, we have$\# \phi^{-1}(Q)$equals the separable degree of$\phi$. (2) For every$P\in E_1\$, the ramification index ...

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