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Let X be the union of two planes in $\mathbb{A}^4$ touching at origin. Blow up X at the origin. Call it $\overline{X}$. It has two disjoint copies of $\mathbb{A}^2$ blown up at the origins. Their exceptional divisors are $E_1$ and $E_2$ say. It is clear that $E_1$ and $E_2$ are isomorphic to $\mathbb{P}^1$. Choose an isomorphism between them and glue $\overline{X}$ via this isomorphism. My question is : Is this Glued scheme quasi-projective?

I think the ample cartier divisor $\mathcal{O}(-E_1-E_2)$ descends to an ample cartier divisor which makes the Glued scheme quasi projective.

Let $X$ is an irreducible surface got by identifying two smooth points of a smooth irreducible surface $Y$. The completion of the local ring at the only singular point of $X$ is the above ring i.e, two planes in $\mathbb{A}^4$ touching at a point. Then we do the same construction. That is Blow up X at the singular point, identify two exceptional divisors and get a new scheme Z. Is it quasi projective??

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    $\begingroup$ I don't understand the question. If $\bar X$ is the blow up of $X$ at the origin, then it's clearly quasi-projective since $X$ is. $\endgroup$ – R. van Dobben de Bruyn Aug 20 '17 at 12:38
  • $\begingroup$ I am saying the glued up scheme is also quasi projective $\endgroup$ – user100841 Aug 20 '17 at 12:43
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Let $Y$ be the total space of the vector bundle $O(-1)\oplus O(-1) \to \mathbb{P}^1$. Then your glued up scheme is isomorphic to the subscheme of $Y$ given by the union of the two closed subschemes given by the total spaces of the two line bundle $O(-1)\to \mathbb{P}^1$. So yes, your scheme is quasi-projective.

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  • $\begingroup$ yes meanwhile i have shown $\mathbb{P}^n\coprod \mathbb{P}^n /\{\mathbb{P}^1=\mathbb{P}^1\}$ is projective (by writing the equation of this variety). Our glued up scheme can be seen as a subvariety of this $\mathbb{P}^n\coprod \mathbb{P}^n /\{\mathbb{P}^1=\mathbb{P}^1\}$ for some suitable large n. So it is quasi projective. $\endgroup$ – user100841 Aug 25 '17 at 16:36
  • $\begingroup$ :Let $X$ is an irreducible surface got by identifying two smooth points of a smooth irreducible surface $Y$. The completion of the local ring at the only singular point of $X$ is the above ring i.e, two planes in $\mathbb{A}^4$ touching at a point. Then we do the same construction. That is Blow up X at the singular point, identify two exceptional divisors and get a new scheme Z. Is it quasi projective?? $\endgroup$ – user100841 Aug 25 '17 at 16:44
  • $\begingroup$ I am guessing this is correct because again X Blown up at the singular point can be embedded in to some $\mathbb{P}^n$. So we are identifying two disjoint lines in that $\mathbb{P}^n$ which have the same degree in $\mathbb{P}^n$, and asking whether it is projective. It is because the $\mathbb{O}_{\mathbb{P}^n}(1)$ descends to the $\mathbb{P}^n$ glued along the lines. $\endgroup$ – user100841 Aug 25 '17 at 17:45

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