bio | website | homepage.sns.it/vistoli |
---|---|---|
location | Italy | |
age | 56 | |
visits | member for | 4 years, 4 months |
seen | Jan 28 at 15:16 | |
stats | profile views | 18,747 |
My name is Angelo Vistoli. I do algebraic geometry, mostly moduli theory.
Normally I don't answer questions from anonymous users.
Normally I don't answer questions from anonymous users.
Jul 14 |
awarded | Nice Answer |
Jul 2 |
awarded | Curious |
May 12 |
awarded | Good Answer |
Mar 21 |
awarded | Yearling |
Jan 13 |
awarded | Enlightened |
Jan 13 |
awarded | Nice Answer |
Dec 26 |
awarded | Enlightened |
Dec 26 |
awarded | Nice Answer |
Sep 16 |
awarded | Guru |
Sep 15 |
comment |
Descent for group actions
Sorry, I had not read the question properly, I thought that $G$ was the Galois group. |
Sep 15 |
comment |
Descent for group actions
It always descends to the space of invariants in $V$. |
Sep 13 |
answered | Determine complex analytic germ along a smooth compact curve via normal bundle? |
Aug 16 |
awarded | Nice Answer |
Aug 9 |
comment |
Can I conclude that a morphism of vector bundles is zero if it is so fiberwise?
Consider the case that $X = Y$, $f = \mathrm{id}_X$, $\cal U = \cal V = \cal O$, and $X$ is not reduced. |
Jul 30 |
awarded | Enlightened |
Jul 30 |
awarded | Nice Answer |
Jul 25 |
comment |
Vanishing of Tor
$I^{n-1}/I^n$ is a free $R/I$-module, so the statement follows from a simple induction. |
Jul 6 |
comment |
Decomposition of algebra over a commutative artinian ring
Every algebra over a finite product of commutative rings is a product of algebras over the factors (just use the idempotents of the ring to split the algebra). |
Jun 28 |
awarded | Enlightened |
Jun 28 |
awarded | Nice Answer |