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Angelo
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revised
Why does a complex linear normalization of a real algebraic surface inherit a real structure?
added 1573 characters in body
answered
Why does a complex linear normalization of a real algebraic surface inherit a real structure?
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Fiber product arising from reductive group action on varieties
A commonly used notation, which avoids the confusion with fibered products, is $G \times^{P}Y$.
answered
Do Henselian extensions have left lifting property with respect to smooth morphisms?
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Zero in colimit of sheaves category
Isn't this colimit always 0?
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Is the completion of an infinitely generated module, again infinitely generated
Completing a module that is not finitely generated is often very bad for its health. That's why derived completion (a gentler kind of completion) was invented (just do a google search).
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Is the completion of an infinitely generated module, again infinitely generated
The $\frak m$-adic completion could even be 0 (for example, when $A$ is domain and $M$ is its fraction field).
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Chern character of coherent sheaf on singular variety
You can not push forward classes from an open subset. The Chern character does not exist, in general; the closest that comes to my mind is the Riemann-Roch map.
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Does there exist a genus $g$ curve over $\mathbb{Q}$ with every type of stable reduction?
Oops, sorry, this was a really moronic comment.
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Does there exist a genus $g$ curve over $\mathbb{Q}$ with every type of stable reduction?
A smooth curve over $\mathbb Q$ only has finitely many points of bad reduction.
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Does the Cantor-Schröder-Bernstein Theorem hold in the category opposite to the category of noetherian commutative rings?
The embedding $\mathbb{Z} \times \mathbb{Z} \subseteq \mathbb{Z} \times \mathbb{Q}$ is such an example.
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Formally smooth maps of schemes and factorization systems
No, they are not. See mathoverflow.net/questions/22015/…. And any case Wikipedia is not very authoritative.
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Formally smooth maps of schemes and factorization systems
Formally smooth morphisms only have the property above with respect to nilpotent embeddings of affine schemes.
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Killing a Brauer class by a flat projective morphism
I'd be happy to, but I really don't like to interact with anonymous users, so could you please send me a private email at <angelo dot vistoli at sns dot it> telling me who you are? Thanks.
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Killing a Brauer class by a flat projective morphism
A Brauer-Severi scheme with a section has a trivial Brauer class.
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Given a zero-dimensional ideal $(f_1,...,f_n)$, is the ideal $(f_1-\alpha_1,...,f_n-\alpha_n)$ also zero-dimensional?
Sorry, I misread the question. The answer should be positive, by semicontinuity of the fiber dimension (EGA IV 13.1.3).
revised
Is $H^i(\mathcal{M}_g,F)$ necessarily finite dimensional for a coherent sheaf $F$?
edited body
answered
Is $H^i(\mathcal{M}_g,F)$ necessarily finite dimensional for a coherent sheaf $F$?
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Is $H^i(\mathcal{M}_g,F)$ necessarily finite dimensional for a coherent sheaf $F$?
Why should that be true? In fact, a quasiprojective variety $X$ with the property that $\mathrm{H}^i(X, F)$ is finite-dimensional for all locally free sheaves $F$ is projective.
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The work of mathematicians outside their professional environment
Work habits differ wildly, but I would imagine that almost all mathematicians work outside their office some, or most, of the time.
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