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seen Feb 7 '13 at 18:18

Feb
3
comment An example of decomposing a projective variety
Have you tried showing your variety is irreducible? You could use the "dictionary" between closed subsets of projective $n$-space and ideals of k[a_0,\ldots,a_n]$. Being irreducible as a closed subset usually boils down to the polynomials being irreducible.
Feb
2
comment Is the number of automorphisms of a hyperelliptic curve bounded
The curve $y^2=x^{2g+1}+1$ admits the automorphism $(x,y)\mapsto (\zeta_{2g+1}^n x, y)$ for all $n=1,\ldots,2g+1$. Thus, the number of automorphisms of this hyperelliptic curve is at least $2g+1$. Is this correct?
Jan
29
comment Elliptic curves over Spec Z
I wanted to be sure myself, but let $E$ be an elliptic curve over $\mathbf Q$ with good reduction over $\mathbf Z$. Then its minimal regular model coincides with its minimal Weierstrass model.
Jan
25
comment Doesn't Abhyankar's lemma contradict faithful flat descent (no, but I'm confused)
Yes, I just wanted to be sure. Thank you for this. It completely cleared up everything.
Jan
25
comment Doesn't Abhyankar's lemma contradict faithful flat descent (no, but I'm confused)
I think I understand what I'm doing wrong now. Starting with $Y\to X$ as in the question, the base change gives $Y\times_X X^\prime \to X^\prime$. This is not necessarily etale, but once you normalize you obtain $Y^\prime\to X^\prime$ and this morphism is etale by Abhyankar. Now, if by some chance $Y^\prime = Y\times_X X^\prime$ (i.e., the base change is normal) then faithfully flat descent implies that the morphism we started out with was already etale. Thank you very much QiL!!
Jan
25
comment Doesn't Abhyankar's lemma contradict faithful flat descent (no, but I'm confused)
By "some maximal ideal" do you mean "any maximal ideal", or does Abhyankar really mean that there is maybe just "one" maximal ideal?
Jan
24
comment Is a group with only finitely many subgroups of index n (for all n) finitely generated
Great! I guess one can also use this to obtain a non-abelian example by taking the product of $\mathbf Q/\mathbf Z$ with a finite non-abelian group $G$. Then $\mathbf Q/\mathbf Z \times G$ is not finitely generated, but it also has only finitely many subgroups of index $n$, right?
Jan
20
comment Is $M_g$ a subvariety of $M_{h}$ for some $h>g$
No, but maybe something like Knudsen clutching can be used?
Jan
19
comment Is the relative rank function with respect to an ample line bundle non-decreasing
Ok. I should have thought of that...I can still save the situation. In my case, $\mathcal L = \Omega^1_{X/K}$. What can we do in this case? Of course, if you take a rational variety this is not going to work, but let's assume $X$ is such that $h^0(\Omega_{X/K}) >0$.
Jan
19
comment Is the relative rank function with respect to an ample line bundle non-decreasing
I fixed some typos and inaccuracies. The answer is still incomplete though. I need to figure out why $\mathcal L \cong \mathcal O_X(D)$ with $D$ effective. Do you know if this is true?
Jan
13
comment Are there moduli spaces of higher-dimensional varieties
When I say moduli stack I mean a Deligne-Mumford stack solving the moduli problem. Sorry for not saying that explicitly.
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
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
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
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
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 $R$.)
Dec
2
comment Are there infinitely many rational functions of bounded degree and given ramification
I've edited the question. I admit the question was poorly written. Thanks for the comments. I hope it's more clear now.
Dec
2
comment Are there infinitely many rational functions of bounded degree and given ramification
I just edited the question. I forgot to put the important condition that the rational function $X\to \mathbf{P}^1(\mathbf C)$ is ramified only inside $R$.