Skip to main content
MathOverflow

Return to Question

added 21 characters in body
Source Link
Panna
Panna
  • 61
  • 3

many books and lecture notes say that it is difficult to prove that the composite of two proper morphisms of rigid analytic varieties are also proper. That is, if two morphisms of rigid analytic varieties $f:X\to Y$ and $g:Y\to Z$ are proper, then their composite $g\circ f$ is also proper. However I could solve this by a straightforward calculation. I guess I've made some mistake somewhere but I don't know. Are there any mistakes? Here's the proof:

We can assume that $Z$ is an affinoid variety $Z = SpA $. Since $g:Y\to SpA$ is proper, there are two admissible affinoid coverings $\{ U_i\}$ and $\{ U'_i \} $ of $Y$ such that $ U_i \Subset_{SpA} U'_i$ for all i. Since properness is stable under base changes, $f^{-1}(U'_i)\to U'_i$ is also proper for each i. Hence we can also take two admissible affinoid coverings $\{V_{ij}\}$ and $\{V'_{ij}\}$ of $f^{-1}(U'_{i})$ such that $ V_{ij} \Subset_{U_i} V'_{ij} $$ V_{ij} \Subset_{U'_i} V'_{ij} $. Let $U_i=SpB_i$, $U'_i= SpB'_i$, $V_{ij} =SpC_{ij}$ and $V'_{ij} =SpC'_{ij}$ and fix i and j. To prove that $ g\circ f$ is proper, it suffices to show that $f^{-1}(U_i) \cap V_{ij} \Subset_{SpA} V'_{ij}$. Since $f^{-1}(U_i) \cap V_{ij}$ is an inverse image of the morphism $X\to X\times_{SpA}Y$, which is the base change of the diagonal morphism $Y\to Y\times_{SpA} Y$ (note that this is a closed immersion since $g$ is proper) by $X\times_{SpA}Y \to Y\times _{SpA} Y$. Since base changes of closed immersions are also closed immersions, hence separated, it follows that $f^{-1}(U_i) \cap V_{ij}$ is an affinoid space. Now, let $g_1,...,g_r\in B'_i$ be the affinoid generating system of $B'_i$ over $A$ such that $U_i \subset \{x\in U'_i ; \forall k, |g_k(x)|<1 \}$ and $h_1,...h_s\in C'_ij$ be that of $C'_ij$ over $B_i$ such that $ V_{ij}\subset\{x\in V'_{ij};\forall l, |h_l(x)|<1\}$.then,$ \{f^*(g_k),h_l\}$ is an affinoid generating system of $C'_{ij}$ over$A$ and \begin{equation} f^{-1}(U_i) \cap V_{ij}\subset \{ x \in V'_{ij}; \forall k\ and\ \forall l,|f^*(g_k)(x)|<1,\ |h_l(x)|<1\} \end{equation} which proves that $f^{-1}(U_i) \cap V_{ij}\Subset_{SpA} V'_{ij}$.

many books and lecture notes say that it is difficult to prove that the composite of two proper morphisms of rigid analytic varieties are also proper. That is, if two morphisms of rigid analytic varieties $f:X\to Y$ and $g:Y\to Z$ are proper, then their composite $g\circ f$ is also proper. However I could solve this by a straightforward calculation. I guess I've made some mistake somewhere but I don't know. Are there any mistakes? Here's the proof:

We can assume that $Z$ is an affinoid variety $Z = SpA $. Since $g:Y\to SpA$ is proper, there are two admissible affinoid coverings $\{ U_i\}$ and $\{ U'_i \} $ of $Y$ such that $ U_i \Subset_{SpA} U'_i$ for all i. Since properness is stable under base changes, $f^{-1}(U'_i)\to U'_i$ is also proper for each i. Hence we can also take two admissible affinoid coverings $\{V_{ij}\}$ and $\{V'_{ij}\}$ such that $ V_{ij} \Subset_{U_i} V'_{ij} $. Let $U_i=SpB_i$, $U'_i= SpB'_i$, $V_{ij} =SpC_{ij}$ and $V'_{ij} =SpC'_{ij}$ and fix i and j. To prove that $ g\circ f$ is proper, it suffices to show that $f^{-1}(U_i) \cap V_{ij} \Subset_{SpA} V'_{ij}$. Since $f^{-1}(U_i) \cap V_{ij}$ is an inverse image of the morphism $X\to X\times_{SpA}Y$, which is the base change of the diagonal morphism $Y\to Y\times_{SpA} Y$ (note that this is a closed immersion since $g$ is proper) by $X\times_{SpA}Y \to Y\times _{SpA} Y$. Since base changes of closed immersions are also closed immersions, hence separated, it follows that $f^{-1}(U_i) \cap V_{ij}$ is an affinoid space. Now, let $g_1,...,g_r\in B'_i$ be the affinoid generating system of $B'_i$ over $A$ such that $U_i \subset \{x\in U'_i ; \forall k, |g_k(x)|<1 \}$ and $h_1,...h_s\in C'_ij$ be that of $C'_ij$ over $B_i$ such that $ V_{ij}\subset\{x\in V'_{ij};\forall l, |h_l(x)|<1\}$.then,$ \{f^*(g_k),h_l\}$ is an affinoid generating system of $C'_{ij}$ over$A$ and \begin{equation} f^{-1}(U_i) \cap V_{ij}\subset \{ x \in V'_{ij}; \forall k\ and\ \forall l,|f^*(g_k)(x)|<1,\ |h_l(x)|<1\} \end{equation} which proves that $f^{-1}(U_i) \cap V_{ij}\Subset_{SpA} V'_{ij}$.

many books and lecture notes say that it is difficult to prove that the composite of two proper morphisms of rigid analytic varieties are also proper. That is, if two morphisms of rigid analytic varieties $f:X\to Y$ and $g:Y\to Z$ are proper, then their composite $g\circ f$ is also proper. However I could solve this by a straightforward calculation. I guess I've made some mistake somewhere but I don't know. Are there any mistakes? Here's the proof:

We can assume that $Z$ is an affinoid variety $Z = SpA $. Since $g:Y\to SpA$ is proper, there are two admissible affinoid coverings $\{ U_i\}$ and $\{ U'_i \} $ of $Y$ such that $ U_i \Subset_{SpA} U'_i$ for all i. Since properness is stable under base changes, $f^{-1}(U'_i)\to U'_i$ is also proper for each i. Hence we can also take two admissible affinoid coverings $\{V_{ij}\}$ and $\{V'_{ij}\}$ of $f^{-1}(U'_{i})$ such that $ V_{ij} \Subset_{U'_i} V'_{ij} $. Let $U_i=SpB_i$, $U'_i= SpB'_i$, $V_{ij} =SpC_{ij}$ and $V'_{ij} =SpC'_{ij}$ and fix i and j. To prove that $ g\circ f$ is proper, it suffices to show that $f^{-1}(U_i) \cap V_{ij} \Subset_{SpA} V'_{ij}$. Since $f^{-1}(U_i) \cap V_{ij}$ is an inverse image of the morphism $X\to X\times_{SpA}Y$, which is the base change of the diagonal morphism $Y\to Y\times_{SpA} Y$ (note that this is a closed immersion since $g$ is proper) by $X\times_{SpA}Y \to Y\times _{SpA} Y$. Since base changes of closed immersions are also closed immersions, hence separated, it follows that $f^{-1}(U_i) \cap V_{ij}$ is an affinoid space. Now, let $g_1,...,g_r\in B'_i$ be the affinoid generating system of $B'_i$ over $A$ such that $U_i \subset \{x\in U'_i ; \forall k, |g_k(x)|<1 \}$ and $h_1,...h_s\in C'_ij$ be that of $C'_ij$ over $B_i$ such that $ V_{ij}\subset\{x\in V'_{ij};\forall l, |h_l(x)|<1\}$.then,$ \{f^*(g_k),h_l\}$ is an affinoid generating system of $C'_{ij}$ over$A$ and \begin{equation} f^{-1}(U_i) \cap V_{ij}\subset \{ x \in V'_{ij}; \forall k\ and\ \forall l,|f^*(g_k)(x)|<1,\ |h_l(x)|<1\} \end{equation} which proves that $f^{-1}(U_i) \cap V_{ij}\Subset_{SpA} V'_{ij}$.

edited body
Source Link
Panna
Panna
  • 61
  • 3

many books and lecture notes say that it is difficult to prove that the composite of two proper morphisms of rigid analytic varieties are also proper. That is, if two morphisms of rigid analytic varieties $f:X\to Y$ and $g:Y\to Z$ are proper, then their composite $g\circ f$ is also proper. However I could solve this by a straightforward calculation. I guess I've made some mistake somewhere but I don't know. Are there any mistakes? Here's the proof:

We can assume that $Z$ is an affinoid variety $Z = SpA $. Since $g:Y\to SpA$ is proper, there are two admissible affinoid coverings $\{ U_i\}$ and $\{ U'_i \} $ of $Y$ such that $ U_i \Subset_{SpA} U'_i$ for all i. Since properness is stable under base changes, $f^{-1}(U'_i)\to U'_i$ is also proper for each i. Hence we can also take two admissible affinoid coverings $\{V_{ij}\}$ and $\{V'_{ij}\}$ such that $ V_{ij} \Subset_{U_i} V'_{ij} $. Let $U_i=SpB_i$, $U'_i= SpB'_i$, $V_{ij} =SpC_{ij}$ and $V'_{ij} =SpC'_{ij}$ and fix i and j. To prove that $ g\circ f$ is proper, it suffices to show that $f^{-1}(U_i) \cap V_{ij} \Subset_{SpA} V'_{ij}$. Since $f^{-1}(U_i) \cap V_{ij}$ is an inverse image of the morphism $X\to X\times_{SpA}Y$, which is the base change of the diagonal morphism $Y\to Y\times_{SpA} Y$ (note that this is a closed immersion since $g$ is proper) by $X\times_{SpA}X \to Y\times _{SpA} Y$$X\times_{SpA}Y \to Y\times _{SpA} Y$. Since base changes of closed immersions are also closed immersions, hence separated, it follows that $f^{-1}(U_i) \cap V_{ij}$ is an affinoid space. Now, let $g_1,...,g_r\in B'_i$ be the affinoid generating system of $B'_i$ over $A$ such that $U_i \subset \{x\in U'_i ; \forall k, |g_k(x)|<1 \}$ and $h_1,...h_s\in C'_ij$ be that of $C'_ij$ over $B_i$ such that $ V_{ij}\subset\{x\in V'_{ij};\forall l, |h_l(x)|<1\}$.then,$ \{f^*(g_k),h_l\}$ is an affinoid generating system of $C'_{ij}$ over$A$ and \begin{equation} f^{-1}(U_i) \cap V_{ij}\subset \{ x \in V'_{ij}; \forall k\ and\ \forall l,|f^*(g_k)(x)|<1,\ |h_l(x)|<1\} \end{equation} which proves that $f^{-1}(U_i) \cap V_{ij}\Subset_{SpA} V'_{ij}$.

many books and lecture notes say that it is difficult to prove that the composite of two proper morphisms of rigid analytic varieties are also proper. That is, if two morphisms of rigid analytic varieties $f:X\to Y$ and $g:Y\to Z$ are proper, then their composite $g\circ f$ is also proper. However I could solve this by a straightforward calculation. I guess I've made some mistake somewhere but I don't know. Are there any mistakes? Here's the proof:

We can assume that $Z$ is an affinoid variety $Z = SpA $. Since $g:Y\to SpA$ is proper, there are two admissible affinoid coverings $\{ U_i\}$ and $\{ U'_i \} $ of $Y$ such that $ U_i \Subset_{SpA} U'_i$ for all i. Since properness is stable under base changes, $f^{-1}(U'_i)\to U'_i$ is also proper for each i. Hence we can also take two admissible affinoid coverings $\{V_{ij}\}$ and $\{V'_{ij}\}$ such that $ V_{ij} \Subset_{U_i} V'_{ij} $. Let $U_i=SpB_i$, $U'_i= SpB'_i$, $V_{ij} =SpC_{ij}$ and $V'_{ij} =SpC'_{ij}$ and fix i and j. To prove that $ g\circ f$ is proper, it suffices to show that $f^{-1}(U_i) \cap V_{ij} \Subset_{SpA} V'_{ij}$. Since $f^{-1}(U_i) \cap V_{ij}$ is an inverse image of the morphism $X\to X\times_{SpA}Y$, which is the base change of the diagonal morphism $Y\to Y\times_{SpA} Y$ (note that this is a closed immersion since $g$ is proper) by $X\times_{SpA}X \to Y\times _{SpA} Y$. Since base changes of closed immersions are also closed immersions, hence separated, it follows that $f^{-1}(U_i) \cap V_{ij}$ is an affinoid space. Now, let $g_1,...,g_r\in B'_i$ be the affinoid generating system of $B'_i$ over $A$ such that $U_i \subset \{x\in U'_i ; \forall k, |g_k(x)|<1 \}$ and $h_1,...h_s\in C'_ij$ be that of $C'_ij$ over $B_i$ such that $ V_{ij}\subset\{x\in V'_{ij};\forall l, |h_l(x)|<1\}$.then,$ \{f^*(g_k),h_l\}$ is an affinoid generating system of $C'_{ij}$ over$A$ and \begin{equation} f^{-1}(U_i) \cap V_{ij}\subset \{ x \in V'_{ij}; \forall k\ and\ \forall l,|f^*(g_k)(x)|<1,\ |h_l(x)|<1\} \end{equation} which proves that $f^{-1}(U_i) \cap V_{ij}\Subset_{SpA} V'_{ij}$.

many books and lecture notes say that it is difficult to prove that the composite of two proper morphisms of rigid analytic varieties are also proper. That is, if two morphisms of rigid analytic varieties $f:X\to Y$ and $g:Y\to Z$ are proper, then their composite $g\circ f$ is also proper. However I could solve this by a straightforward calculation. I guess I've made some mistake somewhere but I don't know. Are there any mistakes? Here's the proof:

We can assume that $Z$ is an affinoid variety $Z = SpA $. Since $g:Y\to SpA$ is proper, there are two admissible affinoid coverings $\{ U_i\}$ and $\{ U'_i \} $ of $Y$ such that $ U_i \Subset_{SpA} U'_i$ for all i. Since properness is stable under base changes, $f^{-1}(U'_i)\to U'_i$ is also proper for each i. Hence we can also take two admissible affinoid coverings $\{V_{ij}\}$ and $\{V'_{ij}\}$ such that $ V_{ij} \Subset_{U_i} V'_{ij} $. Let $U_i=SpB_i$, $U'_i= SpB'_i$, $V_{ij} =SpC_{ij}$ and $V'_{ij} =SpC'_{ij}$ and fix i and j. To prove that $ g\circ f$ is proper, it suffices to show that $f^{-1}(U_i) \cap V_{ij} \Subset_{SpA} V'_{ij}$. Since $f^{-1}(U_i) \cap V_{ij}$ is an inverse image of the morphism $X\to X\times_{SpA}Y$, which is the base change of the diagonal morphism $Y\to Y\times_{SpA} Y$ (note that this is a closed immersion since $g$ is proper) by $X\times_{SpA}Y \to Y\times _{SpA} Y$. Since base changes of closed immersions are also closed immersions, hence separated, it follows that $f^{-1}(U_i) \cap V_{ij}$ is an affinoid space. Now, let $g_1,...,g_r\in B'_i$ be the affinoid generating system of $B'_i$ over $A$ such that $U_i \subset \{x\in U'_i ; \forall k, |g_k(x)|<1 \}$ and $h_1,...h_s\in C'_ij$ be that of $C'_ij$ over $B_i$ such that $ V_{ij}\subset\{x\in V'_{ij};\forall l, |h_l(x)|<1\}$.then,$ \{f^*(g_k),h_l\}$ is an affinoid generating system of $C'_{ij}$ over$A$ and \begin{equation} f^{-1}(U_i) \cap V_{ij}\subset \{ x \in V'_{ij}; \forall k\ and\ \forall l,|f^*(g_k)(x)|<1,\ |h_l(x)|<1\} \end{equation} which proves that $f^{-1}(U_i) \cap V_{ij}\Subset_{SpA} V'_{ij}$.

Source Link
Panna
Panna
  • 61
  • 3

The composition of proper morphisms in rigid geometry

many books and lecture notes say that it is difficult to prove that the composite of two proper morphisms of rigid analytic varieties are also proper. That is, if two morphisms of rigid analytic varieties $f:X\to Y$ and $g:Y\to Z$ are proper, then their composite $g\circ f$ is also proper. However I could solve this by a straightforward calculation. I guess I've made some mistake somewhere but I don't know. Are there any mistakes? Here's the proof:

We can assume that $Z$ is an affinoid variety $Z = SpA $. Since $g:Y\to SpA$ is proper, there are two admissible affinoid coverings $\{ U_i\}$ and $\{ U'_i \} $ of $Y$ such that $ U_i \Subset_{SpA} U'_i$ for all i. Since properness is stable under base changes, $f^{-1}(U'_i)\to U'_i$ is also proper for each i. Hence we can also take two admissible affinoid coverings $\{V_{ij}\}$ and $\{V'_{ij}\}$ such that $ V_{ij} \Subset_{U_i} V'_{ij} $. Let $U_i=SpB_i$, $U'_i= SpB'_i$, $V_{ij} =SpC_{ij}$ and $V'_{ij} =SpC'_{ij}$ and fix i and j. To prove that $ g\circ f$ is proper, it suffices to show that $f^{-1}(U_i) \cap V_{ij} \Subset_{SpA} V'_{ij}$. Since $f^{-1}(U_i) \cap V_{ij}$ is an inverse image of the morphism $X\to X\times_{SpA}Y$, which is the base change of the diagonal morphism $Y\to Y\times_{SpA} Y$ (note that this is a closed immersion since $g$ is proper) by $X\times_{SpA}X \to Y\times _{SpA} Y$. Since base changes of closed immersions are also closed immersions, hence separated, it follows that $f^{-1}(U_i) \cap V_{ij}$ is an affinoid space. Now, let $g_1,...,g_r\in B'_i$ be the affinoid generating system of $B'_i$ over $A$ such that $U_i \subset \{x\in U'_i ; \forall k, |g_k(x)|<1 \}$ and $h_1,...h_s\in C'_ij$ be that of $C'_ij$ over $B_i$ such that $ V_{ij}\subset\{x\in V'_{ij};\forall l, |h_l(x)|<1\}$.then,$ \{f^*(g_k),h_l\}$ is an affinoid generating system of $C'_{ij}$ over$A$ and \begin{equation} f^{-1}(U_i) \cap V_{ij}\subset \{ x \in V'_{ij}; \forall k\ and\ \forall l,|f^*(g_k)(x)|<1,\ |h_l(x)|<1\} \end{equation} which proves that $f^{-1}(U_i) \cap V_{ij}\Subset_{SpA} V'_{ij}$.