Proof check and clarification.
$S$ is a locally Noetherian scheme and $s in S$ . $Y$ is a scheme of finite type over $Spec mathcal O_ {S,s} $ I have to show there exists open $U$ containing $s$ and a scheme of finite type $X rightarrow U$ such that $Y = X times _ {U} Spec mathcal O_ {S,s}$ .
Here's how I proceeded. First assume $S$ and $Y$ are affine. Then say $S= Spec{ } A, s= p in Spec A $ and $Y= Spec { } B$. Then $B$ is a finitely generated $Ap$ algebra and hence $B= A_p[X_1,..., X_k]/<I>$ where $I$ is an ideal of $A[X_1,....,X_k]$ then since localization and quotients commute we have $B= A_p otimes _A A[X_1,..., X_k]/I$ .Thus we have $ Y= Spec mathcal O_ {S,s} times _{Spec A } Spec A[X_1,..., X_k]/I $
So basically I took $U$ to be $S$ itself in the case everything is affine. However i didn't use the Noetherian condition which gives me the doubt there's a flaw somewhere. Kindly help me out.
algebraic-geometry schemes
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$S$ is a locally Noetherian scheme and $s in S$ . $Y$ is a scheme of finite type over $Spec mathcal O_ {S,s} $ I have to show there exists open $U$ containing $s$ and a scheme of finite type $X rightarrow U$ such that $Y = X times _ {U} Spec mathcal O_ {S,s}$ .
Here's how I proceeded. First assume $S$ and $Y$ are affine. Then say $S= Spec{ } A, s= p in Spec A $ and $Y= Spec { } B$. Then $B$ is a finitely generated $Ap$ algebra and hence $B= A_p[X_1,..., X_k]/<I>$ where $I$ is an ideal of $A[X_1,....,X_k]$ then since localization and quotients commute we have $B= A_p otimes _A A[X_1,..., X_k]/I$ .Thus we have $ Y= Spec mathcal O_ {S,s} times _{Spec A } Spec A[X_1,..., X_k]/I $
So basically I took $U$ to be $S$ itself in the case everything is affine. However i didn't use the Noetherian condition which gives me the doubt there's a flaw somewhere. Kindly help me out.
algebraic-geometry schemes
add a comment |
$S$ is a locally Noetherian scheme and $s in S$ . $Y$ is a scheme of finite type over $Spec mathcal O_ {S,s} $ I have to show there exists open $U$ containing $s$ and a scheme of finite type $X rightarrow U$ such that $Y = X times _ {U} Spec mathcal O_ {S,s}$ .
Here's how I proceeded. First assume $S$ and $Y$ are affine. Then say $S= Spec{ } A, s= p in Spec A $ and $Y= Spec { } B$. Then $B$ is a finitely generated $Ap$ algebra and hence $B= A_p[X_1,..., X_k]/<I>$ where $I$ is an ideal of $A[X_1,....,X_k]$ then since localization and quotients commute we have $B= A_p otimes _A A[X_1,..., X_k]/I$ .Thus we have $ Y= Spec mathcal O_ {S,s} times _{Spec A } Spec A[X_1,..., X_k]/I $
So basically I took $U$ to be $S$ itself in the case everything is affine. However i didn't use the Noetherian condition which gives me the doubt there's a flaw somewhere. Kindly help me out.
algebraic-geometry schemes
$S$ is a locally Noetherian scheme and $s in S$ . $Y$ is a scheme of finite type over $Spec mathcal O_ {S,s} $ I have to show there exists open $U$ containing $s$ and a scheme of finite type $X rightarrow U$ such that $Y = X times _ {U} Spec mathcal O_ {S,s}$ .
Here's how I proceeded. First assume $S$ and $Y$ are affine. Then say $S= Spec{ } A, s= p in Spec A $ and $Y= Spec { } B$. Then $B$ is a finitely generated $Ap$ algebra and hence $B= A_p[X_1,..., X_k]/<I>$ where $I$ is an ideal of $A[X_1,....,X_k]$ then since localization and quotients commute we have $B= A_p otimes _A A[X_1,..., X_k]/I$ .Thus we have $ Y= Spec mathcal O_ {S,s} times _{Spec A } Spec A[X_1,..., X_k]/I $
So basically I took $U$ to be $S$ itself in the case everything is affine. However i didn't use the Noetherian condition which gives me the doubt there's a flaw somewhere. Kindly help me out.
algebraic-geometry schemes
algebraic-geometry schemes
edited Dec 10 '18 at 21:30
asked Dec 10 '18 at 21:13
Soumik Ghosh
19610
19610
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