Mixing
Hartshorne II.3.22c following the hint
$begingroup$
I know this exercise can be solved in several ways, such as Ravi Vakil's proof or this answer, but I'd like to try to follow the given hint if possible.
The exercise says the following:
Let $f:X→Y$ be a dominant morphism of integral schemes of finite type over a field $k$. Show there is an open dense subset $U⊆X$ s.t. $dim U_y=dim X−dim Y$ for all $y$ in the image of $U$.
Hint: First reduce to the case where $X$ and $Y$ are affine, say $X=SpecA$ and $Y=SpecB$. Then $A$ is a finitely generated $B$-algebra. Take $t_1,dots,t_e∈A$ which form a transcendence base of $K(X)$ over $K(Y)$, and let $X_1=SpecB[t_1,...,t_e]$. Then $X_1$ is isomorphic to affine $e$-space over $Y$, and the morphism $X→X_1$ is generically finite. Now use (Ex. 3.7) above.
I've done most of the process, so I'll go directly to the place I'm stuck. Let $g:Xto X_1$ the morphism of the hint, which I've proved that is dominant, generically finite and of finite type.
By Ex II.3.7 there exists an open dense subset $U_1subseteq X_1$ such that, takin $U=g^{-1}(U_1)$, the induced morphism $g|_U:U→U_1$ is finite. I claim that $U⊆X$ is the desired set. Certainly it is a dense open set, since $X$ integral implies $X$ is irreducible (Proposition 3.1) and $U$ is nonempty, so that it is automatically dense. Thus, we need only check that $y∈f(U)$ implies $dim U_y=e$.
I think using (b) I can show that $dim U_ygeq e$, by I don't know how to proof the other inequality. In addition I haven't (at least explicitily) used that $X_1$ is an affine $e$-space, so that might be the key.
Part (b), under the same hypothesis as (c)
Let $e=dim X-dim Y$. For any point $yin f(X)$, show that every irreducible component of the fibre $X_y$ has dimension $geq e$.
algebraic-geometry dimension-theory fibre-product
asked Feb 1 at 23:19
JaviJavi
3,15321032
$endgroup$
add a comment |
$begingroup$
I know this exercise can be solved in several ways, such as Ravi Vakil's proof or this answer, but I'd like to try to follow the given hint if possible.
The exercise says the following:
Let $f:X→Y$ be a dominant morphism of integral schemes of finite type over a field $k$. Show there is an open dense subset $U⊆X$ s.t. $dim U_y=dim X−dim Y$ for all $y$ in the image of $U$.
Hint: First reduce to the case where $X$ and $Y$ are affine, say $X=SpecA$ and $Y=SpecB$. Then $A$ is a finitely generated $B$-algebra. Take $t_1,dots,t_e∈A$ which form a transcendence base of $K(X)$ over $K(Y)$, and let $X_1=SpecB[t_1,...,t_e]$. Then $X_1$ is isomorphic to affine $e$-space over $Y$, and the morphism $X→X_1$ is generically finite. Now use (Ex. 3.7) above.
I've done most of the process, so I'll go directly to the place I'm stuck. Let $g:Xto X_1$ the morphism of the hint, which I've proved that is dominant, generically finite and of finite type.
By Ex II.3.7 there exists an open dense subset $U_1subseteq X_1$ such that, takin $U=g^{-1}(U_1)$, the induced morphism $g|_U:U→U_1$ is finite. I claim that $U⊆X$ is the desired set. Certainly it is a dense open set, since $X$ integral implies $X$ is irreducible (Proposition 3.1) and $U$ is nonempty, so that it is automatically dense. Thus, we need only check that $y∈f(U)$ implies $dim U_y=e$.
I think using (b) I can show that $dim U_ygeq e$, by I don't know how to proof the other inequality. In addition I haven't (at least explicitily) used that $X_1$ is an affine $e$-space, so that might be the key.
Part (b), under the same hypothesis as (c)
Let $e=dim X-dim Y$. For any point $yin f(X)$, show that every irreducible component of the fibre $X_y$ has dimension $geq e$.
algebraic-geometry dimension-theory fibre-product
asked Feb 1 at 23:19
JaviJavi
3,15321032
$endgroup$
3
$begingroup$
Please include the statement of the exercise in your post. Posts should be self contained. Thank you!
$endgroup$
– Pedro Tamaroff♦
Feb 1 at 23:30
1
$begingroup$
It should follow from finiteness+that $X_1$ is affine space over $Y$. Roughly speaking, we should have that the fiber in $U_1$ is $e$ dimensional, since $X_1$ is affine $e$-space, and then since $Uto U_1$ is finite, the fiber in $U$ will be at most $e$-dimensional, since if $f:Xto Y$ is integral, then $dim Yge dim X$. See Stacks.
$endgroup$
– jgon
Feb 2 at 3:20
$begingroup$
Thank you @jgon Perhaps you want to write an answer
$endgroup$
– Javi
Feb 2 at 9:24
$begingroup$
By the way @jgon I never said $f$ was integral, $X$ and $Y$ are integral schemes, does that somehow imply that $f$ must be integral?
$endgroup$
– Javi
Feb 3 at 11:51
1
$begingroup$
@Javi finite implies integral, so $g$ restricted to $U$ is integral. $f$ is not integral, I phrased the result I was trying to quote poorly in my prior comment by reusing your variables thoughtlessly. My bad.
$endgroup$
– jgon
Feb 3 at 13:41
add a comment |
$begingroup$
I know this exercise can be solved in several ways, such as Ravi Vakil's proof or this answer, but I'd like to try to follow the given hint if possible.
The exercise says the following:
Let $f:X→Y$ be a dominant morphism of integral schemes of finite type over a field $k$. Show there is an open dense subset $U⊆X$ s.t. $dim U_y=dim X−dim Y$ for all $y$ in the image of $U$.
Hint: First reduce to the case where $X$ and $Y$ are affine, say $X=SpecA$ and $Y=SpecB$. Then $A$ is a finitely generated $B$-algebra. Take $t_1,dots,t_e∈A$ which form a transcendence base of $K(X)$ over $K(Y)$, and let $X_1=SpecB[t_1,...,t_e]$. Then $X_1$ is isomorphic to affine $e$-space over $Y$, and the morphism $X→X_1$ is generically finite. Now use (Ex. 3.7) above.
I've done most of the process, so I'll go directly to the place I'm stuck. Let $g:Xto X_1$ the morphism of the hint, which I've proved that is dominant, generically finite and of finite type.
By Ex II.3.7 there exists an open dense subset $U_1subseteq X_1$ such that, takin $U=g^{-1}(U_1)$, the induced morphism $g|_U:U→U_1$ is finite. I claim that $U⊆X$ is the desired set. Certainly it is a dense open set, since $X$ integral implies $X$ is irreducible (Proposition 3.1) and $U$ is nonempty, so that it is automatically dense. Thus, we need only check that $y∈f(U)$ implies $dim U_y=e$.
I think using (b) I can show that $dim U_ygeq e$, by I don't know how to proof the other inequality. In addition I haven't (at least explicitily) used that $X_1$ is an affine $e$-space, so that might be the key.
Part (b), under the same hypothesis as (c)
Let $e=dim X-dim Y$. For any point $yin f(X)$, show that every irreducible component of the fibre $X_y$ has dimension $geq e$.
algebraic-geometry dimension-theory fibre-product
asked Feb 1 at 23:19
JaviJavi
3,15321032
$endgroup$
I know this exercise can be solved in several ways, such as Ravi Vakil's proof or this answer, but I'd like to try to follow the given hint if possible.
The exercise says the following:
Let $f:X→Y$ be a dominant morphism of integral schemes of finite type over a field $k$. Show there is an open dense subset $U⊆X$ s.t. $dim U_y=dim X−dim Y$ for all $y$ in the image of $U$.
Hint: First reduce to the case where $X$ and $Y$ are affine, say $X=SpecA$ and $Y=SpecB$. Then $A$ is a finitely generated $B$-algebra. Take $t_1,dots,t_e∈A$ which form a transcendence base of $K(X)$ over $K(Y)$, and let $X_1=SpecB[t_1,...,t_e]$. Then $X_1$ is isomorphic to affine $e$-space over $Y$, and the morphism $X→X_1$ is generically finite. Now use (Ex. 3.7) above.
I've done most of the process, so I'll go directly to the place I'm stuck. Let $g:Xto X_1$ the morphism of the hint, which I've proved that is dominant, generically finite and of finite type.
By Ex II.3.7 there exists an open dense subset $U_1subseteq X_1$ such that, takin $U=g^{-1}(U_1)$, the induced morphism $g|_U:U→U_1$ is finite. I claim that $U⊆X$ is the desired set. Certainly it is a dense open set, since $X$ integral implies $X$ is irreducible (Proposition 3.1) and $U$ is nonempty, so that it is automatically dense. Thus, we need only check that $y∈f(U)$ implies $dim U_y=e$.
I think using (b) I can show that $dim U_ygeq e$, by I don't know how to proof the other inequality. In addition I haven't (at least explicitily) used that $X_1$ is an affine $e$-space, so that might be the key.
Part (b), under the same hypothesis as (c)
Let $e=dim X-dim Y$. For any point $yin f(X)$, show that every irreducible component of the fibre $X_y$ has dimension $geq e$.
algebraic-geometry dimension-theory fibre-product
algebraic-geometry dimension-theory fibre-product
asked Feb 1 at 23:19
JaviJavi
3,15321032
asked Feb 1 at 23:19
JaviJavi
3,15321032
edited Feb 1 at 23:41
Javi
asked Feb 1 at 23:19
JaviJavi
3,15321032
asked Feb 1 at 23:19
JaviJavi
3,15321032
asked Feb 1 at 23:19
JaviJavi
3,15321032
3,15321032
3
$begingroup$
Please include the statement of the exercise in your post. Posts should be self contained. Thank you!
$endgroup$
– Pedro Tamaroff♦
Feb 1 at 23:30
1
$begingroup$
It should follow from finiteness+that $X_1$ is affine space over $Y$. Roughly speaking, we should have that the fiber in $U_1$ is $e$ dimensional, since $X_1$ is affine $e$-space, and then since $Uto U_1$ is finite, the fiber in $U$ will be at most $e$-dimensional, since if $f:Xto Y$ is integral, then $dim Yge dim X$. See Stacks.
$endgroup$
– jgon
Feb 2 at 3:20
$begingroup$
Thank you @jgon Perhaps you want to write an answer
$endgroup$
– Javi
Feb 2 at 9:24
$begingroup$
By the way @jgon I never said $f$ was integral, $X$ and $Y$ are integral schemes, does that somehow imply that $f$ must be integral?
$endgroup$
– Javi
Feb 3 at 11:51
1
$begingroup$
@Javi finite implies integral, so $g$ restricted to $U$ is integral. $f$ is not integral, I phrased the result I was trying to quote poorly in my prior comment by reusing your variables thoughtlessly. My bad.
$endgroup$
– jgon
Feb 3 at 13:41
add a comment |
3
$begingroup$
Please include the statement of the exercise in your post. Posts should be self contained. Thank you!
$endgroup$
– Pedro Tamaroff♦
Feb 1 at 23:30
1
$begingroup$
It should follow from finiteness+that $X_1$ is affine space over $Y$. Roughly speaking, we should have that the fiber in $U_1$ is $e$ dimensional, since $X_1$ is affine $e$-space, and then since $Uto U_1$ is finite, the fiber in $U$ will be at most $e$-dimensional, since if $f:Xto Y$ is integral, then $dim Yge dim X$. See Stacks.
$endgroup$
– jgon
Feb 2 at 3:20
$begingroup$
Thank you @jgon Perhaps you want to write an answer
$endgroup$
– Javi
Feb 2 at 9:24
$begingroup$
By the way @jgon I never said $f$ was integral, $X$ and $Y$ are integral schemes, does that somehow imply that $f$ must be integral?
$endgroup$
– Javi
Feb 3 at 11:51
1
$begingroup$
@Javi finite implies integral, so $g$ restricted to $U$ is integral. $f$ is not integral, I phrased the result I was trying to quote poorly in my prior comment by reusing your variables thoughtlessly. My bad.
$endgroup$
– jgon
Feb 3 at 13:41
3
3
$begingroup$
Please include the statement of the exercise in your post. Posts should be self contained. Thank you!
$endgroup$
– Pedro Tamaroff♦
Feb 1 at 23:30
$begingroup$
Please include the statement of the exercise in your post. Posts should be self contained. Thank you!
$endgroup$
– Pedro Tamaroff♦
Feb 1 at 23:30
1
1
$begingroup$
It should follow from finiteness+that $X_1$ is affine space over $Y$. Roughly speaking, we should have that the fiber in $U_1$ is $e$ dimensional, since $X_1$ is affine $e$-space, and then since $Uto U_1$ is finite, the fiber in $U$ will be at most $e$-dimensional, since if $f:Xto Y$ is integral, then $dim Yge dim X$. See Stacks.
$endgroup$
– jgon
Feb 2 at 3:20
$begingroup$
It should follow from finiteness+that $X_1$ is affine space over $Y$. Roughly speaking, we should have that the fiber in $U_1$ is $e$ dimensional, since $X_1$ is affine $e$-space, and then since $Uto U_1$ is finite, the fiber in $U$ will be at most $e$-dimensional, since if $f:Xto Y$ is integral, then $dim Yge dim X$. See Stacks.
$endgroup$
– jgon
Feb 2 at 3:20
$begingroup$
Thank you @jgon Perhaps you want to write an answer
$endgroup$
– Javi
Feb 2 at 9:24
$begingroup$
Thank you @jgon Perhaps you want to write an answer
$endgroup$
– Javi
Feb 2 at 9:24
$begingroup$
By the way @jgon I never said $f$ was integral, $X$ and $Y$ are integral schemes, does that somehow imply that $f$ must be integral?
$endgroup$
– Javi
Feb 3 at 11:51
$begingroup$
By the way @jgon I never said $f$ was integral, $X$ and $Y$ are integral schemes, does that somehow imply that $f$ must be integral?
$endgroup$
– Javi
Feb 3 at 11:51
1
1
$begingroup$
@Javi finite implies integral, so $g$ restricted to $U$ is integral. $f$ is not integral, I phrased the result I was trying to quote poorly in my prior comment by reusing your variables thoughtlessly. My bad.
$endgroup$
– jgon
Feb 3 at 13:41
$begingroup$
@Javi finite implies integral, so $g$ restricted to $U$ is integral. $f$ is not integral, I phrased the result I was trying to quote poorly in my prior comment by reusing your variables thoughtlessly. My bad.
$endgroup$
– jgon
Feb 3 at 13:41
add a comment |
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3
$begingroup$
Please include the statement of the exercise in your post. Posts should be self contained. Thank you!
$endgroup$
– Pedro Tamaroff♦
Feb 1 at 23:30
1
$begingroup$
It should follow from finiteness+that $X_1$ is affine space over $Y$. Roughly speaking, we should have that the fiber in $U_1$ is $e$ dimensional, since $X_1$ is affine $e$-space, and then since $Uto U_1$ is finite, the fiber in $U$ will be at most $e$-dimensional, since if $f:Xto Y$ is integral, then $dim Yge dim X$. See Stacks.
$endgroup$
– jgon
Feb 2 at 3:20
$begingroup$
Thank you @jgon Perhaps you want to write an answer
$endgroup$
– Javi
Feb 2 at 9:24
$begingroup$
By the way @jgon I never said $f$ was integral, $X$ and $Y$ are integral schemes, does that somehow imply that $f$ must be integral?
$endgroup$
– Javi
Feb 3 at 11:51
1
$begingroup$
@Javi finite implies integral, so $g$ restricted to $U$ is integral. $f$ is not integral, I phrased the result I was trying to quote poorly in my prior comment by reusing your variables thoughtlessly. My bad.
$endgroup$
– jgon
Feb 3 at 13:41