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The answer is no. This is a corrected version of Nicolast's comment. Let $E$ be an elliptic curve, let $f: E \to E$ be an endomorphism and let $H_1(f) : H_1(E) \to H_1(E)$ be the induced map on $H_1$. …

$\def\CC{\mathbb{C}}\def\HH{\mathbb{H}}\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}\def\Id{\mathrm{Id}}$These geodesics are always algebraic. We can understand their equations using the classical modular cu …

Any construction along these lines is going to run into an obstruction pointed out by Serre. Consider the elliptic curve $E = \{ y^2 = x^3+x \}$ over $\mathbb{Z}[i]$, and let $p$ be a prime which is $ …

I don't know if this is the kind of answer which will satisfy. Write $f$ for the function on $\mathbb{C} - \Lambda$ and $z$ for the coordinate on $\mathbb{C}$. Write $D^{\ast}$ for the punctured disc. …

In case you don't know the general context: There is a curve of genus $g$ with endomorphism group contained in $G$ if and only if $G$ can be generated by elements $g_1$, $g_2$, ..., $g_k$ with orders …

I wrote up notes for the 4 lectures I did going through a completely algebraic proof at the end of a Shavarevich based algebraic geometry course. I think this is a nice approach in that it introduces …

Joe Harris, as recorded in his course notes here, gives the following slick proof when both $D$ and $K-D$ are effective; it has the advantage of never mentioning $H^1$. See lecture 1 for this argument …

This answer is basically a longer version of Felipe Voloch's, but maybe it will be useful. Both proofs take a class in $H^1(E)$, pull it back to $H^1(\mathbb{P}^1)$ and note that $H^1(\mathbb{P}^1)$ i …

No. Letting $\sigma$ and $\tau$ denote the two involutions, $\sigma \circ \tau$ is a translation by an element of $\mathrm{Pic}^0(C)$. In general, this translation will not be torsion, so its orbit th …

I think this will be a needlessly confusing example. In algebraic geometry over an algebraically closed field, there are two basic examples of nonnormal curves: the node and the cusp. Explicit equatio …

I'd look into numerical homotopy software, such as Bertini or PHCPack. Numerical homotopy software attempts to solve problems of the following sort: Suppose we have a family of polynomial equations $f …

Every fiber of $A$ is a projective space. If $V = H^0(X, \mathcal{O}(D))$, then $A^{-1}(A(D))$ is naturally identified with $\mathbb{P}(V)$. A divisor $E$ in $A^{-1}(A(D))$ is the zero locus of a nonz …

In a comment above, marco asks whether this is true for larger $n$: That is to say, $M$ a complex $n$-fold, $R$ a totally real sub-real-$n$-fold and $\alpha$ a $(1,1)$-form on $R$. The answer is no fo …

Here is the most algebraic way I can see to compute this. Let $Q_1$ and $Q_2$ be two quadratic polynomials in four variables. Let $R$ be the graded ring $k[x_1, x_2, x_3, x_4]/(Q_1, Q_2)$. Let $V_d$ b …

Here is a particularly simple example. We will take a cylinder and glue its ends together in two different ways. The cylinder will be $$C:=\{ z: 0 \leq \mathrm{Im}(z) \leq 1 \} / \mathbb{Z},$$ where …