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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 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 …

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 …

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 …

I think that you have understood the analogy correctly, and you have pinpointed one of its weaknesses. Although number fields are like one dimensional functional fields in many ways, one of the differ …

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 …