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I am following the hint given in Exercise 9.3.H of Ravi Vakil’s notes. It can be found on page 261, here. The exercise states: any finitely presented morphism $\pi:X\to\operatorname{Spec} B$ is a pullback of a finite type morphism $\pi’:X’\to\operatorname{Spec} \mathbb{Z}[x_1,\cdots,x_N]$ for some integer $N$. It is easy when $X$ is affine, but I am stuck considering more general cases.

Any hints or comments would be appreciated. Thank you very much!

KReiser
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Fiona
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  • How did you get on when considering the first part of Vakil's hint (consider $X$ to be the union of two affine open schemes whose intersection is again an affine open subset)? Separately, some typographical advice: using $\operatorname{Spec}$ to format $\operatorname{Spec}$ produces better spacing - I've upgraded your post with this change. – KReiser Aug 14 '21 at 06:50
  • @KReiser I don’t know how to deal with this case, either. I have trouble gluing the two schemes in the position of $X’$ for the two affine open schemes. – Fiona Aug 14 '21 at 13:01
  • Let $X_1,X_2$ be the two affine open subsets which have intersection $X_{12}$. Then the gluing is described by the inclusion maps $X_{12}\to X_1$ and $X_{12}\to X_1$. Since everything in sight is affine, this gives you ring maps - can you check that these depend on a finite amount of data, and therefore can be pulled back from some morphism of finite presentation? – KReiser Aug 14 '21 at 20:27
  • @KReiser Are you suggesting the idea in the answer posted by Evans Gambit? You can see where I got stuck from my comment below. – Fiona Aug 15 '21 at 12:57

1 Answers1

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First step:

Let $X=\operatorname{Spec} A_1\cup \operatorname{Spec} A_2$ such that $\operatorname{Spec} A_1\cap \operatorname{Spec} A_2=\operatorname{Spec} A_{12}$.

Let $A_{12}=B[z_1,\dots, z_n]/(f_1,\dots, f_r)$.

Write $A_1=B[\lambda_1,\dots, \lambda_n]/(g_{11},\cdots, g_{1m})$ and $A_2=B[\mu_1,\dots, \mu_n]/(g_{21},\cdots, g_{2m}).$

Consider the map $A_1\to A_{12}$, and the images of $\lambda_i\mapsto p_i(\underline{z})$.

Similarly consider the map $A_2\to A_{12}$, and the images of $\mu_i\mapsto q_i(\underline{z})$.

Also consider $g_{1j}(p_1(\underline{z}),\dots, p_n(\underline{z}))=\sum_r c_rf_r$ in $B[z_1,\dots, z_n]$.

Similarly $g_{2j}(q_1(\underline{z}),\dots, q_n(\underline{z}))=\sum_r d_rf_r$ in $B[z_1,\dots, z_n]$

Let $C=\mathbb{Z}[\text{coef. of} ~f_i| i][\text{coef. of} ~g_{1i}| i][\text{coef. of} ~g_{2i}| i][\text{coef. of} ~p_i| i][\text{coef. of} ~q_i| i][\text{coef. of} ~c_r| r][\text{coef. of} ~d_r| r]$

Now write $R_1=C[\lambda_1,\dots, \lambda_n]/(g_{11},\cdots, g_{1m})$ and $R_2=C[\mu_1,\dots, \mu_n]/(g_{21},\cdots, g_{2m})$ and $R_{12}= C[z_1,\dots, z_n]/(f_1,\dots, f_r)$.

We observe that the map $\operatorname{Spec} R_{12}\to \operatorname{Spec} R_1$ is an open immersion, similarly the map $\operatorname{Spec} R_{12}\to \operatorname{Spec} R_2$ is an open immersion.

We now glue $\operatorname{Spec} R_1$ and $\operatorname{Spec} R_2$ along $\operatorname{Spec} R_{12}$ to get $X'$.

So we have $X\to \operatorname{Spec} B$ as a pullback of a finitely presented map $X'\to \operatorname{Spec} C$.

Evans Gambit
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  • I think you need to add to $C$ the coefficients of $g_{kl},(k=1,2;,l=1,\cdots,m)$ and that of $h_i^{kl}$ ($g_{1l}(p_1(z),\cdots,p_n(z))=h_1^{1l}f_1+\cdots+h_r^{1l}f_r$ and similarly for $h_i^{2l}$). Besides, how can you show $\operatorname{Spec}R_{12}$ is an open subscheme of $\operatorname{Spec}R_1$? – Fiona Aug 14 '21 at 13:12
  • @Fiona: You are right. Thanks for spotting that. I was sort of overwhelmed by all the coefficients involved :-). About the rest, let me get back to you. – Evans Gambit Aug 14 '21 at 13:45