# Tag Info

1

What makes Fermat's equation really hard is the presence of one of the unknown variables, n, as an exponent. This makes it an example of an "exponential Diophantine equation". If one has a "polynomial Diophantine equation", (polynomial function in some number of variables) = 0, then finding rational solutions is still rather difficult, but not usually quite ...

0

I was disposed to solve your question but the point $R=(63,19)$ does not belong to the curve $E$ so have no sense neither, the addition with $R$ nor the duplication point $2R$. If you edit your post, I could give an answer maybe. I wonder a related problem could be find the point $R=(x,y)$ in order to have your relations (not forgetting that $nP$ is ...

1

Since $+$ is a rational map $E \times E \to E$, and this map is the restriction of $+$ to $E \times \{P\}$, it is a rational map.

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It is probably two years too late to matter, but here is an answer to your question: Let us denote the four points that you found by $P_{-\infty}=(1 : -1 : 0), P_{+\infty}=(1 : 1 : 0), P_1=(-1 : -1 : 3)$ and $P_2=(-1 : 1 : 3)$. Note that the points as given are in weighted homogenous space, and are NOT the coordinates for $x$ and $y$. What you want to ...

6

Let $m$ and $n$ be relatively prime integers and let$$A = |(m^3 + 32n^3)m|,\text{ }B = |4(m^3 - 4n^3)n|.$$Denote by $D$ the greatest common divisor of $A$ and $B$. In order to show the inequality in question, it suffices to show that $D$ is a divisor of $144$. For, if that is the case and if the $x$-coordinate of $P$ is given by $m/n$ ($n \neq 0$) in lowest ...

3

Because it doesn't induce an automorphism of the function field of the curve. Automorphisms of an elliptic curve $E$ are endomorphisms $\varphi \colon E\to E$ which have degree $1$. Recall that the degree is defined as $[k(E)\colon \varphi^*(k(E)) ]$, where $k(E)$ is the function field of the curve and $\varphi^*$ is the induced map on $k(E)$. The Frobenius ...

3

Here’s how to do it for an elliptic curve over $\Bbb Q$: Start with $y^2=x^3+ax+b$, with the neutral point $\Bbb O$ up at infinity. So you homogenize and then set $y=1$ to get $z=x^3+axz^2+bz^3$. I like to think of this as a recursive schema for expanding $z$ as a series in $x$, but you can do it in various ways, to get  z = x^3 + ax^7 + bx^9 + ...

0

An easier read than Silverman's Arithmetic is Silverman-Tate's Rational Points on Elliptic Curves which is aimed at the undergraduate level but still touches on more advanced topics like Galois representations which is heavily researched, along with looking at elliptic curves over finite fields. This also includes a chapter on complex multiplication and an ...

0

Here’s a worm’s-eye view, perhaps not much help to you. You may always start with a supersingular curve in characteristic $p$, like $p=13$, where $y^2=x^3+x+4$ is supersingular. I choose $13$, because neither $y^2=x^3-x$ nor $y^2=x^3+1$ is supersingular there. Then you may always take your constant supersingular curve and jiggle it algebraically, here you ...

3

Elliptic curves are very famous examples of algebraic varieties. You seem to have forgotten that the function can have more than one variable; take the function $f(x,y)=y^2-x(x^2+5)$, then $E$ is just the variety $V(f)$.

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I think that the best alternative to Silverman's book is: Dale Husemoller's "Elliptic Curves". It is a bit softer, but still comprehensive and rigorous. It also includes a brief introduction to research topics: Birch-Swinnerton-Dyer, modularity, Calabi-Yau varieties, cryptography, topological modular forms... Also, the references are quite extensive.

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Let's lift the curves into projective space. We then have the map $\phi:E \rightarrow E$, $\phi([x:y:z])=[x:-y:z]$. Assume that $[x:y:z] \neq [0:1:0]$, the point at infinity. This means $z \neq 0$ so we may divide by it. Then we can rewrite the map as $\phi([x:y:z])=[x/z:-y/z:z/z]=[x/z:y/z:1]$. Now we see that $\Phi_x=x/z$ which is of the form $g/h$ where ...

1

Let $E_1, E_2$ be elliptic curves. Fix a point $Q \in E_2$. Then the map $\phi : E_1 \rightarrow E_2$ defined by $P \mapsto Q$ is a constant morphism. This is because it is a rational map (it is just effectively a constant polynomial) and is defined everywhere trivially.

4

Yes, they are. In fact, note that if $E,E'$ are $k$-isogenous, then their $l$-adic Tate modules are isomorphic as $Gal(\overline{k}/k)$-modules, where $l$ is any prime number not lying below $v$. Since $v$ is a place of good reduction, the Tate modules of the reduced curves are isomorphic as $Gal(\overline{\mathbb F_v}/\mathbb F_v)$-modules. Now a theorem of ...

2

$|n-(p+1)| \le 2 \sqrt{p}$ is a stronger statement than $n - (p+1) \le 2 \sqrt{p}$. Rephrasing the theorem in the way you suggest would weaken the conclusion. The point of Hasse's theorem is that the curve has "roughly" $p+1$ points. It gives a bound on how far the actual number of points, $n$, can deviate in either direction from $p+1$ (by no more than $2 ... 0 There is a group morphism$\rho : E_m(\Bbb Q) \to E_m(\Bbb Z_5)$obtained by reduction modulo$5$. If$\rho(P) \notin 2E_m(\Bbb Z_5)$, then you can deduce that$P \notin 2E_m(\Bbb Q)$. You have to check this for all$4$possible choices of$m$mod$5\$.

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