Image of base change of immersion












1












$begingroup$


I'm revising for my exams now and struggling with the following exercise:



Let $f: X to S$ be an open or closed immersion and $g: S' to S$ another morphism where $X,S,S'$ are schemes. Then the base-change morphism $f' : X times_S S' to S'$ is such that $text{Im}(f') = g^{-1}(text{Im}(f))$.



If anyone could prove this for me I'd be really grateful!






algebraic-geometry schemes







share|cite|improve this question









$endgroup$












  • $begingroup$
    Is $operatorname{Im}$ the scheme-theoretic or set-theoretic image?
    $endgroup$
    – Hoot
    May 27 '16 at 18:04












  • $begingroup$
    Scheme-theoretic, I think...
    $endgroup$
    – CameronJWhitehead
    May 27 '16 at 19:30
















1












$begingroup$


I'm revising for my exams now and struggling with the following exercise:



Let $f: X to S$ be an open or closed immersion and $g: S' to S$ another morphism where $X,S,S'$ are schemes. Then the base-change morphism $f' : X times_S S' to S'$ is such that $text{Im}(f') = g^{-1}(text{Im}(f))$.



If anyone could prove this for me I'd be really grateful!






algebraic-geometry schemes







share|cite|improve this question









$endgroup$












  • $begingroup$
    Is $operatorname{Im}$ the scheme-theoretic or set-theoretic image?
    $endgroup$
    – Hoot
    May 27 '16 at 18:04












  • $begingroup$
    Scheme-theoretic, I think...
    $endgroup$
    – CameronJWhitehead
    May 27 '16 at 19:30














1












1








1





$begingroup$


I'm revising for my exams now and struggling with the following exercise:



Let $f: X to S$ be an open or closed immersion and $g: S' to S$ another morphism where $X,S,S'$ are schemes. Then the base-change morphism $f' : X times_S S' to S'$ is such that $text{Im}(f') = g^{-1}(text{Im}(f))$.



If anyone could prove this for me I'd be really grateful!






algebraic-geometry schemes







share|cite|improve this question









$endgroup$




I'm revising for my exams now and struggling with the following exercise:



Let $f: X to S$ be an open or closed immersion and $g: S' to S$ another morphism where $X,S,S'$ are schemes. Then the base-change morphism $f' : X times_S S' to S'$ is such that $text{Im}(f') = g^{-1}(text{Im}(f))$.



If anyone could prove this for me I'd be really grateful!






algebraic-geometry schemes




algebraic-geometry schemes






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked May 27 '16 at 16:35









CameronJWhiteheadCameronJWhitehead

1,436518




1,436518












  • $begingroup$
    Is $operatorname{Im}$ the scheme-theoretic or set-theoretic image?
    $endgroup$
    – Hoot
    May 27 '16 at 18:04












  • $begingroup$
    Scheme-theoretic, I think...
    $endgroup$
    – CameronJWhitehead
    May 27 '16 at 19:30


















  • $begingroup$
    Is $operatorname{Im}$ the scheme-theoretic or set-theoretic image?
    $endgroup$
    – Hoot
    May 27 '16 at 18:04












  • $begingroup$
    Scheme-theoretic, I think...
    $endgroup$
    – CameronJWhitehead
    May 27 '16 at 19:30
















$begingroup$
Is $operatorname{Im}$ the scheme-theoretic or set-theoretic image?
$endgroup$
– Hoot
May 27 '16 at 18:04






$begingroup$
Is $operatorname{Im}$ the scheme-theoretic or set-theoretic image?
$endgroup$
– Hoot
May 27 '16 at 18:04














$begingroup$
Scheme-theoretic, I think...
$endgroup$
– CameronJWhitehead
May 27 '16 at 19:30




$begingroup$
Scheme-theoretic, I think...
$endgroup$
– CameronJWhitehead
May 27 '16 at 19:30










1 Answer
1






active

oldest

votes


















0












$begingroup$

In this answer, $X$(resp. $Xotimes_S S'$), and $mathrm{Im} f$(resp. $mathrm{Im} f'$) are identified. (This is not problematic at all, especially about $Xotimes_S S'$, and $mathrm{Im} f'$, thanks to the following fact. This is a standard result in algebraic geometry, and I think you have proved it in class. However, if you do not know it, please see Section 7.3, Proposition 13 of S. Bosch, Algebraic Geometry and Commutative Algebra, Springer, 2013 for a proof.)



Fact $f'$ is an open (resp. closed) immersion if $f$ is an open (resp. closed) immersion.



When $f$ is an open immersion, it is clear from the pull-back diagram
$$
begin{array}{ccccccccc}
g^{-1}(X)& xrightarrow{f'} & S' \
downarrow& & downarrow{g} \
X& xrightarrow{f} & S,
end{array}
$$
which owes to the universal property of open immersion.



Prop (The Universal Property of Open Immersion)
If $f:Xto Y$ is a morphism between schemes, $Vsubseteq Y$ an open subset(, or subscheme), and $f(X)subseteq V$, then there exists a unique morphism $f':Xto V$ whose composition with the open immersion $Vto X$ is $f$.



Proof You can do it through looking at the structure sheaf of V, etc.



When $f$ is a closed immersion, first assume that $S$, and $S'$(, and so $X$) are affine schemes. Let $S = mathrm{Spec} A, X = mathrm{Spec} A/mathfrak{a}$, and $S' = mathrm{Spec} B$. Then, $Xtimes_S S'=mathrm{Spec} A/mathfrak{a}otimes_A B=mathrm{Spec} B/mathfrak{a}B = V(mathfrak{a}B) = g^{-1}(V(mathfrak{a})) = g^{-1}(X)$. For the second last equality, if $phi: Ato B$ is the corresponding homomorphism between rings to $g$, then $mathfrak{q}in g^{-1}(V(mathfrak{a}))Leftrightarrow phi^{-1}(mathfrak{p})in V(mathfrak{a})Leftrightarrow phi^{-1}(mathfrak{p})supseteqmathfrak{a}Leftrightarrowmathfrak{p}supseteqphi(mathfrak{a})(=mathfrak{a}B)Leftrightarrow mathfrak{p}in V(mathfrak{a}B)$.



In the general case, take an affine open covering $S = cup_{jin J} V_j$, and for each $jin J$, $g^{-1}(V_j) = cup_{iin I_j} U_{i,j}$. Let $g_{i,j}:U_{i,j}to V_j$, and $g_{j}:g^{-1}(V_j)to V_j$ be the restriction of f. Then, $Xtimes_S S' = cup_{jin J}cup_{iin I_j} f^{-1}(V_j)times_{V_j} U_{i,j} = cupcup g_{i,j}^{-1} (V_j) = cup g_{j}^{-1} (V_j) = g^{-1}(V)$.






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    $begingroup$

    In this answer, $X$(resp. $Xotimes_S S'$), and $mathrm{Im} f$(resp. $mathrm{Im} f'$) are identified. (This is not problematic at all, especially about $Xotimes_S S'$, and $mathrm{Im} f'$, thanks to the following fact. This is a standard result in algebraic geometry, and I think you have proved it in class. However, if you do not know it, please see Section 7.3, Proposition 13 of S. Bosch, Algebraic Geometry and Commutative Algebra, Springer, 2013 for a proof.)



    Fact $f'$ is an open (resp. closed) immersion if $f$ is an open (resp. closed) immersion.



    When $f$ is an open immersion, it is clear from the pull-back diagram
    $$
    begin{array}{ccccccccc}
    g^{-1}(X)& xrightarrow{f'} & S' \
    downarrow& & downarrow{g} \
    X& xrightarrow{f} & S,
    end{array}
    $$
    which owes to the universal property of open immersion.



    Prop (The Universal Property of Open Immersion)
    If $f:Xto Y$ is a morphism between schemes, $Vsubseteq Y$ an open subset(, or subscheme), and $f(X)subseteq V$, then there exists a unique morphism $f':Xto V$ whose composition with the open immersion $Vto X$ is $f$.



    Proof You can do it through looking at the structure sheaf of V, etc.



    When $f$ is a closed immersion, first assume that $S$, and $S'$(, and so $X$) are affine schemes. Let $S = mathrm{Spec} A, X = mathrm{Spec} A/mathfrak{a}$, and $S' = mathrm{Spec} B$. Then, $Xtimes_S S'=mathrm{Spec} A/mathfrak{a}otimes_A B=mathrm{Spec} B/mathfrak{a}B = V(mathfrak{a}B) = g^{-1}(V(mathfrak{a})) = g^{-1}(X)$. For the second last equality, if $phi: Ato B$ is the corresponding homomorphism between rings to $g$, then $mathfrak{q}in g^{-1}(V(mathfrak{a}))Leftrightarrow phi^{-1}(mathfrak{p})in V(mathfrak{a})Leftrightarrow phi^{-1}(mathfrak{p})supseteqmathfrak{a}Leftrightarrowmathfrak{p}supseteqphi(mathfrak{a})(=mathfrak{a}B)Leftrightarrow mathfrak{p}in V(mathfrak{a}B)$.



    In the general case, take an affine open covering $S = cup_{jin J} V_j$, and for each $jin J$, $g^{-1}(V_j) = cup_{iin I_j} U_{i,j}$. Let $g_{i,j}:U_{i,j}to V_j$, and $g_{j}:g^{-1}(V_j)to V_j$ be the restriction of f. Then, $Xtimes_S S' = cup_{jin J}cup_{iin I_j} f^{-1}(V_j)times_{V_j} U_{i,j} = cupcup g_{i,j}^{-1} (V_j) = cup g_{j}^{-1} (V_j) = g^{-1}(V)$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      In this answer, $X$(resp. $Xotimes_S S'$), and $mathrm{Im} f$(resp. $mathrm{Im} f'$) are identified. (This is not problematic at all, especially about $Xotimes_S S'$, and $mathrm{Im} f'$, thanks to the following fact. This is a standard result in algebraic geometry, and I think you have proved it in class. However, if you do not know it, please see Section 7.3, Proposition 13 of S. Bosch, Algebraic Geometry and Commutative Algebra, Springer, 2013 for a proof.)



      Fact $f'$ is an open (resp. closed) immersion if $f$ is an open (resp. closed) immersion.



      When $f$ is an open immersion, it is clear from the pull-back diagram
      $$
      begin{array}{ccccccccc}
      g^{-1}(X)& xrightarrow{f'} & S' \
      downarrow& & downarrow{g} \
      X& xrightarrow{f} & S,
      end{array}
      $$
      which owes to the universal property of open immersion.



      Prop (The Universal Property of Open Immersion)
      If $f:Xto Y$ is a morphism between schemes, $Vsubseteq Y$ an open subset(, or subscheme), and $f(X)subseteq V$, then there exists a unique morphism $f':Xto V$ whose composition with the open immersion $Vto X$ is $f$.



      Proof You can do it through looking at the structure sheaf of V, etc.



      When $f$ is a closed immersion, first assume that $S$, and $S'$(, and so $X$) are affine schemes. Let $S = mathrm{Spec} A, X = mathrm{Spec} A/mathfrak{a}$, and $S' = mathrm{Spec} B$. Then, $Xtimes_S S'=mathrm{Spec} A/mathfrak{a}otimes_A B=mathrm{Spec} B/mathfrak{a}B = V(mathfrak{a}B) = g^{-1}(V(mathfrak{a})) = g^{-1}(X)$. For the second last equality, if $phi: Ato B$ is the corresponding homomorphism between rings to $g$, then $mathfrak{q}in g^{-1}(V(mathfrak{a}))Leftrightarrow phi^{-1}(mathfrak{p})in V(mathfrak{a})Leftrightarrow phi^{-1}(mathfrak{p})supseteqmathfrak{a}Leftrightarrowmathfrak{p}supseteqphi(mathfrak{a})(=mathfrak{a}B)Leftrightarrow mathfrak{p}in V(mathfrak{a}B)$.



      In the general case, take an affine open covering $S = cup_{jin J} V_j$, and for each $jin J$, $g^{-1}(V_j) = cup_{iin I_j} U_{i,j}$. Let $g_{i,j}:U_{i,j}to V_j$, and $g_{j}:g^{-1}(V_j)to V_j$ be the restriction of f. Then, $Xtimes_S S' = cup_{jin J}cup_{iin I_j} f^{-1}(V_j)times_{V_j} U_{i,j} = cupcup g_{i,j}^{-1} (V_j) = cup g_{j}^{-1} (V_j) = g^{-1}(V)$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        In this answer, $X$(resp. $Xotimes_S S'$), and $mathrm{Im} f$(resp. $mathrm{Im} f'$) are identified. (This is not problematic at all, especially about $Xotimes_S S'$, and $mathrm{Im} f'$, thanks to the following fact. This is a standard result in algebraic geometry, and I think you have proved it in class. However, if you do not know it, please see Section 7.3, Proposition 13 of S. Bosch, Algebraic Geometry and Commutative Algebra, Springer, 2013 for a proof.)



        Fact $f'$ is an open (resp. closed) immersion if $f$ is an open (resp. closed) immersion.



        When $f$ is an open immersion, it is clear from the pull-back diagram
        $$
        begin{array}{ccccccccc}
        g^{-1}(X)& xrightarrow{f'} & S' \
        downarrow& & downarrow{g} \
        X& xrightarrow{f} & S,
        end{array}
        $$
        which owes to the universal property of open immersion.



        Prop (The Universal Property of Open Immersion)
        If $f:Xto Y$ is a morphism between schemes, $Vsubseteq Y$ an open subset(, or subscheme), and $f(X)subseteq V$, then there exists a unique morphism $f':Xto V$ whose composition with the open immersion $Vto X$ is $f$.



        Proof You can do it through looking at the structure sheaf of V, etc.



        When $f$ is a closed immersion, first assume that $S$, and $S'$(, and so $X$) are affine schemes. Let $S = mathrm{Spec} A, X = mathrm{Spec} A/mathfrak{a}$, and $S' = mathrm{Spec} B$. Then, $Xtimes_S S'=mathrm{Spec} A/mathfrak{a}otimes_A B=mathrm{Spec} B/mathfrak{a}B = V(mathfrak{a}B) = g^{-1}(V(mathfrak{a})) = g^{-1}(X)$. For the second last equality, if $phi: Ato B$ is the corresponding homomorphism between rings to $g$, then $mathfrak{q}in g^{-1}(V(mathfrak{a}))Leftrightarrow phi^{-1}(mathfrak{p})in V(mathfrak{a})Leftrightarrow phi^{-1}(mathfrak{p})supseteqmathfrak{a}Leftrightarrowmathfrak{p}supseteqphi(mathfrak{a})(=mathfrak{a}B)Leftrightarrow mathfrak{p}in V(mathfrak{a}B)$.



        In the general case, take an affine open covering $S = cup_{jin J} V_j$, and for each $jin J$, $g^{-1}(V_j) = cup_{iin I_j} U_{i,j}$. Let $g_{i,j}:U_{i,j}to V_j$, and $g_{j}:g^{-1}(V_j)to V_j$ be the restriction of f. Then, $Xtimes_S S' = cup_{jin J}cup_{iin I_j} f^{-1}(V_j)times_{V_j} U_{i,j} = cupcup g_{i,j}^{-1} (V_j) = cup g_{j}^{-1} (V_j) = g^{-1}(V)$.






        share|cite|improve this answer









        $endgroup$



        In this answer, $X$(resp. $Xotimes_S S'$), and $mathrm{Im} f$(resp. $mathrm{Im} f'$) are identified. (This is not problematic at all, especially about $Xotimes_S S'$, and $mathrm{Im} f'$, thanks to the following fact. This is a standard result in algebraic geometry, and I think you have proved it in class. However, if you do not know it, please see Section 7.3, Proposition 13 of S. Bosch, Algebraic Geometry and Commutative Algebra, Springer, 2013 for a proof.)



        Fact $f'$ is an open (resp. closed) immersion if $f$ is an open (resp. closed) immersion.



        When $f$ is an open immersion, it is clear from the pull-back diagram
        $$
        begin{array}{ccccccccc}
        g^{-1}(X)& xrightarrow{f'} & S' \
        downarrow& & downarrow{g} \
        X& xrightarrow{f} & S,
        end{array}
        $$
        which owes to the universal property of open immersion.



        Prop (The Universal Property of Open Immersion)
        If $f:Xto Y$ is a morphism between schemes, $Vsubseteq Y$ an open subset(, or subscheme), and $f(X)subseteq V$, then there exists a unique morphism $f':Xto V$ whose composition with the open immersion $Vto X$ is $f$.



        Proof You can do it through looking at the structure sheaf of V, etc.



        When $f$ is a closed immersion, first assume that $S$, and $S'$(, and so $X$) are affine schemes. Let $S = mathrm{Spec} A, X = mathrm{Spec} A/mathfrak{a}$, and $S' = mathrm{Spec} B$. Then, $Xtimes_S S'=mathrm{Spec} A/mathfrak{a}otimes_A B=mathrm{Spec} B/mathfrak{a}B = V(mathfrak{a}B) = g^{-1}(V(mathfrak{a})) = g^{-1}(X)$. For the second last equality, if $phi: Ato B$ is the corresponding homomorphism between rings to $g$, then $mathfrak{q}in g^{-1}(V(mathfrak{a}))Leftrightarrow phi^{-1}(mathfrak{p})in V(mathfrak{a})Leftrightarrow phi^{-1}(mathfrak{p})supseteqmathfrak{a}Leftrightarrowmathfrak{p}supseteqphi(mathfrak{a})(=mathfrak{a}B)Leftrightarrow mathfrak{p}in V(mathfrak{a}B)$.



        In the general case, take an affine open covering $S = cup_{jin J} V_j$, and for each $jin J$, $g^{-1}(V_j) = cup_{iin I_j} U_{i,j}$. Let $g_{i,j}:U_{i,j}to V_j$, and $g_{j}:g^{-1}(V_j)to V_j$ be the restriction of f. Then, $Xtimes_S S' = cup_{jin J}cup_{iin I_j} f^{-1}(V_j)times_{V_j} U_{i,j} = cupcup g_{i,j}^{-1} (V_j) = cup g_{j}^{-1} (V_j) = g^{-1}(V)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '18 at 4:03









        neanderneander

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