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5

Take two linearly independent rational differential forms $\omega_1, \omega_2\in L(K)=\Omega^1(C)$ with divisors $\operatorname {div} (\omega_1)=P_1+P'_1$ and $\operatorname {div} (\omega_2)=P_2+P'_2$ . The rational function $f=\frac {\omega_1}{\omega_2}\in \operatorname {Rat}(C)$ then has a divisor of the required form $$\operatorname ...


4

One way to think about this is via the modular curves parametrizing elliptic curves $E$ with either $E[3]$ or $\Delta^{1/3}$ rational. Note that $\Delta^{1/3}$ is rational iff $j^{1/3}$ is rational, because $j = E_4^3 / \Delta$. Assume for simplicity that $K$ contains the cube roots of unity (because $K(E[3])$ contains them in any case thanks to the Weil ...


3

There are fairly detailed notes by Douglas Ulmer available online. They cover the important case of function fields over finite fields. I couldn't tell if this is the case that interest you.


3

The question reduces to show that if $\mathfrak a$ is a radical ideal, then its extension is also radical. Equivalently, if $K[X]/\mathfrak a$ is reduced, then $K[X]/\mathfrak a\otimes_K\overline K$ is also reduced. If $K$ is a perfect field (and this is an assumption on page 1 of Silverman's book), then this holds as it is pointed out in this answer and ...


3

HINT: You can parametrise in several ways. How about $$x=c_1t^2$$ and $$y=c_2t^3$$ where $c_1^3=c_2^2$ ?


3

It is very easy to find an example. The curve $E\colon y^2=x^3+x+2$ has $j$-invariant $432/7$, which is not an alegbraic integer. Thus $E$ has no CM. In general, if $\Lambda$ is of the form $\mathbb Z+\mathbb Z\tau$ for some $\tau\in \mathbb C$, then $\mathbb C/\Lambda$ has CM if and only if $\tau$ is imaginary quadratic. This makes it easy to construct ...


3

The discriminant of a polynomial $f$ of degree $n$ and leading coefficient $a_n$ is $$\operatorname{disc}f=\frac{(-1)^{\tfrac{n(n-1)}2}}{a_n}\operatorname{Res}(f,f').$$ ($\operatorname{Res}(f,g)$ denotes the resultant of the polynomials $f$ and $g$). Thus it is $0$ if and only if $f$ has a multiple root. For any polynomials $f,g$, it can be shown there ...


3

The equations which define $P_3$ have coefficients in $k$, so $Gal(\bar k/k)$ preserves the locus of these equations i.e it fixes the coordinates of $P_3$ since it is the unique point of this locus.


2

Yes what you are doing seems correct. Take $x=t$ where $t \in \Bbb Q$. Then $y=x^2=t^2 \Rightarrow y \in \Bbb Q$. Also it is not true that '$x^2$ is rational $\Rightarrow x$ is rational'.(Take $x^2=2$, then $x=\pm \sqrt 2 \notin \Bbb Q$) Hence $\{(t,t^2):t \in \Bbb Q\}$ is the set of all rational points on the parabola $y=x^2$.


2

There is Elliptic Curves: Number Theory and Criptography by Lawrence, where on chapter 8 the author treats this subject.


1

As is suggested in the question, in order to find a free generator of $\Omega_A^1$, it suffices to find $P,Q \in A$ satisfying $P(3x^2-1) + Qy = 1$. Indeed, a generator is then given by $2P dx + Qdy$. The solution below finds such $P$ and $Q$, but it is rather ad hoc and I would be very interested in seeing a systematic method, if one exists. Let's first ...


1

The problem is reduced to the Euclidean algorithm in $\mathbb C[x]$. This technique works in the case of a nonsingular plane curve given by an equation of the form $y^n = f(x)$, and in particular for an elliptic curve in Weierstrass form. We wish to write $$1 = P\cdot f' + Q\cdot ny^{n-1}$$ for some $P,Q \in \mathbb C[x,y]/(y^n-f)$. As in msteve's answer, ...


1

I found a reference which has the required result with proof (which I've copied here). I claim no originality. Lemma (38.9.9): Let $k$ be a field. Let $A$ be an abelian variety over $k$. Then $[d]:A\to A$ is étale if and only if $d$ is invertible in $k$. Proof: Observe that $[d](x+y)=[d](x)+[d](y)$. Since translation by a point is an automorphism of $A$, ...


1

If you start with $x = X$, $y = Y$, i.e. simply renaming the variables, then trivially $$ X^3+Y^3−X^2+Y−1 = 0 $$ Now you multiply each term by either $Z^3, Z^2, Z^1$ or $Z^0 = 1$ so that the sum of the exponents of each term is $3$. So $X^3$ and $Y^3$ gets multiplied by $1$, $X^2$ gets multiplied by $Z$ and so on. You end up with $$ X^3+Y^3−ZX^2+Z^2Y−Z^3 ...


1

You are almost right: since $\phi\widehat{\phi}=2$ by definition of dual and $\phi^2=-2$, you have $\widehat{\phi}=-\phi$. Now you should get the right result in (b), your argument is correct!


1

LONG SOLUTION. The equations defining the parabola can be rewritten as parametric equations as follows: $$ \cases{ x=t \cr \displaystyle y={1\over2}t^2-{1\over2}\cr \displaystyle z={1\over2}t^2+{1\over2}\cr } $$ The parametric equation of the line passing through the origin and a point $P(t)=(t,\ t^2/2-{1/2},\ t^2/2+{1/2})$ of the parabola is just ...


1

If $E$ is an elliptic curve over $\mathbb{F}_q$, then the Weil pairing $E(\mathbb{F}_q)\times E(\mathbb{F}_q)\rightarrow \mathbb{F}_q^*$ shows that there exist positive integers $m_1,m_2$ such that $$ E(\mathbb{F}_q)\cong \mathbb{Z}/m_1 \mathbb{Z} \times \mathbb{Z}/m_2 \mathbb{Z}, $$ with $m_1\mid gcd(m_2,q-1)$, see Chapter III, Corollary $8.1.1$ in ...



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