Mixing
Definition of $going-up$ map
$begingroup$
The exercise 5.10 of Atiyah's Commutative Algebra gives the definition of
$going-up$ map:
A ring homomorphism $f:Arightarrow B$ is said to have the $going-up$ (resp. the $going-down$ property) if the conclusion of going-up theorem (resp. the going-down theorem) holds for $B$ and its subring $f(A)$.
I think this definition is unnatural and I think the definition should be
....... if the conclusion of going-up theorem (resp. the going-down theorem) holds for $B$ and $A$.
It seems these two definitions are different. Why the book chooses the "unnatural" definition? Also one part of exercise is
i) $f$ has the going-down property
ii) For any prime ideal $q$ of $B$, if $p=q^{c}$, then $f^{*}: Spec(B_{q})to Spec(A_{p})$ is surjective
prove that i)$Leftrightarrow$ ii)
It is trivial if using the definition which I think. But I do not know how to prove if using the book's definition. I am very confused now.
algebraic-geometry commutative-algebra schemes
asked Jan 12 at 9:29
MikeMike
6417
$endgroup$
add a comment |
$begingroup$
The exercise 5.10 of Atiyah's Commutative Algebra gives the definition of
$going-up$ map:
A ring homomorphism $f:Arightarrow B$ is said to have the $going-up$ (resp. the $going-down$ property) if the conclusion of going-up theorem (resp. the going-down theorem) holds for $B$ and its subring $f(A)$.
I think this definition is unnatural and I think the definition should be
....... if the conclusion of going-up theorem (resp. the going-down theorem) holds for $B$ and $A$.
It seems these two definitions are different. Why the book chooses the "unnatural" definition? Also one part of exercise is
i) $f$ has the going-down property
ii) For any prime ideal $q$ of $B$, if $p=q^{c}$, then $f^{*}: Spec(B_{q})to Spec(A_{p})$ is surjective
prove that i)$Leftrightarrow$ ii)
It is trivial if using the definition which I think. But I do not know how to prove if using the book's definition. I am very confused now.
algebraic-geometry commutative-algebra schemes
asked Jan 12 at 9:29
MikeMike
6417
$endgroup$
3
$begingroup$
Your alternative definition has nothing to do with the map $f$, so how could it possibly be a definition of a property of $f$? Besides, the setting you (wrongly) assume you have is $A subseteq B$, only then can you talk about the going-up property for $B vert A$. Now we generalize this setting naturally to rings that are related via a ring homomorphism $fcolon A to B$, by talking about the ring extension $f(A) subseteq B$. This even gives the prior definition in the special case $A subseteq B$ by taking $f$ to be the inclusion of $A$ into $B$.
$endgroup$
– Layer Cake
Jan 12 at 10:37
add a comment |
$begingroup$
The exercise 5.10 of Atiyah's Commutative Algebra gives the definition of
$going-up$ map:
A ring homomorphism $f:Arightarrow B$ is said to have the $going-up$ (resp. the $going-down$ property) if the conclusion of going-up theorem (resp. the going-down theorem) holds for $B$ and its subring $f(A)$.
I think this definition is unnatural and I think the definition should be
....... if the conclusion of going-up theorem (resp. the going-down theorem) holds for $B$ and $A$.
It seems these two definitions are different. Why the book chooses the "unnatural" definition? Also one part of exercise is
i) $f$ has the going-down property
ii) For any prime ideal $q$ of $B$, if $p=q^{c}$, then $f^{*}: Spec(B_{q})to Spec(A_{p})$ is surjective
prove that i)$Leftrightarrow$ ii)
It is trivial if using the definition which I think. But I do not know how to prove if using the book's definition. I am very confused now.
algebraic-geometry commutative-algebra schemes
asked Jan 12 at 9:29
MikeMike
6417
$endgroup$
The exercise 5.10 of Atiyah's Commutative Algebra gives the definition of
$going-up$ map:
A ring homomorphism $f:Arightarrow B$ is said to have the $going-up$ (resp. the $going-down$ property) if the conclusion of going-up theorem (resp. the going-down theorem) holds for $B$ and its subring $f(A)$.
I think this definition is unnatural and I think the definition should be
....... if the conclusion of going-up theorem (resp. the going-down theorem) holds for $B$ and $A$.
It seems these two definitions are different. Why the book chooses the "unnatural" definition? Also one part of exercise is
i) $f$ has the going-down property
ii) For any prime ideal $q$ of $B$, if $p=q^{c}$, then $f^{*}: Spec(B_{q})to Spec(A_{p})$ is surjective
prove that i)$Leftrightarrow$ ii)
It is trivial if using the definition which I think. But I do not know how to prove if using the book's definition. I am very confused now.
algebraic-geometry commutative-algebra schemes
algebraic-geometry commutative-algebra schemes
asked Jan 12 at 9:29
MikeMike
6417
asked Jan 12 at 9:29
MikeMike
6417
asked Jan 12 at 9:29
MikeMike
6417
asked Jan 12 at 9:29
MikeMike
6417
asked Jan 12 at 9:29
MikeMike
6417
6417
3
$begingroup$
Your alternative definition has nothing to do with the map $f$, so how could it possibly be a definition of a property of $f$? Besides, the setting you (wrongly) assume you have is $A subseteq B$, only then can you talk about the going-up property for $B vert A$. Now we generalize this setting naturally to rings that are related via a ring homomorphism $fcolon A to B$, by talking about the ring extension $f(A) subseteq B$. This even gives the prior definition in the special case $A subseteq B$ by taking $f$ to be the inclusion of $A$ into $B$.
$endgroup$
– Layer Cake
Jan 12 at 10:37
add a comment |
3
$begingroup$
Your alternative definition has nothing to do with the map $f$, so how could it possibly be a definition of a property of $f$? Besides, the setting you (wrongly) assume you have is $A subseteq B$, only then can you talk about the going-up property for $B vert A$. Now we generalize this setting naturally to rings that are related via a ring homomorphism $fcolon A to B$, by talking about the ring extension $f(A) subseteq B$. This even gives the prior definition in the special case $A subseteq B$ by taking $f$ to be the inclusion of $A$ into $B$.
$endgroup$
– Layer Cake
Jan 12 at 10:37
3
3
$begingroup$
Your alternative definition has nothing to do with the map $f$, so how could it possibly be a definition of a property of $f$? Besides, the setting you (wrongly) assume you have is $A subseteq B$, only then can you talk about the going-up property for $B vert A$. Now we generalize this setting naturally to rings that are related via a ring homomorphism $fcolon A to B$, by talking about the ring extension $f(A) subseteq B$. This even gives the prior definition in the special case $A subseteq B$ by taking $f$ to be the inclusion of $A$ into $B$.
$endgroup$
– Layer Cake
Jan 12 at 10:37
$begingroup$
Your alternative definition has nothing to do with the map $f$, so how could it possibly be a definition of a property of $f$? Besides, the setting you (wrongly) assume you have is $A subseteq B$, only then can you talk about the going-up property for $B vert A$. Now we generalize this setting naturally to rings that are related via a ring homomorphism $fcolon A to B$, by talking about the ring extension $f(A) subseteq B$. This even gives the prior definition in the special case $A subseteq B$ by taking $f$ to be the inclusion of $A$ into $B$.
$endgroup$
– Layer Cake
Jan 12 at 10:37
add a comment |
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$begingroup$
Your alternative definition has nothing to do with the map $f$, so how could it possibly be a definition of a property of $f$? Besides, the setting you (wrongly) assume you have is $A subseteq B$, only then can you talk about the going-up property for $B vert A$. Now we generalize this setting naturally to rings that are related via a ring homomorphism $fcolon A to B$, by talking about the ring extension $f(A) subseteq B$. This even gives the prior definition in the special case $A subseteq B$ by taking $f$ to be the inclusion of $A$ into $B$.
$endgroup$
– Layer Cake
Jan 12 at 10:37