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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 ...

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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!

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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 ...

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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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