About the kernel of the structure map of a morphism of schemes

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Let $f:Xto Y$ be a morphism of schemes, let $mathcalK$ be the kernel of the structure map $mathcalO_Yto f_*mathcalO_X$. Do we have

$$mathrmSupp(mathcalO_Y/mathcalK)=overlinef(X)?$$

Recall that the support of a sheaf (of abelian groups) is the set of points at which the stalk of that sheaf is nonzero.

I can prove $subset$ as follows: If $yin Y$ is in the support, then $(f_*mathcalO_X)_yneq 0$ (since the image of $1$ is $1$), so for any open neighborhood $V$ of $y$, we have $mathcalO_X(f^-1V)neq 0$, so $f^-1Vneq emptyset$, thus $yin overlinef(X)$. How about the convese? As $mathcalO_Y/mathcalK$ is of finite type, the support is closed, so one possible way is to show $f(X)subsetmathrmSupp(mathcalO_Y/mathcalK)$.

Somebody told me it's always true but you can add some mild conditions if you need.

edited Jul 30 at 6:04

asked Jul 30 at 5:23

Lao-tzu

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up vote
2
down vote

favorite

Let $f:Xto Y$ be a morphism of schemes, let $mathcalK$ be the kernel of the structure map $mathcalO_Yto f_*mathcalO_X$. Do we have

$$mathrmSupp(mathcalO_Y/mathcalK)=overlinef(X)?$$

Recall that the support of a sheaf (of abelian groups) is the set of points at which the stalk of that sheaf is nonzero.

Somebody told me it's always true but you can add some mild conditions if you need.

edited Jul 30 at 6:04

asked Jul 30 at 5:23

Lao-tzu

1,004923

add a commentÂ |Â

up vote
2
down vote

favorite

Let $f:Xto Y$ be a morphism of schemes, let $mathcalK$ be the kernel of the structure map $mathcalO_Yto f_*mathcalO_X$. Do we have

$$mathrmSupp(mathcalO_Y/mathcalK)=overlinef(X)?$$

Recall that the support of a sheaf (of abelian groups) is the set of points at which the stalk of that sheaf is nonzero.

Somebody told me it's always true but you can add some mild conditions if you need.

edited Jul 30 at 6:04

asked Jul 30 at 5:23

Lao-tzu

1,004923

Let $f:Xto Y$ be a morphism of schemes, let $mathcalK$ be the kernel of the structure map $mathcalO_Yto f_*mathcalO_X$. Do we have

$$mathrmSupp(mathcalO_Y/mathcalK)=overlinef(X)?$$

Recall that the support of a sheaf (of abelian groups) is the set of points at which the stalk of that sheaf is nonzero.

Somebody told me it's always true but you can add some mild conditions if you need.

edited Jul 30 at 6:04

asked Jul 30 at 5:23

Lao-tzu

1,004923

edited Jul 30 at 6:04

asked Jul 30 at 5:23

Lao-tzu

1,004923

asked Jul 30 at 5:23

Lao-tzu

1,004923

asked Jul 30 at 5:23

Lao-tzu

1,004923

add a commentÂ |Â

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I think I can prove "=" in one stroke (without extra assumption), in detail as follows: the key observation is that any ring hom maps $0$ to $0$ and $1$ to $1$, and for the structure sheaf of a scheme $Xneq emptyset$, we have $1neq 0inmathcalO_X(U)Leftrightarrow mathcalO_X(U)neq 0Leftrightarrow Uneq emptyset$. Thus
$$yin mathrmSupp(mathcalO_Y/mathcalK)Leftrightarrow (mathcalO_Y/mathcalK)_yneq 0Leftrightarrow (f_*mathcalO_X)_yneq 0Leftrightarrow 1neq 0in(f_*mathcalO_X)_yLeftrightarrow 1neq 0in(f_*mathcalO_X)(V), forall VLeftrightarrow mathcalO_X(f^-1V)neq 0, forall VLeftrightarrow f^-1Vneq emptyset, forall VLeftrightarrow Vcap f(X)neq emptyset, forall VLeftrightarrow yin overlinef(X),$$
where $forall V$ means for all open neighborhoods $V$ of $y$ in $Y$; the 2nd $Leftrightarrow$ is because we have an injective ring hom $(mathcalO_Y/mathcalK)_yrightarrow(f_*mathcalO_X)_y$; the 4th $Leftrightarrow$ is because the stalk is a filtred colimit over all open neighborhoods. Thus
$$mathrmSupp(mathcalO_Y/mathcalK)=overlinef(X).$$

edited Jul 30 at 23:48

answered Jul 30 at 23:40

Lao-tzu

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If you find an unreasonable point, please make a comment.
â€“Â Lao-tzu
Jul 30 at 23:45

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Here's one way to see this if we assume that $f$ is quasi-compact:

First note that the statement is true if both $X,Y$ are affine, say $X=operatornameSpec(B)$ and $Y=operatornameSpec(A)$. Indeed in this case, we have that $overlinef(X) = V(I)$, where $I$ is the kernel of the map $Arightarrow B$. Of course here we have that $operatornameSupp(A/I) = V(I)$.

Now let's suppose that $f$ is quasi-compact. We can assume that $Y$ is affine. Since $f$ is quasi-compact, we can cover $X$ by finitely many affine opens $U_i$, and we have that $overlinef(X) = overline bigcup_i f(U_i) = bigcup_i overlinef(U_i)$ (we used finiteness for the last equality).

By the above, we have that $overlinef(U_i) = operatornameSupp(mathcalO_Y/K_U_i)$, where $K_U_i$ is the kernel of the map $mathcalO_Y rightarrow (f|_U_i)_* mathcalO_U_i$. We claim that $operatornameSupp(mathcalO_Y/K_U_i) subseteq operatornameSupp(mathcalO_Y/K_X)$, where $K_X$ is the kernel of the map $mathcalO_Y rightarrow f_* mathcalO_X$. This follows from the fact that $K_X subseteq K_U_i$.

This implies that $overlinef(X) subseteq operatornameSupp(mathcalO_Y/K_X)$.

answered Jul 30 at 22:29

loch

1,403169

Don't you need $Y$ compact for the reduction to $Y$ affine to work?
â€“Â user347489
Jul 30 at 22:41

@user347489 ah I think you're right
â€“Â loch
Jul 30 at 22:50

I don't see why one needs $Y$ affine for the reduction to work - we have $overlinef(X)^X cap U = overlineU cap f(X)^U$ (here I use a supscript to indicate in which ambient space the closure is taken), so the whole statement is local on $Y$. Possibly interesting sidenote: Under your assumptions, you can replace $X$ by $coprod U_i$ without changing $K$. Then your morphism becomes qcqs, so $f_* O_X$ and thus $K$ is quasi-coherent, which gives a scheme structure on $overlinef(X)$, known as scheme-theoretic image.
â€“Â Johann Haas
Jul 31 at 13:43

@JohannHaas The statement is local on $Y$, hence I can assume that $Y$ is affine. If you're asking if this whole thing is necessary - then I suppose not - see OP's answer.
â€“Â loch
Jul 31 at 13:58

@loch Indeed. I just wanted to correct the confusion between you and user347389. I still think your argument has some merit: The OP's is more general and works for arbitrary locally ringed spaces; but yours can be extended to obtain a refinement of the OP's statement involving subscheme structures.
â€“Â Johann Haas
Jul 31 at 18:13

Â |Â
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edited Jul 30 at 23:48

answered Jul 30 at 23:40

Lao-tzu

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If you find an unreasonable point, please make a comment.
â€“Â Lao-tzu
Jul 30 at 23:45

add a commentÂ |Â

up vote
2
down vote

edited Jul 30 at 23:48

answered Jul 30 at 23:40

Lao-tzu

1,004923

If you find an unreasonable point, please make a comment.
â€“Â Lao-tzu
Jul 30 at 23:45

add a commentÂ |Â

up vote
2
down vote

edited Jul 30 at 23:48

answered Jul 30 at 23:40

Lao-tzu

1,004923

edited Jul 30 at 23:48

answered Jul 30 at 23:40

Lao-tzu

1,004923

edited Jul 30 at 23:48

answered Jul 30 at 23:40

Lao-tzu

1,004923

answered Jul 30 at 23:40

Lao-tzu

1,004923

answered Jul 30 at 23:40

Lao-tzu

1,004923

If you find an unreasonable point, please make a comment.
â€“Â Lao-tzu
Jul 30 at 23:45

add a commentÂ |Â

If you find an unreasonable point, please make a comment.
â€“Â Lao-tzu
Jul 30 at 23:45

If you find an unreasonable point, please make a comment.
â€“Â Lao-tzu
Jul 30 at 23:45

add a commentÂ |Â

up vote
1
down vote

Here's one way to see this if we assume that $f$ is quasi-compact:

This implies that $overlinef(X) subseteq operatornameSupp(mathcalO_Y/K_X)$.

answered Jul 30 at 22:29

loch

1,403169

Don't you need $Y$ compact for the reduction to $Y$ affine to work?
â€“Â user347489
Jul 30 at 22:41

@user347489 ah I think you're right
â€“Â loch
Jul 30 at 22:50

I don't see why one needs $Y$ affine for the reduction to work - we have $overlinef(X)^X cap U = overlineU cap f(X)^U$ (here I use a supscript to indicate in which ambient space the closure is taken), so the whole statement is local on $Y$. Possibly interesting sidenote: Under your assumptions, you can replace $X$ by $coprod U_i$ without changing $K$. Then your morphism becomes qcqs, so $f_* O_X$ and thus $K$ is quasi-coherent, which gives a scheme structure on $overlinef(X)$, known as scheme-theoretic image.
â€“Â Johann Haas
Jul 31 at 13:43

@JohannHaas The statement is local on $Y$, hence I can assume that $Y$ is affine. If you're asking if this whole thing is necessary - then I suppose not - see OP's answer.
â€“Â loch
Jul 31 at 13:58

@loch Indeed. I just wanted to correct the confusion between you and user347389. I still think your argument has some merit: The OP's is more general and works for arbitrary locally ringed spaces; but yours can be extended to obtain a refinement of the OP's statement involving subscheme structures.
â€“Â Johann Haas
Jul 31 at 18:13

Â |Â
show 3 more comments

up vote
1
down vote

Here's one way to see this if we assume that $f$ is quasi-compact:

This implies that $overlinef(X) subseteq operatornameSupp(mathcalO_Y/K_X)$.

answered Jul 30 at 22:29

loch

1,403169

Don't you need $Y$ compact for the reduction to $Y$ affine to work?
â€“Â user347489
Jul 30 at 22:41

@user347489 ah I think you're right
â€“Â loch
Jul 30 at 22:50

I don't see why one needs $Y$ affine for the reduction to work - we have $overlinef(X)^X cap U = overlineU cap f(X)^U$ (here I use a supscript to indicate in which ambient space the closure is taken), so the whole statement is local on $Y$. Possibly interesting sidenote: Under your assumptions, you can replace $X$ by $coprod U_i$ without changing $K$. Then your morphism becomes qcqs, so $f_* O_X$ and thus $K$ is quasi-coherent, which gives a scheme structure on $overlinef(X)$, known as scheme-theoretic image.
â€“Â Johann Haas
Jul 31 at 13:43

@JohannHaas The statement is local on $Y$, hence I can assume that $Y$ is affine. If you're asking if this whole thing is necessary - then I suppose not - see OP's answer.
â€“Â loch
Jul 31 at 13:58

@loch Indeed. I just wanted to correct the confusion between you and user347389. I still think your argument has some merit: The OP's is more general and works for arbitrary locally ringed spaces; but yours can be extended to obtain a refinement of the OP's statement involving subscheme structures.
â€“Â Johann Haas
Jul 31 at 18:13

Â |Â
show 3 more comments

up vote
1
down vote

Here's one way to see this if we assume that $f$ is quasi-compact:

This implies that $overlinef(X) subseteq operatornameSupp(mathcalO_Y/K_X)$.

answered Jul 30 at 22:29

loch

1,403169

Here's one way to see this if we assume that $f$ is quasi-compact:

This implies that $overlinef(X) subseteq operatornameSupp(mathcalO_Y/K_X)$.

answered Jul 30 at 22:29

loch

1,403169

answered Jul 30 at 22:29

loch

1,403169

answered Jul 30 at 22:29

loch

1,403169

answered Jul 30 at 22:29

loch

1,403169

Don't you need $Y$ compact for the reduction to $Y$ affine to work?
â€“Â user347489
Jul 30 at 22:41

@user347489 ah I think you're right
â€“Â loch
Jul 30 at 22:50

I don't see why one needs $Y$ affine for the reduction to work - we have $overlinef(X)^X cap U = overlineU cap f(X)^U$ (here I use a supscript to indicate in which ambient space the closure is taken), so the whole statement is local on $Y$. Possibly interesting sidenote: Under your assumptions, you can replace $X$ by $coprod U_i$ without changing $K$. Then your morphism becomes qcqs, so $f_* O_X$ and thus $K$ is quasi-coherent, which gives a scheme structure on $overlinef(X)$, known as scheme-theoretic image.
â€“Â Johann Haas
Jul 31 at 13:43

@JohannHaas The statement is local on $Y$, hence I can assume that $Y$ is affine. If you're asking if this whole thing is necessary - then I suppose not - see OP's answer.
â€“Â loch
Jul 31 at 13:58

@loch Indeed. I just wanted to correct the confusion between you and user347389. I still think your argument has some merit: The OP's is more general and works for arbitrary locally ringed spaces; but yours can be extended to obtain a refinement of the OP's statement involving subscheme structures.
â€“Â Johann Haas
Jul 31 at 18:13

Â |Â
show 3 more comments

Don't you need $Y$ compact for the reduction to $Y$ affine to work?
â€“Â user347489
Jul 30 at 22:41

@user347489 ah I think you're right
â€“Â loch
Jul 30 at 22:50

I don't see why one needs $Y$ affine for the reduction to work - we have $overlinef(X)^X cap U = overlineU cap f(X)^U$ (here I use a supscript to indicate in which ambient space the closure is taken), so the whole statement is local on $Y$. Possibly interesting sidenote: Under your assumptions, you can replace $X$ by $coprod U_i$ without changing $K$. Then your morphism becomes qcqs, so $f_* O_X$ and thus $K$ is quasi-coherent, which gives a scheme structure on $overlinef(X)$, known as scheme-theoretic image.
â€“Â Johann Haas
Jul 31 at 13:43

@JohannHaas The statement is local on $Y$, hence I can assume that $Y$ is affine. If you're asking if this whole thing is necessary - then I suppose not - see OP's answer.
â€“Â loch
Jul 31 at 13:58

@loch Indeed. I just wanted to correct the confusion between you and user347389. I still think your argument has some merit: The OP's is more general and works for arbitrary locally ringed spaces; but yours can be extended to obtain a refinement of the OP's statement involving subscheme structures.
â€“Â Johann Haas
Jul 31 at 18:13

Don't you need $Y$ compact for the reduction to $Y$ affine to work?
â€“Â user347489
Jul 30 at 22:41

@user347489 ah I think you're right
â€“Â loch
Jul 30 at 22:50

I don't see why one needs $Y$ affine for the reduction to work - we have $overlinef(X)^X cap U = overlineU cap f(X)^U$ (here I use a supscript to indicate in which ambient space the closure is taken), so the whole statement is local on $Y$. Possibly interesting sidenote: Under your assumptions, you can replace $X$ by $coprod U_i$ without changing $K$. Then your morphism becomes qcqs, so $f_* O_X$ and thus $K$ is quasi-coherent, which gives a scheme structure on $overlinef(X)$, known as scheme-theoretic image.
â€“Â Johann Haas
Jul 31 at 13:43

@JohannHaas The statement is local on $Y$, hence I can assume that $Y$ is affine. If you're asking if this whole thing is necessary - then I suppose not - see OP's answer.
â€“Â loch
Jul 31 at 13:58

@loch Indeed. I just wanted to correct the confusion between you and user347389. I still think your argument has some merit: The OP's is more general and works for arbitrary locally ringed spaces; but yours can be extended to obtain a refinement of the OP's statement involving subscheme structures.
â€“Â Johann Haas
Jul 31 at 18:13

Â |Â
show 3 more comments

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