Let $k$ be a field, and consider $X$ the spectrum of the dual numbers $k[x]/(x^2)$. This affine scheme is a closed subscheme of the affine line. Moreover, the affine line is open in the projective line, making $X$ a quasi-projective scheme. Does $X$ have any chance of being projective over k?
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2Why do you think that $X \to \Bbb P^1$ is not a closed embedding? – LurchiDerLurch Oct 17 '22 at 12:33
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Using this, we conclude that it suffices to verify the claim with respect to the standard open affine cover of $\Bbb{P}^1$, namely by $\mathrm{Spec}\:k[x] \cong \mathbb{A}^1$ and $\mathrm{Spec}\:k[x^{-1}] \cong \mathbb{A}^1$. $\mathrm{Spec}\:k[x]/(x^2)$ embeds into $\mathrm{Spec}\:k[x]$ using the quotient map $k[x]\to k[x]/(x^2)$ and hence is a closed immersion. This defines a map to $\mathbb{P}^1$. The image of the unique point of $\mathrm{Spec}\:k[x]/(x^2)$ is $(x)\in \mathrm{Spec}\:k[x]$, and in particular does not lie in $\mathrm{Spec}\:k[x^{-1}]$. So, the condition of being a closed immersion is satisfied vacuously over $\mathrm{Spec}\:k[x^{-1}]$ and we are done.
Alekos Robotis
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