I want to prove that given $A \to B$ a ring homomorphism of finite type, then the induced morphism of schemes $X \to Y$ is quasi-projective.
A morphism is quasi-projective if it factors into an open immersion followed by a projective morphism.
My attempt is as follows:
$B= A[X_0,\dots,X_n ] / I$, so there is a closed immersion $Proj B \to \mathbb{P}^n_{A}$; composing it with the projection $\mathbb{P}_A^n \to Spec A$, we have a projective morphism $Proj B \to SpecA$. I tried to find an open immersion $Spec B \to Proj B$ but I couldn't. Is this the right approach for this problem ?
Best Answer
From $B=A[X_1,\ldots,X_n]/I$, you can conclude that you have a closed immersion $\mathrm{Spec}(B)\hookrightarrow \mathbb A^n_A$. You have the standard open immersion $\mathbb A^n_A\hookrightarrow\mathbb P^n_A$. The morphism $\mathrm{Spec}(B)\to\mathrm{Spec}(A)$ now factors as $$\mathrm{Spec}(B)\hookrightarrow \mathbb{A}^n_A \hookrightarrow \mathbb{P}^n_A \twoheadrightarrow \mathrm{Spec}(A)$$ We have here a closed immersion followed by an open immersion and you can write that as an open immersion followed by a closed one. That's exactly what you want.