Mixing
Hartshorne II-3.22(b)
$begingroup$
Let $f:Xto Y$ be a dominant morphism of integral schemes of finite type over a field $k$.
Let $e=dim(X)-dim(Y)$. For any point $yin f(X)$, show that every irreducible component of the fibre $X_y$ has dimension $geq e$. [Hint: Let $Y'=overline{{y}}$ and use (a) and Ex. 3.20b]
I think there is a typo and we have to use Ex 3.20d, which says that if $Y$ is a closed subset of $X$ then $dim(Y)+mathrm{codim}(Y,X)=dim(X)$. In addition, (a) says that if $Y'$ is a closed irreducible subset of $Y$ whose generic point $eta'$ is contained in $f(X)$ and $Z$ is any irreducible componet of $f^{-1}(Y')$ such that $eta'in f(Z)$, then $mathrm{codim}(Z,X)leqmathrm{codim}(Y',Y)$.
What I have done
By exercise 3.10a, $X_y$ is homeomorphic to $f^{-1}(y)$ with the induced topology, so I'm trying to show the result for $f^{-1}(y)$ since it looks easier, but using the hint I've only managed to prove it for $f^{-1}(Y')$. This is my proof:
Since $y$ is obviously the genereic point of $Y'$, if we take $Z$ to be an irreducible component of $f^{-1}(Y')$ containing $y$, by (a) we have
$$operatorname{codim}(Z,X) leq operatorname{codim}(Y',Y)$$
Now Ex 3.20d implies
$$dim (X) - dim (Z) leq dim (Y) - dim (Y')$$
and thus
$$e = dim (X) - dim (Y) leq dim (Z) - dim (Y') leq dim (Z)$$
But this only shows that any irreducible component of $f^{-1}(Y')$ has dimension $geq e$. How can I conclude that this is true also for the irreducible components of the fibre $f^{-1}(y)$?
general-topology algebraic-geometry dimension-theory
asked Jan 27 at 0:02
JaviJavi
3,0212832
$endgroup$
add a comment |
$begingroup$
Let $f:Xto Y$ be a dominant morphism of integral schemes of finite type over a field $k$.
Let $e=dim(X)-dim(Y)$. For any point $yin f(X)$, show that every irreducible component of the fibre $X_y$ has dimension $geq e$. [Hint: Let $Y'=overline{{y}}$ and use (a) and Ex. 3.20b]
I think there is a typo and we have to use Ex 3.20d, which says that if $Y$ is a closed subset of $X$ then $dim(Y)+mathrm{codim}(Y,X)=dim(X)$. In addition, (a) says that if $Y'$ is a closed irreducible subset of $Y$ whose generic point $eta'$ is contained in $f(X)$ and $Z$ is any irreducible componet of $f^{-1}(Y')$ such that $eta'in f(Z)$, then $mathrm{codim}(Z,X)leqmathrm{codim}(Y',Y)$.
What I have done
By exercise 3.10a, $X_y$ is homeomorphic to $f^{-1}(y)$ with the induced topology, so I'm trying to show the result for $f^{-1}(y)$ since it looks easier, but using the hint I've only managed to prove it for $f^{-1}(Y')$. This is my proof:
Since $y$ is obviously the genereic point of $Y'$, if we take $Z$ to be an irreducible component of $f^{-1}(Y')$ containing $y$, by (a) we have
$$operatorname{codim}(Z,X) leq operatorname{codim}(Y',Y)$$
Now Ex 3.20d implies
$$dim (X) - dim (Z) leq dim (Y) - dim (Y')$$
and thus
$$e = dim (X) - dim (Y) leq dim (Z) - dim (Y') leq dim (Z)$$
But this only shows that any irreducible component of $f^{-1}(Y')$ has dimension $geq e$. How can I conclude that this is true also for the irreducible components of the fibre $f^{-1}(y)$?
general-topology algebraic-geometry dimension-theory
asked Jan 27 at 0:02
JaviJavi
3,0212832
$endgroup$
add a comment |
$begingroup$
Let $f:Xto Y$ be a dominant morphism of integral schemes of finite type over a field $k$.
Let $e=dim(X)-dim(Y)$. For any point $yin f(X)$, show that every irreducible component of the fibre $X_y$ has dimension $geq e$. [Hint: Let $Y'=overline{{y}}$ and use (a) and Ex. 3.20b]
I think there is a typo and we have to use Ex 3.20d, which says that if $Y$ is a closed subset of $X$ then $dim(Y)+mathrm{codim}(Y,X)=dim(X)$. In addition, (a) says that if $Y'$ is a closed irreducible subset of $Y$ whose generic point $eta'$ is contained in $f(X)$ and $Z$ is any irreducible componet of $f^{-1}(Y')$ such that $eta'in f(Z)$, then $mathrm{codim}(Z,X)leqmathrm{codim}(Y',Y)$.
What I have done
By exercise 3.10a, $X_y$ is homeomorphic to $f^{-1}(y)$ with the induced topology, so I'm trying to show the result for $f^{-1}(y)$ since it looks easier, but using the hint I've only managed to prove it for $f^{-1}(Y')$. This is my proof:
Since $y$ is obviously the genereic point of $Y'$, if we take $Z$ to be an irreducible component of $f^{-1}(Y')$ containing $y$, by (a) we have
$$operatorname{codim}(Z,X) leq operatorname{codim}(Y',Y)$$
Now Ex 3.20d implies
$$dim (X) - dim (Z) leq dim (Y) - dim (Y')$$
and thus
$$e = dim (X) - dim (Y) leq dim (Z) - dim (Y') leq dim (Z)$$
But this only shows that any irreducible component of $f^{-1}(Y')$ has dimension $geq e$. How can I conclude that this is true also for the irreducible components of the fibre $f^{-1}(y)$?
general-topology algebraic-geometry dimension-theory
asked Jan 27 at 0:02
JaviJavi
3,0212832
$endgroup$
Let $f:Xto Y$ be a dominant morphism of integral schemes of finite type over a field $k$.
Let $e=dim(X)-dim(Y)$. For any point $yin f(X)$, show that every irreducible component of the fibre $X_y$ has dimension $geq e$. [Hint: Let $Y'=overline{{y}}$ and use (a) and Ex. 3.20b]
I think there is a typo and we have to use Ex 3.20d, which says that if $Y$ is a closed subset of $X$ then $dim(Y)+mathrm{codim}(Y,X)=dim(X)$. In addition, (a) says that if $Y'$ is a closed irreducible subset of $Y$ whose generic point $eta'$ is contained in $f(X)$ and $Z$ is any irreducible componet of $f^{-1}(Y')$ such that $eta'in f(Z)$, then $mathrm{codim}(Z,X)leqmathrm{codim}(Y',Y)$.
What I have done
By exercise 3.10a, $X_y$ is homeomorphic to $f^{-1}(y)$ with the induced topology, so I'm trying to show the result for $f^{-1}(y)$ since it looks easier, but using the hint I've only managed to prove it for $f^{-1}(Y')$. This is my proof:
Since $y$ is obviously the genereic point of $Y'$, if we take $Z$ to be an irreducible component of $f^{-1}(Y')$ containing $y$, by (a) we have
$$operatorname{codim}(Z,X) leq operatorname{codim}(Y',Y)$$
Now Ex 3.20d implies
$$dim (X) - dim (Z) leq dim (Y) - dim (Y')$$
and thus
$$e = dim (X) - dim (Y) leq dim (Z) - dim (Y') leq dim (Z)$$
But this only shows that any irreducible component of $f^{-1}(Y')$ has dimension $geq e$. How can I conclude that this is true also for the irreducible components of the fibre $f^{-1}(y)$?
general-topology algebraic-geometry dimension-theory
general-topology algebraic-geometry dimension-theory
asked Jan 27 at 0:02
JaviJavi
3,0212832
asked Jan 27 at 0:02
JaviJavi
3,0212832
asked Jan 27 at 0:02
JaviJavi
3,0212832
asked Jan 27 at 0:02
JaviJavi
3,0212832
asked Jan 27 at 0:02
JaviJavi
3,0212832
3,0212832
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $W$ be an irreducible component of $X_y$, and let $k$, $k'$ be the codimensions of $overline{W}$ (closure in $X$) in $X$ and $Y'$ in $Y$, respectively.
One argues that $overline{W}$ is an irreducible component of $f^{-1}(Y’)$. Thus $k leq k'$ by part (a).
Applying Ex. 3.20b and using the additivity of transcendence degrees in towers, we see that $$text{dim}(W)=e+k'-k geq e.
$$
answered Jan 27 at 3:02
OvenOven
1416
$endgroup$
$begingroup$
In general the preimage of the closure is not the closure of the preimage. Why do we know that $overline{f^{-1}(y)}=f^{-1}(Y')$? In principle we only have inclusion to the right since $f^{-1}(Y')$ is a closed subset containing $f^{-1}(y)$. Equality only holds if $f$ is closed.
$endgroup$
– Javi
Jan 27 at 12:16
1
$begingroup$
I took that part out (it was not important).
$endgroup$
– Oven
Jan 27 at 14:41
$begingroup$
Ok I'll try to fill the details
$endgroup$
– Javi
Jan 27 at 17:38
$begingroup$
I have a doubt about the transcendence degrees. I actually don't know which tower you're using to get the last equality.
$endgroup$
– Javi
Feb 1 at 17:57
1
$begingroup$
The tower is $F/K/k$, where $F$ is the function field of $W$ and $K$ is the residue field of $y$.
$endgroup$
– Oven
Feb 2 at 0:10
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
Let $W$ be an irreducible component of $X_y$, and let $k$, $k'$ be the codimensions of $overline{W}$ (closure in $X$) in $X$ and $Y'$ in $Y$, respectively.
One argues that $overline{W}$ is an irreducible component of $f^{-1}(Y’)$. Thus $k leq k'$ by part (a).
Applying Ex. 3.20b and using the additivity of transcendence degrees in towers, we see that $$text{dim}(W)=e+k'-k geq e.
$$
answered Jan 27 at 3:02
OvenOven
1416
$endgroup$
$begingroup$
In general the preimage of the closure is not the closure of the preimage. Why do we know that $overline{f^{-1}(y)}=f^{-1}(Y')$? In principle we only have inclusion to the right since $f^{-1}(Y')$ is a closed subset containing $f^{-1}(y)$. Equality only holds if $f$ is closed.
$endgroup$
– Javi
Jan 27 at 12:16
1
$begingroup$
I took that part out (it was not important).
$endgroup$
– Oven
Jan 27 at 14:41
$begingroup$
Ok I'll try to fill the details
$endgroup$
– Javi
Jan 27 at 17:38
$begingroup$
I have a doubt about the transcendence degrees. I actually don't know which tower you're using to get the last equality.
$endgroup$
– Javi
Feb 1 at 17:57
1
$begingroup$
The tower is $F/K/k$, where $F$ is the function field of $W$ and $K$ is the residue field of $y$.
$endgroup$
– Oven
Feb 2 at 0:10
add a comment |
$begingroup$
Let $W$ be an irreducible component of $X_y$, and let $k$, $k'$ be the codimensions of $overline{W}$ (closure in $X$) in $X$ and $Y'$ in $Y$, respectively.
One argues that $overline{W}$ is an irreducible component of $f^{-1}(Y’)$. Thus $k leq k'$ by part (a).
Applying Ex. 3.20b and using the additivity of transcendence degrees in towers, we see that $$text{dim}(W)=e+k'-k geq e.
$$
answered Jan 27 at 3:02
OvenOven
1416
$endgroup$
$begingroup$
In general the preimage of the closure is not the closure of the preimage. Why do we know that $overline{f^{-1}(y)}=f^{-1}(Y')$? In principle we only have inclusion to the right since $f^{-1}(Y')$ is a closed subset containing $f^{-1}(y)$. Equality only holds if $f$ is closed.
$endgroup$
– Javi
Jan 27 at 12:16
1
$begingroup$
I took that part out (it was not important).
$endgroup$
– Oven
Jan 27 at 14:41
$begingroup$
Ok I'll try to fill the details
$endgroup$
– Javi
Jan 27 at 17:38
$begingroup$
I have a doubt about the transcendence degrees. I actually don't know which tower you're using to get the last equality.
$endgroup$
– Javi
Feb 1 at 17:57
1
$begingroup$
The tower is $F/K/k$, where $F$ is the function field of $W$ and $K$ is the residue field of $y$.
$endgroup$
– Oven
Feb 2 at 0:10
add a comment |
$begingroup$
Let $W$ be an irreducible component of $X_y$, and let $k$, $k'$ be the codimensions of $overline{W}$ (closure in $X$) in $X$ and $Y'$ in $Y$, respectively.
One argues that $overline{W}$ is an irreducible component of $f^{-1}(Y’)$. Thus $k leq k'$ by part (a).
Applying Ex. 3.20b and using the additivity of transcendence degrees in towers, we see that $$text{dim}(W)=e+k'-k geq e.
$$
answered Jan 27 at 3:02
OvenOven
1416
$endgroup$
Let $W$ be an irreducible component of $X_y$, and let $k$, $k'$ be the codimensions of $overline{W}$ (closure in $X$) in $X$ and $Y'$ in $Y$, respectively.
One argues that $overline{W}$ is an irreducible component of $f^{-1}(Y’)$. Thus $k leq k'$ by part (a).
Applying Ex. 3.20b and using the additivity of transcendence degrees in towers, we see that $$text{dim}(W)=e+k'-k geq e.
$$
answered Jan 27 at 3:02
OvenOven
1416
edited Jan 27 at 14:38
answered Jan 27 at 3:02
OvenOven
1416
answered Jan 27 at 3:02
OvenOven
1416
answered Jan 27 at 3:02
OvenOven
1416
1416
$begingroup$
In general the preimage of the closure is not the closure of the preimage. Why do we know that $overline{f^{-1}(y)}=f^{-1}(Y')$? In principle we only have inclusion to the right since $f^{-1}(Y')$ is a closed subset containing $f^{-1}(y)$. Equality only holds if $f$ is closed.
$endgroup$
– Javi
Jan 27 at 12:16
1
$begingroup$
I took that part out (it was not important).
$endgroup$
– Oven
Jan 27 at 14:41
$begingroup$
Ok I'll try to fill the details
$endgroup$
– Javi
Jan 27 at 17:38
$begingroup$
I have a doubt about the transcendence degrees. I actually don't know which tower you're using to get the last equality.
$endgroup$
– Javi
Feb 1 at 17:57
1
$begingroup$
The tower is $F/K/k$, where $F$ is the function field of $W$ and $K$ is the residue field of $y$.
$endgroup$
– Oven
Feb 2 at 0:10
add a comment |
$begingroup$
In general the preimage of the closure is not the closure of the preimage. Why do we know that $overline{f^{-1}(y)}=f^{-1}(Y')$? In principle we only have inclusion to the right since $f^{-1}(Y')$ is a closed subset containing $f^{-1}(y)$. Equality only holds if $f$ is closed.
$endgroup$
– Javi
Jan 27 at 12:16
1
$begingroup$
I took that part out (it was not important).
$endgroup$
– Oven
Jan 27 at 14:41
$begingroup$
Ok I'll try to fill the details
$endgroup$
– Javi
Jan 27 at 17:38
$begingroup$
I have a doubt about the transcendence degrees. I actually don't know which tower you're using to get the last equality.
$endgroup$
– Javi
Feb 1 at 17:57
1
$begingroup$
The tower is $F/K/k$, where $F$ is the function field of $W$ and $K$ is the residue field of $y$.
$endgroup$
– Oven
Feb 2 at 0:10
$begingroup$
In general the preimage of the closure is not the closure of the preimage. Why do we know that $overline{f^{-1}(y)}=f^{-1}(Y')$? In principle we only have inclusion to the right since $f^{-1}(Y')$ is a closed subset containing $f^{-1}(y)$. Equality only holds if $f$ is closed.
$endgroup$
– Javi
Jan 27 at 12:16
$begingroup$
In general the preimage of the closure is not the closure of the preimage. Why do we know that $overline{f^{-1}(y)}=f^{-1}(Y')$? In principle we only have inclusion to the right since $f^{-1}(Y')$ is a closed subset containing $f^{-1}(y)$. Equality only holds if $f$ is closed.
$endgroup$
– Javi
Jan 27 at 12:16
1
1
$begingroup$
I took that part out (it was not important).
$endgroup$
– Oven
Jan 27 at 14:41
$begingroup$
I took that part out (it was not important).
$endgroup$
– Oven
Jan 27 at 14:41
$begingroup$
Ok I'll try to fill the details
$endgroup$
– Javi
Jan 27 at 17:38
$begingroup$
Ok I'll try to fill the details
$endgroup$
– Javi
Jan 27 at 17:38
$begingroup$
I have a doubt about the transcendence degrees. I actually don't know which tower you're using to get the last equality.
$endgroup$
– Javi
Feb 1 at 17:57
$begingroup$
I have a doubt about the transcendence degrees. I actually don't know which tower you're using to get the last equality.
$endgroup$
– Javi
Feb 1 at 17:57
1
1
$begingroup$
The tower is $F/K/k$, where $F$ is the function field of $W$ and $K$ is the residue field of $y$.
$endgroup$
– Oven
Feb 2 at 0:10
$begingroup$
The tower is $F/K/k$, where $F$ is the function field of $W$ and $K$ is the residue field of $y$.
$endgroup$
– Oven
Feb 2 at 0:10
add a comment |
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