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The stronger version of the Nullstellensatz states that a maximal ideal $I$ of $R=k[x_1,\ldots,x_n]$ is the kernel of a $k$-homomorphism from $R$ to $L$ where $L/k$ is a finite extension. Let $a_1,\ld …

Yes. In zero characteristic the image of an isogeny of elliptic curves is determined up to isomorphism by its kernel. Your isogeny has the same kernel as the doubling map from $E$ to itself.

There is a whole industry devoted to this. The basic method is by descent, which is a formalized version of the infinite descent proofs of Fermat and Euler. It helps if there are rational 2-torsion po …

The compactification is the usual one coming up in the theory of modular forms, with the cusps being orbits of $\Gamma$ on $\mathbb{Q}\cup\{\infty\}$. As for the proof, I like this paper by Bernhard …

The standard example is to let $X$ be the affine plane over a field, and $U=X-\{(0,0)\}$.

I'm not sure about what "functorial" would entail here, but at least when $p\ge5$ from a naive point of view things are quite simple. Once one knows that the $j$-invariant of a supersingular elliptic …

The geometric genus (the dimension of the space of global sections of the canonical sheaf) of projective $n$-space is zero. See Hartshorne II.8.

One thing to do is to try to express these in terms of squares. Note that $$12x(3x-1)=36x^2-12x=(6x-1)^2-1$$ so that your equations become $$a_1^2+b_1^2=c_1^2+1$$ and $$a_1^2-b_1^2=d_1^2-1$$ where $a_ …

Under the assumption that $E[n]\subseteq E(\mathbb{F}_q)$, to compute $E[n]$ I'd avoid using division polynomials, as they rapidly become cumbersome. Rather I would generate random elements of $E[n]$ …

No, $div(f)$ is an element of the free abelian group generated by symbols $[P]$ as $P$ runs through the points of the curve. So a function $f$ having a simple pole at a point $P$ and a simple zero at …

You might consult the following paper John Brillhart & Patrick Morton 'Class numbers of quadratic fields, Hasse invariants of elliptic curves, and the supersingular polynomial', Journal of Number The …

For (b) recall that a prime ideal is the kernel of a map from $G$ into a field. Such a field must be an extension field of $H$ and the image of $G$ is generated by $H$ and the $p^{r_i}$-th roots of ce …

The subgroup $j_* H_2(Q)$ must be generated by twice the generator of $H_2(P^2(\mathbb{C}))$ (I'm dropping the coefficient group from my notation). To see this, your map $\psi$ decomposes as the embed …

Wikipedia calls a module over a commutative Noetherian ring with only one associated prime a coprimary module. I don't recall hearing this terminology elsewhere, but it is certainly common to call a s …