When does a quasifinite surjective flat morphism have constant fiber multiplicity near a point?
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Let $V subset mathbb{A}_{mathbb{C}}^2 = operatorname{Spec}mathbb{C}[x,t]$ be a closed subscheme containing the point $x = t = 0$, and suppose we have a quasifinite flat surjective morphism $pi colon V to mathbb{A}_{mathbb{C}}^1 = operatorname{Spec} mathbb{C}[t]$ that sends the point $x = t = 0$ to the point $t = 0$. Choose a sufficiently small analytic neighborhood $U subset mathbb{A}_{mathbb{C}}^2$ containing the point $x = t = 0$ such that every component of $U cap V$ passes through the point $x = t = 0$. Does there exist an analytic neighborhood $U' subset mathbb{A}_{mathbb{C}}^1$ of the point $t = 0$ such that the multiplicity of the fiber $U cap pi^{-1}(p)$ is independent of the choice of $p in U'$?
What I know so far: it is certainly possible for this to fail if we drop the condition that $V$ is a closed subscheme of $mathbb{A}_{mathbb{C}}^2$. Indeed, take $$V = operatorname{Spec} mathbb{C}[[x]][t]/(x(x-t))$$ and take $pi$ to be the morphism of affine schemes induced by the obvious ring map $mathbb{C}[t] to mathbb{C}[[x]][t]/(x(x-t))$. Then $pi$ is quasifinite because it has finite fibers (multiplicity $1$ for $tneq 0$ and multiplicity $2$ for $t =0$), and it is also flat, because it is obtained by precomposing the flat morphism $operatorname{Spec} mathbb{C}[x][t]/(x(x-t)) to operatorname{Spec} mathbb{C}[t]$ with the morphism of affine schemes induced by the localization $mathbb{C}[t] to mathbb{C}[[t]]$. However, the fiber multiplicity jumps from $1$ to $2$ at $t=0$.
What appears to go wrong in the above example is that the subscheme $V$ picks up some additional fuzz in the $x$-direction at $t=0$, but intuitively, it seems like this can't happen if $V$ is a closed subscheme to begin with.
algebraic-geometry commutative-algebra schemes affine-schemes flatness
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show 1 more comment
$begingroup$
Let $V subset mathbb{A}_{mathbb{C}}^2 = operatorname{Spec}mathbb{C}[x,t]$ be a closed subscheme containing the point $x = t = 0$, and suppose we have a quasifinite flat surjective morphism $pi colon V to mathbb{A}_{mathbb{C}}^1 = operatorname{Spec} mathbb{C}[t]$ that sends the point $x = t = 0$ to the point $t = 0$. Choose a sufficiently small analytic neighborhood $U subset mathbb{A}_{mathbb{C}}^2$ containing the point $x = t = 0$ such that every component of $U cap V$ passes through the point $x = t = 0$. Does there exist an analytic neighborhood $U' subset mathbb{A}_{mathbb{C}}^1$ of the point $t = 0$ such that the multiplicity of the fiber $U cap pi^{-1}(p)$ is independent of the choice of $p in U'$?
What I know so far: it is certainly possible for this to fail if we drop the condition that $V$ is a closed subscheme of $mathbb{A}_{mathbb{C}}^2$. Indeed, take $$V = operatorname{Spec} mathbb{C}[[x]][t]/(x(x-t))$$ and take $pi$ to be the morphism of affine schemes induced by the obvious ring map $mathbb{C}[t] to mathbb{C}[[x]][t]/(x(x-t))$. Then $pi$ is quasifinite because it has finite fibers (multiplicity $1$ for $tneq 0$ and multiplicity $2$ for $t =0$), and it is also flat, because it is obtained by precomposing the flat morphism $operatorname{Spec} mathbb{C}[x][t]/(x(x-t)) to operatorname{Spec} mathbb{C}[t]$ with the morphism of affine schemes induced by the localization $mathbb{C}[t] to mathbb{C}[[t]]$. However, the fiber multiplicity jumps from $1$ to $2$ at $t=0$.
What appears to go wrong in the above example is that the subscheme $V$ picks up some additional fuzz in the $x$-direction at $t=0$, but intuitively, it seems like this can't happen if $V$ is a closed subscheme to begin with.
algebraic-geometry commutative-algebra schemes affine-schemes flatness
$endgroup$
2
$begingroup$
What about $V=operatorname{Spec}mathbb{C}[x,t]/(tx^2-x)tooperatorname{Spec}mathbb{C}[t]$. Then $V$ is the union of two irreducible components : $x=0$ and $xt=1$. Both are flat. The first one is onto, hence so is $V$. The second one is not finite, hence neither is $V$.
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– Roland
Dec 16 '18 at 10:12
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@Roland Ah, that's a great counterexample! Now suppose we stipulate that no irreducible components of $V$ go off to infinity as $t to 0$. More precisely, suppose the closure of $V$ in $mathbb{P}_{mathbb{C}}^2$ is not supported at the point defined by $t = 0$ in the line at infinity. Is it then true that $pi$ is finite?
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– Ashvin Swaminathan
Dec 16 '18 at 17:59
2
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Well the problem does not happen only at $t=0$. Take $V=operatorname{Spec}mathbb{C}[x,t]/((t-1)x^2-x)$. Then there is a problem at $t=1$...
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– Roland
Dec 16 '18 at 18:49
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@Roland Right, this problem could indeed occur at any $t$. I guess I'm more interested in what happens locally near a point of $V$, and I've rewritten my question in that context.
$endgroup$
– Ashvin Swaminathan
Dec 16 '18 at 19:50
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I'm still thinking about it. I honestly didn't understand the example with $xy-x-t = 0$ mapping to $operatorname{Spec} mathbb{C}[t]$ via $(x,y,t) mapsto t^2 - t$. Why is the fiber above $0$ one point? I also didn't understand what condition you're saying is insufficient.
$endgroup$
– Ashvin Swaminathan
Jan 10 at 5:17
|
show 1 more comment
$begingroup$
Let $V subset mathbb{A}_{mathbb{C}}^2 = operatorname{Spec}mathbb{C}[x,t]$ be a closed subscheme containing the point $x = t = 0$, and suppose we have a quasifinite flat surjective morphism $pi colon V to mathbb{A}_{mathbb{C}}^1 = operatorname{Spec} mathbb{C}[t]$ that sends the point $x = t = 0$ to the point $t = 0$. Choose a sufficiently small analytic neighborhood $U subset mathbb{A}_{mathbb{C}}^2$ containing the point $x = t = 0$ such that every component of $U cap V$ passes through the point $x = t = 0$. Does there exist an analytic neighborhood $U' subset mathbb{A}_{mathbb{C}}^1$ of the point $t = 0$ such that the multiplicity of the fiber $U cap pi^{-1}(p)$ is independent of the choice of $p in U'$?
What I know so far: it is certainly possible for this to fail if we drop the condition that $V$ is a closed subscheme of $mathbb{A}_{mathbb{C}}^2$. Indeed, take $$V = operatorname{Spec} mathbb{C}[[x]][t]/(x(x-t))$$ and take $pi$ to be the morphism of affine schemes induced by the obvious ring map $mathbb{C}[t] to mathbb{C}[[x]][t]/(x(x-t))$. Then $pi$ is quasifinite because it has finite fibers (multiplicity $1$ for $tneq 0$ and multiplicity $2$ for $t =0$), and it is also flat, because it is obtained by precomposing the flat morphism $operatorname{Spec} mathbb{C}[x][t]/(x(x-t)) to operatorname{Spec} mathbb{C}[t]$ with the morphism of affine schemes induced by the localization $mathbb{C}[t] to mathbb{C}[[t]]$. However, the fiber multiplicity jumps from $1$ to $2$ at $t=0$.
What appears to go wrong in the above example is that the subscheme $V$ picks up some additional fuzz in the $x$-direction at $t=0$, but intuitively, it seems like this can't happen if $V$ is a closed subscheme to begin with.
algebraic-geometry commutative-algebra schemes affine-schemes flatness
$endgroup$
Let $V subset mathbb{A}_{mathbb{C}}^2 = operatorname{Spec}mathbb{C}[x,t]$ be a closed subscheme containing the point $x = t = 0$, and suppose we have a quasifinite flat surjective morphism $pi colon V to mathbb{A}_{mathbb{C}}^1 = operatorname{Spec} mathbb{C}[t]$ that sends the point $x = t = 0$ to the point $t = 0$. Choose a sufficiently small analytic neighborhood $U subset mathbb{A}_{mathbb{C}}^2$ containing the point $x = t = 0$ such that every component of $U cap V$ passes through the point $x = t = 0$. Does there exist an analytic neighborhood $U' subset mathbb{A}_{mathbb{C}}^1$ of the point $t = 0$ such that the multiplicity of the fiber $U cap pi^{-1}(p)$ is independent of the choice of $p in U'$?
What I know so far: it is certainly possible for this to fail if we drop the condition that $V$ is a closed subscheme of $mathbb{A}_{mathbb{C}}^2$. Indeed, take $$V = operatorname{Spec} mathbb{C}[[x]][t]/(x(x-t))$$ and take $pi$ to be the morphism of affine schemes induced by the obvious ring map $mathbb{C}[t] to mathbb{C}[[x]][t]/(x(x-t))$. Then $pi$ is quasifinite because it has finite fibers (multiplicity $1$ for $tneq 0$ and multiplicity $2$ for $t =0$), and it is also flat, because it is obtained by precomposing the flat morphism $operatorname{Spec} mathbb{C}[x][t]/(x(x-t)) to operatorname{Spec} mathbb{C}[t]$ with the morphism of affine schemes induced by the localization $mathbb{C}[t] to mathbb{C}[[t]]$. However, the fiber multiplicity jumps from $1$ to $2$ at $t=0$.
What appears to go wrong in the above example is that the subscheme $V$ picks up some additional fuzz in the $x$-direction at $t=0$, but intuitively, it seems like this can't happen if $V$ is a closed subscheme to begin with.
algebraic-geometry commutative-algebra schemes affine-schemes flatness
algebraic-geometry commutative-algebra schemes affine-schemes flatness
edited Dec 16 '18 at 19:46
Ashvin Swaminathan
asked Dec 16 '18 at 6:44
Ashvin SwaminathanAshvin Swaminathan
1,571520
1,571520
2
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What about $V=operatorname{Spec}mathbb{C}[x,t]/(tx^2-x)tooperatorname{Spec}mathbb{C}[t]$. Then $V$ is the union of two irreducible components : $x=0$ and $xt=1$. Both are flat. The first one is onto, hence so is $V$. The second one is not finite, hence neither is $V$.
$endgroup$
– Roland
Dec 16 '18 at 10:12
$begingroup$
@Roland Ah, that's a great counterexample! Now suppose we stipulate that no irreducible components of $V$ go off to infinity as $t to 0$. More precisely, suppose the closure of $V$ in $mathbb{P}_{mathbb{C}}^2$ is not supported at the point defined by $t = 0$ in the line at infinity. Is it then true that $pi$ is finite?
$endgroup$
– Ashvin Swaminathan
Dec 16 '18 at 17:59
2
$begingroup$
Well the problem does not happen only at $t=0$. Take $V=operatorname{Spec}mathbb{C}[x,t]/((t-1)x^2-x)$. Then there is a problem at $t=1$...
$endgroup$
– Roland
Dec 16 '18 at 18:49
$begingroup$
@Roland Right, this problem could indeed occur at any $t$. I guess I'm more interested in what happens locally near a point of $V$, and I've rewritten my question in that context.
$endgroup$
– Ashvin Swaminathan
Dec 16 '18 at 19:50
$begingroup$
I'm still thinking about it. I honestly didn't understand the example with $xy-x-t = 0$ mapping to $operatorname{Spec} mathbb{C}[t]$ via $(x,y,t) mapsto t^2 - t$. Why is the fiber above $0$ one point? I also didn't understand what condition you're saying is insufficient.
$endgroup$
– Ashvin Swaminathan
Jan 10 at 5:17
|
show 1 more comment
2
$begingroup$
What about $V=operatorname{Spec}mathbb{C}[x,t]/(tx^2-x)tooperatorname{Spec}mathbb{C}[t]$. Then $V$ is the union of two irreducible components : $x=0$ and $xt=1$. Both are flat. The first one is onto, hence so is $V$. The second one is not finite, hence neither is $V$.
$endgroup$
– Roland
Dec 16 '18 at 10:12
$begingroup$
@Roland Ah, that's a great counterexample! Now suppose we stipulate that no irreducible components of $V$ go off to infinity as $t to 0$. More precisely, suppose the closure of $V$ in $mathbb{P}_{mathbb{C}}^2$ is not supported at the point defined by $t = 0$ in the line at infinity. Is it then true that $pi$ is finite?
$endgroup$
– Ashvin Swaminathan
Dec 16 '18 at 17:59
2
$begingroup$
Well the problem does not happen only at $t=0$. Take $V=operatorname{Spec}mathbb{C}[x,t]/((t-1)x^2-x)$. Then there is a problem at $t=1$...
$endgroup$
– Roland
Dec 16 '18 at 18:49
$begingroup$
@Roland Right, this problem could indeed occur at any $t$. I guess I'm more interested in what happens locally near a point of $V$, and I've rewritten my question in that context.
$endgroup$
– Ashvin Swaminathan
Dec 16 '18 at 19:50
$begingroup$
I'm still thinking about it. I honestly didn't understand the example with $xy-x-t = 0$ mapping to $operatorname{Spec} mathbb{C}[t]$ via $(x,y,t) mapsto t^2 - t$. Why is the fiber above $0$ one point? I also didn't understand what condition you're saying is insufficient.
$endgroup$
– Ashvin Swaminathan
Jan 10 at 5:17
2
2
$begingroup$
What about $V=operatorname{Spec}mathbb{C}[x,t]/(tx^2-x)tooperatorname{Spec}mathbb{C}[t]$. Then $V$ is the union of two irreducible components : $x=0$ and $xt=1$. Both are flat. The first one is onto, hence so is $V$. The second one is not finite, hence neither is $V$.
$endgroup$
– Roland
Dec 16 '18 at 10:12
$begingroup$
What about $V=operatorname{Spec}mathbb{C}[x,t]/(tx^2-x)tooperatorname{Spec}mathbb{C}[t]$. Then $V$ is the union of two irreducible components : $x=0$ and $xt=1$. Both are flat. The first one is onto, hence so is $V$. The second one is not finite, hence neither is $V$.
$endgroup$
– Roland
Dec 16 '18 at 10:12
$begingroup$
@Roland Ah, that's a great counterexample! Now suppose we stipulate that no irreducible components of $V$ go off to infinity as $t to 0$. More precisely, suppose the closure of $V$ in $mathbb{P}_{mathbb{C}}^2$ is not supported at the point defined by $t = 0$ in the line at infinity. Is it then true that $pi$ is finite?
$endgroup$
– Ashvin Swaminathan
Dec 16 '18 at 17:59
$begingroup$
@Roland Ah, that's a great counterexample! Now suppose we stipulate that no irreducible components of $V$ go off to infinity as $t to 0$. More precisely, suppose the closure of $V$ in $mathbb{P}_{mathbb{C}}^2$ is not supported at the point defined by $t = 0$ in the line at infinity. Is it then true that $pi$ is finite?
$endgroup$
– Ashvin Swaminathan
Dec 16 '18 at 17:59
2
2
$begingroup$
Well the problem does not happen only at $t=0$. Take $V=operatorname{Spec}mathbb{C}[x,t]/((t-1)x^2-x)$. Then there is a problem at $t=1$...
$endgroup$
– Roland
Dec 16 '18 at 18:49
$begingroup$
Well the problem does not happen only at $t=0$. Take $V=operatorname{Spec}mathbb{C}[x,t]/((t-1)x^2-x)$. Then there is a problem at $t=1$...
$endgroup$
– Roland
Dec 16 '18 at 18:49
$begingroup$
@Roland Right, this problem could indeed occur at any $t$. I guess I'm more interested in what happens locally near a point of $V$, and I've rewritten my question in that context.
$endgroup$
– Ashvin Swaminathan
Dec 16 '18 at 19:50
$begingroup$
@Roland Right, this problem could indeed occur at any $t$. I guess I'm more interested in what happens locally near a point of $V$, and I've rewritten my question in that context.
$endgroup$
– Ashvin Swaminathan
Dec 16 '18 at 19:50
$begingroup$
I'm still thinking about it. I honestly didn't understand the example with $xy-x-t = 0$ mapping to $operatorname{Spec} mathbb{C}[t]$ via $(x,y,t) mapsto t^2 - t$. Why is the fiber above $0$ one point? I also didn't understand what condition you're saying is insufficient.
$endgroup$
– Ashvin Swaminathan
Jan 10 at 5:17
$begingroup$
I'm still thinking about it. I honestly didn't understand the example with $xy-x-t = 0$ mapping to $operatorname{Spec} mathbb{C}[t]$ via $(x,y,t) mapsto t^2 - t$. Why is the fiber above $0$ one point? I also didn't understand what condition you're saying is insufficient.
$endgroup$
– Ashvin Swaminathan
Jan 10 at 5:17
|
show 1 more comment
1 Answer
1
active
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$begingroup$
Let $V subset mathbb{A}_{mathbb{C}}^2 = operatorname{Spec}mathbb{C}[x,t]$ be a closed subscheme containing the point $x = t = 0$, and suppose we have a quasifinite flat surjective morphism $pi colon V to mathbb{A}_{mathbb{C}}^1 = operatorname{Spec} mathbb{C}[t]$ that sends the point $x = t = 0$ to the point $t = 0$. Choose a sufficiently small analytic neighborhood $U subset mathbb{A}_{mathbb{C}}^2$ containing the point $x = t = 0$ such that every component of $U cap V$ passes through the point $x = t = 0$. Does there exist an analytic neighborhood $U' subset mathbb{A}_{mathbb{C}}^1$ of the point $t = 0$ such that the multiplicity of the fiber $U cap pi^{-1}(p)$ is independent of the choice of $p in U'$?
I think that this holds for sufficiently small $U$, but the condition you impose is not sufficient, e.g. let $U = V$ be defined by $xy = x + t$ and $pi$ be defined by $t^2 - t$. Then the fiber above $0$ is one point---reduced---and the other fibers above closed points consist of $2$ points.
By standard arguments, for some open neighborhoods $U$ and $U'$ we have an induced finite flat map $U to U'$---cf. Lemma in Fischer's Complex Analytic Geometry, page 132---such that the point $x = t = 0$ is the only point mapping to the point $t = 0$. Then for $U'$ connected the direct image of the structure sheaf is a locally free coherent sheaf of some constant rank $d$, and $U_1 subset U$ is an open neighborhood of the origin let $U'_1$ be the complement in $U'$ of the image of $U - U_1$, then for $p$ in $U'_1$ the fiber of $U_1 to U'$ above $p$ is of degree $d$.
What I know so far: it is certainly possible for this to fail if we drop the condition that $V$ is a closed subscheme of $mathbb{A}_{mathbb{C}}^2$. Indeed, take $$V = operatorname{Spec} mathbb{C}[[x]][t]/(x(x-t))$$ and take $pi$ to be the morphism of affine schemes induced by the obvious ring map $mathbb{C}[t] to mathbb{C}[[x]][t]/(x(x-t))$. Then $pi$ is quasifinite$ldots$
It is not of finite type, so not quasifinite in the sense of EGA.
$endgroup$
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$begingroup$
Let $V subset mathbb{A}_{mathbb{C}}^2 = operatorname{Spec}mathbb{C}[x,t]$ be a closed subscheme containing the point $x = t = 0$, and suppose we have a quasifinite flat surjective morphism $pi colon V to mathbb{A}_{mathbb{C}}^1 = operatorname{Spec} mathbb{C}[t]$ that sends the point $x = t = 0$ to the point $t = 0$. Choose a sufficiently small analytic neighborhood $U subset mathbb{A}_{mathbb{C}}^2$ containing the point $x = t = 0$ such that every component of $U cap V$ passes through the point $x = t = 0$. Does there exist an analytic neighborhood $U' subset mathbb{A}_{mathbb{C}}^1$ of the point $t = 0$ such that the multiplicity of the fiber $U cap pi^{-1}(p)$ is independent of the choice of $p in U'$?
I think that this holds for sufficiently small $U$, but the condition you impose is not sufficient, e.g. let $U = V$ be defined by $xy = x + t$ and $pi$ be defined by $t^2 - t$. Then the fiber above $0$ is one point---reduced---and the other fibers above closed points consist of $2$ points.
By standard arguments, for some open neighborhoods $U$ and $U'$ we have an induced finite flat map $U to U'$---cf. Lemma in Fischer's Complex Analytic Geometry, page 132---such that the point $x = t = 0$ is the only point mapping to the point $t = 0$. Then for $U'$ connected the direct image of the structure sheaf is a locally free coherent sheaf of some constant rank $d$, and $U_1 subset U$ is an open neighborhood of the origin let $U'_1$ be the complement in $U'$ of the image of $U - U_1$, then for $p$ in $U'_1$ the fiber of $U_1 to U'$ above $p$ is of degree $d$.
What I know so far: it is certainly possible for this to fail if we drop the condition that $V$ is a closed subscheme of $mathbb{A}_{mathbb{C}}^2$. Indeed, take $$V = operatorname{Spec} mathbb{C}[[x]][t]/(x(x-t))$$ and take $pi$ to be the morphism of affine schemes induced by the obvious ring map $mathbb{C}[t] to mathbb{C}[[x]][t]/(x(x-t))$. Then $pi$ is quasifinite$ldots$
It is not of finite type, so not quasifinite in the sense of EGA.
$endgroup$
add a comment |
$begingroup$
Let $V subset mathbb{A}_{mathbb{C}}^2 = operatorname{Spec}mathbb{C}[x,t]$ be a closed subscheme containing the point $x = t = 0$, and suppose we have a quasifinite flat surjective morphism $pi colon V to mathbb{A}_{mathbb{C}}^1 = operatorname{Spec} mathbb{C}[t]$ that sends the point $x = t = 0$ to the point $t = 0$. Choose a sufficiently small analytic neighborhood $U subset mathbb{A}_{mathbb{C}}^2$ containing the point $x = t = 0$ such that every component of $U cap V$ passes through the point $x = t = 0$. Does there exist an analytic neighborhood $U' subset mathbb{A}_{mathbb{C}}^1$ of the point $t = 0$ such that the multiplicity of the fiber $U cap pi^{-1}(p)$ is independent of the choice of $p in U'$?
I think that this holds for sufficiently small $U$, but the condition you impose is not sufficient, e.g. let $U = V$ be defined by $xy = x + t$ and $pi$ be defined by $t^2 - t$. Then the fiber above $0$ is one point---reduced---and the other fibers above closed points consist of $2$ points.
By standard arguments, for some open neighborhoods $U$ and $U'$ we have an induced finite flat map $U to U'$---cf. Lemma in Fischer's Complex Analytic Geometry, page 132---such that the point $x = t = 0$ is the only point mapping to the point $t = 0$. Then for $U'$ connected the direct image of the structure sheaf is a locally free coherent sheaf of some constant rank $d$, and $U_1 subset U$ is an open neighborhood of the origin let $U'_1$ be the complement in $U'$ of the image of $U - U_1$, then for $p$ in $U'_1$ the fiber of $U_1 to U'$ above $p$ is of degree $d$.
What I know so far: it is certainly possible for this to fail if we drop the condition that $V$ is a closed subscheme of $mathbb{A}_{mathbb{C}}^2$. Indeed, take $$V = operatorname{Spec} mathbb{C}[[x]][t]/(x(x-t))$$ and take $pi$ to be the morphism of affine schemes induced by the obvious ring map $mathbb{C}[t] to mathbb{C}[[x]][t]/(x(x-t))$. Then $pi$ is quasifinite$ldots$
It is not of finite type, so not quasifinite in the sense of EGA.
$endgroup$
add a comment |
$begingroup$
Let $V subset mathbb{A}_{mathbb{C}}^2 = operatorname{Spec}mathbb{C}[x,t]$ be a closed subscheme containing the point $x = t = 0$, and suppose we have a quasifinite flat surjective morphism $pi colon V to mathbb{A}_{mathbb{C}}^1 = operatorname{Spec} mathbb{C}[t]$ that sends the point $x = t = 0$ to the point $t = 0$. Choose a sufficiently small analytic neighborhood $U subset mathbb{A}_{mathbb{C}}^2$ containing the point $x = t = 0$ such that every component of $U cap V$ passes through the point $x = t = 0$. Does there exist an analytic neighborhood $U' subset mathbb{A}_{mathbb{C}}^1$ of the point $t = 0$ such that the multiplicity of the fiber $U cap pi^{-1}(p)$ is independent of the choice of $p in U'$?
I think that this holds for sufficiently small $U$, but the condition you impose is not sufficient, e.g. let $U = V$ be defined by $xy = x + t$ and $pi$ be defined by $t^2 - t$. Then the fiber above $0$ is one point---reduced---and the other fibers above closed points consist of $2$ points.
By standard arguments, for some open neighborhoods $U$ and $U'$ we have an induced finite flat map $U to U'$---cf. Lemma in Fischer's Complex Analytic Geometry, page 132---such that the point $x = t = 0$ is the only point mapping to the point $t = 0$. Then for $U'$ connected the direct image of the structure sheaf is a locally free coherent sheaf of some constant rank $d$, and $U_1 subset U$ is an open neighborhood of the origin let $U'_1$ be the complement in $U'$ of the image of $U - U_1$, then for $p$ in $U'_1$ the fiber of $U_1 to U'$ above $p$ is of degree $d$.
What I know so far: it is certainly possible for this to fail if we drop the condition that $V$ is a closed subscheme of $mathbb{A}_{mathbb{C}}^2$. Indeed, take $$V = operatorname{Spec} mathbb{C}[[x]][t]/(x(x-t))$$ and take $pi$ to be the morphism of affine schemes induced by the obvious ring map $mathbb{C}[t] to mathbb{C}[[x]][t]/(x(x-t))$. Then $pi$ is quasifinite$ldots$
It is not of finite type, so not quasifinite in the sense of EGA.
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Let $V subset mathbb{A}_{mathbb{C}}^2 = operatorname{Spec}mathbb{C}[x,t]$ be a closed subscheme containing the point $x = t = 0$, and suppose we have a quasifinite flat surjective morphism $pi colon V to mathbb{A}_{mathbb{C}}^1 = operatorname{Spec} mathbb{C}[t]$ that sends the point $x = t = 0$ to the point $t = 0$. Choose a sufficiently small analytic neighborhood $U subset mathbb{A}_{mathbb{C}}^2$ containing the point $x = t = 0$ such that every component of $U cap V$ passes through the point $x = t = 0$. Does there exist an analytic neighborhood $U' subset mathbb{A}_{mathbb{C}}^1$ of the point $t = 0$ such that the multiplicity of the fiber $U cap pi^{-1}(p)$ is independent of the choice of $p in U'$?
I think that this holds for sufficiently small $U$, but the condition you impose is not sufficient, e.g. let $U = V$ be defined by $xy = x + t$ and $pi$ be defined by $t^2 - t$. Then the fiber above $0$ is one point---reduced---and the other fibers above closed points consist of $2$ points.
By standard arguments, for some open neighborhoods $U$ and $U'$ we have an induced finite flat map $U to U'$---cf. Lemma in Fischer's Complex Analytic Geometry, page 132---such that the point $x = t = 0$ is the only point mapping to the point $t = 0$. Then for $U'$ connected the direct image of the structure sheaf is a locally free coherent sheaf of some constant rank $d$, and $U_1 subset U$ is an open neighborhood of the origin let $U'_1$ be the complement in $U'$ of the image of $U - U_1$, then for $p$ in $U'_1$ the fiber of $U_1 to U'$ above $p$ is of degree $d$.
What I know so far: it is certainly possible for this to fail if we drop the condition that $V$ is a closed subscheme of $mathbb{A}_{mathbb{C}}^2$. Indeed, take $$V = operatorname{Spec} mathbb{C}[[x]][t]/(x(x-t))$$ and take $pi$ to be the morphism of affine schemes induced by the obvious ring map $mathbb{C}[t] to mathbb{C}[[x]][t]/(x(x-t))$. Then $pi$ is quasifinite$ldots$
It is not of finite type, so not quasifinite in the sense of EGA.
answered Jan 8 at 21:30
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What about $V=operatorname{Spec}mathbb{C}[x,t]/(tx^2-x)tooperatorname{Spec}mathbb{C}[t]$. Then $V$ is the union of two irreducible components : $x=0$ and $xt=1$. Both are flat. The first one is onto, hence so is $V$. The second one is not finite, hence neither is $V$.
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– Roland
Dec 16 '18 at 10:12
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@Roland Ah, that's a great counterexample! Now suppose we stipulate that no irreducible components of $V$ go off to infinity as $t to 0$. More precisely, suppose the closure of $V$ in $mathbb{P}_{mathbb{C}}^2$ is not supported at the point defined by $t = 0$ in the line at infinity. Is it then true that $pi$ is finite?
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– Ashvin Swaminathan
Dec 16 '18 at 17:59
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Well the problem does not happen only at $t=0$. Take $V=operatorname{Spec}mathbb{C}[x,t]/((t-1)x^2-x)$. Then there is a problem at $t=1$...
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– Roland
Dec 16 '18 at 18:49
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@Roland Right, this problem could indeed occur at any $t$. I guess I'm more interested in what happens locally near a point of $V$, and I've rewritten my question in that context.
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– Ashvin Swaminathan
Dec 16 '18 at 19:50
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I'm still thinking about it. I honestly didn't understand the example with $xy-x-t = 0$ mapping to $operatorname{Spec} mathbb{C}[t]$ via $(x,y,t) mapsto t^2 - t$. Why is the fiber above $0$ one point? I also didn't understand what condition you're saying is insufficient.
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– Ashvin Swaminathan
Jan 10 at 5:17