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2d

comment 
Choosing a group action to do GIT of hypersurfaces
@John: Not every linear representation of $\textbf{SL}_{n+1}$ factors through $\textbf{PGL}_{n+1}$. Thus, for an action of $\textbf{PGL}_{n+1}$ on $X$, for the induced action of $\textbf{SL}_{n+1}$, a $\textbf{SL}_{n+1}$linearized ample invertible sheaf $\mathcal{O}(1)$ may not factor through a $\textbf{PGL}_{n+1}$linearization. Abstractly, we could just pass from $\mathcal{O}(1)$ to $\mathcal{O}(n+1)$. However, this is analogous to passing from the ideal of a projective variety to the ideal of its Veronese image, already nontrivial for projective space. 
Oct
1 
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commutable matrices
I do not understand your question. What mathematical operations do you want to perform? 
Oct
1 
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Grothendieck problem
I cannot say immediately that your formulation is the same as the standard formulation of the GrothendieckKatz conjecture, but they look quite similar. 
Oct
1 
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Grothendieck problem
Are you talking about the GrothendieckKatz conjecture? Here is the link to the Wikipedia page: en.wikipedia.org/wiki/…. 
Oct
1 
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Open cover not containing a certain subcover
You are right. Countably infinite spaces do not work. 
Oct
1 
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Open cover not containing a certain subcover
@JeremyRickard: Actually, I think your example works for countably infinite spaces as well. 
Sep
30 
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Embedding of a proper scheme into a smooth one
Also, this new question has strong motivation: such immersions are used in the main construction of Todd homomorphisms in the singular GrothendieckRiemannRoch theorem of BaumFultonMacPherson. 
Sep
30 
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Embedding of a proper scheme into a smooth one
Francesco's answer is completely correct. It does raise another question, that has been much studied: does there exist an immersion from $X$ into a smooth algebraic space? There are still counterexamples if $X$ is allowed to be nonseparated, cf. EdidinHassettKreschVistoli. However, to the best of my knowledge, this question is open for separated $X$ (there is a lovely article of Totaro on this question). 
Sep
30 
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Scheme of irreducible components
You might read Raynaud's article on specialization of the Picard functor. For a specializing family of curves (as in the previous two comments), the nonseparated scheme of irreducible components does play a role in the analysis of the nonseparated scheme that is the closure of the zero section in the nonseparated Picard scheme. 
Sep
28 
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Torsion ideal in symmetric algebra
@AlgirdasRugys: "So in the symmetric algebra $S(M)$ the exists the unique prime ideal ..." I do not understand how this relates to the previous discussion. 
Sep
28 
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Torsion ideal in symmetric algebra
@AlgirdasRugys: Yes, the symmetric algebra of a vector space $V$ with a specified basis $B$ over a field $F$ is isomorphic to a ring of polynomials with coefficients in $F$ and indeterminate set $B$. This is an integral domain. 
Sep
28 
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compact almost complex submanifolds of complex Lie groups
What do you mean by an "almost complex submanifold" of a complex manifold such as $G$? 
Sep
28 
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Torsion ideal in symmetric algebra
The image of $S_R^\bullet(M)$ in $U^{1}S_R^\bullet(M)$ is precisely the quotient of $S_R^\bullet(M)$ by the ideal you mention. Finally, since $M\otimes_R F$ is a vector space over the field $F$, the symmetric algebra $S_F^\bullet(M\otimes_R F)$ is an integral domain. 
Sep
28 
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Torsion ideal in symmetric algebra
Yes, that is true. I refuse to refer to a ring by $D$, since that is too often used for both (a) division algebras, and (b) rings of differential operators. Let $R$ be a commutative ring that is an integral domain. Let $U$ be the multiplicative system of all nonzero elements of $R$. Let $R\to F$ be the localization of $R$ at $U$, i.e., the fraction field. Then for any $R$modules $M$ and $N$, $(M\otimes_R N)\otimes_R F$ is canonically isomorphic to $(M\otimes_R F)\otimes_F (N\otimes_R F)$. Thus the natural map $U^{1} S_R^\bullet(M)\to S_F^\bullet(M\otimes_R F)$ is an isomorphism . . . 
Sep
28 
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Rational connectedness of certain subvarieties of the linear series
Since smooth surfaces in $\mathbb{P}^3$ have trivial $\text{Pic}^0$, every irreducible component $V_i$ of $V$ is the image of the product of the complete linear systems for the irreducible components of a curve in $X$ parameterized by a general point of $V_i$. Thus the component is unirational (possibly even rational). The example of curves in a smooth quadric illustrates this, for instance. 
Sep
28 
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Dual involution on the $Ext^1$
"If we suppose that $F$ is stable then $E$ is stable for all non trivial extension ..." Unfortunately this is not correct, for instance, on $\mathbb{P}^1$. The semistable bundle $\mathcal{O}^{\oplus 2}$ is a nontrivial extension of $\mathcal{O}(+1)$ by $\mathcal{O}(1)$. 
Sep
28 
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Rational connectedness of certain subvarieties of the linear series
The connected components may be reducible. For instance, for $X$ a smooth quadric surface and $a=2$, your variety $V$ will have three irreducible components. The normalizations will be isomorphic to $\mathbb{P}^1\times \mathbb{P}^5$, $\text{Sym}^2(\mathbb{P}^3)$ and $\mathbb{P}^5\times \mathbb{P}^1$, respectively. The intersection of the components is the image of $\mathbb{P}^1\times \mathbb{P}^3\times \mathbb{P}^1$. Did you want to restrict to an irreducible component of $V$? 
Sep
28 
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Dual involution on the $Ext^1$
One correction: actually, the fixed points will be those short exact sequences such that $E$ admits a nondegenerate pairing such that $F^*$ is isotropic and such that the induced pairing of $F^*$ with $F=E/F^*$ is the standard pairing. However, there is no reason this pairing must be skewsymmetric. It might be symmetric, for example. 
Sep
28 
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Dual involution on the $Ext^1$
I am not sure of an explicit involution, but certainly the involution need not be the identity (it seems that the sequence will be fixed by the involution if and only if there exists a skewsymmetric, nondegenerate pairing on $E$ such that the image of $F^*$ is Lagrangian). One nonfixed example is the short exact sequence on $\mathbb{P}^1$, $$0\to \mathcal{O}(3)\oplus \mathcal{O}(1)\to \mathcal{O}(1)^{\oplus 3}\oplus \mathcal{O}(3) \to \mathcal{O}(1)\oplus \mathcal{O}(3) \to 0.$$ 
Sep
27 
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Standard techniques on rationally connected varieties
I am saying that a curve $C$ in $\mathbb{P}^3$, whether or not it is a complete intersection, is singular if and only if it contains a closed subscheme that is the image of $A$ under a projective automorphism of $\mathbb{P}^3$. Indeed, being singular at $p$ precisely means that the curve contains a $2$dimensional subspace of the Zariski tangent space of $\mathbb{P}^3$ at $p$. 