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14h
comment How to understand $d^2=0$ in differential form?
Dear @Najib: you are certainly right, but I confess that I didn't check the original source and only transcribed Cartan's quotation from the epigraph to a book by Gelfand-Manin .
17h
answered How to understand $d^2=0$ in differential form?
20h
revised Cl(Spec A) = 0 implies A is a UFD
added 444 characters in body
21h
answered Cl(Spec A) = 0 implies A is a UFD
1d
answered When is a vector field on a manifold restricted to a submanifold $X$ a vector field on $X$?
1d
comment Biholomorphic in two Complex Variables
If you think that the second reference Beispiel zweier ganzer Funktionen zweier komplexer Variablen, welche eine schlichte volumtreue Abbildung des $\mathcal{R}_4$ auf einen Teil seiner selbst vermitteln is in French, then I have big news for you, mon ami.
1d
comment About Plücker embedding
Dear Brenin: I don't think there is a canonical isomorphism $\mathbb P(V)\cong \mathbb P(V^\vee)$ and I dont understand how you can describe a hyperplane by a map $ V\to H\to 0 $ .
Nov
22
comment Do two isomorphic finite field extensions have the same dimension?
Isomorphic as what?
Nov
22
revised Intrepretation of $H^i(X, \mathcal{O}_X) = 0$
added 421 characters in body
Nov
22
comment Intrepretation of $H^i(X, \mathcal{O}_X) = 0$
Dear @Alex, yes there is more to say of course: I hope you will add some of your insights and I might come back to this answer. As for the Picard variety, it is smooth in characteristic zero ( Cartier's thesis) so that indeed the dimension of the Picard variety is the same as the dimension of its tangent space at the origin, namely $h^1(X,\mathcal O_X)$. [Igusa gave an example of a non-reduced Picard scheme in char. $p$] I didn't want to enter these technicalities in my post, which I tried to make readable also by beginning algebraic geometers.
Nov
22
answered Intrepretation of $H^i(X, \mathcal{O}_X) = 0$
Nov
22
comment Intrepretation of $H^i(X, \mathcal{O}_X) = 0$
@Martin Brandenburg: For quasi-coherent sheaves (like here $\mathcal O_X$ ), étale cohomology coincides with Zariski cohomology.
Nov
21
awarded  Enlightened
Nov
21
awarded  Nice Answer
Nov
21
comment Let $p: E \to B$ be a covering map. If $B$ is a completely regular space then prove that (edited) $E$ is completely regular space.
Here is a complete answer by David White on MathOverflow.
Nov
21
comment Let $p: E \to B$ be a covering map. If $B$ is a completely regular space then prove that (edited) $E$ is completely regular space.
Such a modification will obviously not be continuous. Anyway, this site is not for long dialogues trying to extract some precise answer from users: unless you furnish what mathematicians agree is a proof, I consider you haven't answered the question and will no longer comment on your "hints".
Nov
21
comment Let $p: E \to B$ be a covering map. If $B$ is a completely regular space then prove that (edited) $E$ is completely regular space.
I don't think this has any chance to work because all connected components of $p^{-1}(p(N)) $ except $N$ might be included in $V$ and then the lift $f\circ p$ of your function $f$ downstairs will take the value $0$ at all points $ y\in p^{-1}(p(x))$ of the fiber of $p(x)$ with $y\neq x$. How are you going to modify $f\circ p$ in order to obtain a function taking the value $1$ at all those $y$'s, given that you don't know whether $V$ is completely regular?
Nov
21
comment Submanifold associated to blow up.
You are welcome, ted, and I'm happy that all's well that ends well!
Nov
21
comment Submanifold associated to blow up.
Dear Ted: oh but yes, the $X\cap U_i$'s are open in $X$ : you should review the concept of induced topology.
Nov
21
comment Let $p: E \to B$ be a covering map. If $B$ is a completely regular space then prove that (edited) $E$ is completely regular space.
Dear aes, I don't think this works at all. If you disagree, please post a complete solution instead of alluding to undefined modifications (every function is a modification of any other function, after all ). And by the way, it doesn't make sense to say that $N$ satisfies the covering condition.