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seen Feb 7 '13 at 18:18

Dec
23
accepted Is the complement of an ample divisor always affine
Dec
23
asked Is the complement of an ample divisor always affine
Dec
23
comment Relationship between complex number and vectors
If $z$ is a complex number, then $(Re(z),Im(z))$ is a vector in $\mathbf{R}^2$.
Dec
22
comment Which algebraic variety can become a algebraic group?
The Euler characteristic of an algebraic group should be zero. This follows from the trace formula. In fact, for any non-trivial element $a$, the translation $t_a$ by $a$ has no fixed points. By the trace formula, the trace of $t_a$ on the cohomology of your algebraic group should be zero. The latter equals the euler characteristic. In particular, this shows that in dimension $1$, the genus has to be zero because the Euler characteristic equals $2g-2$. Moreover, note that the Euler characteristic of a torus $\mathbf C^g/Lambda$ is indeed zero thus it all works out.
Dec
22
comment Intuition behind isomorphism of algebraic varieties
Say $V$ and $W$ are isomorphic algebraic sets. Intuitively, it should mean that the set of equations defining $V$ can be transformed into the set of equations defining $W$. For instance, consider two curves $f(x,y) =0$ and $g(x,y)=0$ in $\mathbf{A}^2$. These being isomorphic means that there is an isomorphism $k[x,y]/(f)\to k[x,y]/(g)$.
Dec
22
accepted Does this equation have integer solutions
Dec
21
asked Does this equation have integer solutions
Dec
21
accepted Why is the rank of $f_\ast L$ the degree of $f$
Dec
21
comment If $X\times C$ is isomorphic to $Y\times C$, does it follow that $X$ is isomorphic to $Y$
Thank you! What an amazing result by Fujita!
Dec
21
asked Why is the rank of $f_\ast L$ the degree of $f$
Dec
21
comment If $X\times C$ is isomorphic to $Y\times C$, does it follow that $X$ is isomorphic to $Y$
Thank you very much! Your example did convince me that this is not going to be true. A counterexample shouldn't be hard to find, but it will be cumbersome.
Dec
21
revised If $X\times C$ is isomorphic to $Y\times C$, does it follow that $X$ is isomorphic to $Y$
added 60 characters in body
Dec
21
comment If $X\times C$ is isomorphic to $Y\times C$, does it follow that $X$ is isomorphic to $Y$
What if I take $X$ and $Y$ to be of dimension at least $2$? Then this phenomenon can't happen.
Dec
21
asked If $X\times C$ is isomorphic to $Y\times C$, does it follow that $X$ is isomorphic to $Y$
Dec
20
accepted Abelian subvarieties of a principally polarized abelian variety are principally polarized
Dec
20
revised Abelian subvarieties of a principally polarized abelian variety are principally polarized
added 81 characters in body
Dec
20
revised Abelian subvarieties of a principally polarized abelian variety are principally polarized
added 366 characters in body
Dec
20
asked Abelian subvarieties of a principally polarized abelian variety are principally polarized
Dec
6
answered Shrinking the base field of a scheme of finite type over a field
Dec
3
comment Are there infinitely many rational functions of bounded degree and given ramification
@QiL No. I want to fix $X$. Then I want to find an integer $n$ and a finite set of points $R$ such that there are infinitely many $X\to \mathbf P^1(\mathbf C)$ of degree at most $n$ and with ramification inside $R$. For instance, if $g>1$, then $n=2$ won't work. (There is a unique hyperelliptic map $X\to \mathbf{P}^1$ if it exists at all.) You'll need to take $n$ "big enough" and $R$ also "big enough". (If I take $R$ small there aren't any $X\to \mathbf{P}^1(\mathbf{C})$ with ram locus inside $R$.)