Mixing
Flatness of Residue Field
$begingroup$
My question refers to a step of in the proof of Corollary 8.5.17 in Bosch's "Commutative Algebra and Algebraic Geometry"; see page 395
See the red tagged line below:
We consider the exact sequence
$$0 to I/J to R[t_1, ..., t_n]/J to R[t_1, ..., t_n]/I to 0$$
and we tensor it with $k(s) = mathcal{O}_{S,s}/m_s$.
Why does it stay exact? Indeed, by assumption $mathcal{O}_{S,s}$ is flat so it's ok to tensor it with $mathcal{O}_{S,s}$ but what about $k(s)$? Why does it conserve the exactness?
In addition: Lemma 5.2.9:
algebraic-geometry commutative-algebra
asked Jan 10 at 0:14
KarlPeterKarlPeter
5461315
$endgroup$
|
show 1 more comment
$begingroup$
My question refers to a step of in the proof of Corollary 8.5.17 in Bosch's "Commutative Algebra and Algebraic Geometry"; see page 395
See the red tagged line below:
We consider the exact sequence
$$0 to I/J to R[t_1, ..., t_n]/J to R[t_1, ..., t_n]/I to 0$$
and we tensor it with $k(s) = mathcal{O}_{S,s}/m_s$.
Why does it stay exact? Indeed, by assumption $mathcal{O}_{S,s}$ is flat so it's ok to tensor it with $mathcal{O}_{S,s}$ but what about $k(s)$? Why does it conserve the exactness?
In addition: Lemma 5.2.9:
algebraic-geometry commutative-algebra
asked Jan 10 at 0:14
KarlPeterKarlPeter
5461315
$endgroup$
$begingroup$
Isn't the assumption $R[t_1,ldots, t_n]/I$ flat over $R$? Then, tensoring over $R$ with $k(s)=R/M$, $M$ a maximal ideal, stays exact. Easiest way is to say that the kernel on the left is $Tor^1_R(k(s), R[t_1,ldots.t_n]/I)=0$ by flatness.
$endgroup$
– Mohan
Jan 10 at 1:05
$begingroup$
@Mohan: I'm not sure why $R$ should be flat over $R[t_1,ldots, t_n]/I$. Well, firstly we can indeed assume $S= Spec(R)$ and $X= Spec(R[t_1,ldots, t_n]/I)$ since problem is local. The assumption (ii) says that $f$ is only flat in $x$ not everywhere, so we know that $mathcal{O}_{X,x}$ is flat over $mathcal{O}_{S,s}$, right? I don't see why flatness is inherited for $R[t_1,ldots, t_n]/I$ over $R$. Please, correct me if I have overseen some argument.
$endgroup$
– KarlPeter
Jan 10 at 1:21
$begingroup$
What does 5.2/9 say?
$endgroup$
– Ben
Jan 10 at 1:32
$begingroup$
@Ben: I added it above. So the the core problem stays to see why $R[t_1,ldots, t_n]/I$ is flat over $R$ althought we only know that $f$ is flat in $x$.
$endgroup$
– KarlPeter
Jan 10 at 1:48
$begingroup$
Localize the sequence at $m_x$, which is still exact (localization is exact). Then by assumption the third term is flat over $R_s$, so apply 5.2/9 to the module $k(s) = R_s/m_s$. Does that do it?
$endgroup$
– Ben
Jan 10 at 2:40
|
show 1 more comment
$begingroup$
My question refers to a step of in the proof of Corollary 8.5.17 in Bosch's "Commutative Algebra and Algebraic Geometry"; see page 395
See the red tagged line below:
We consider the exact sequence
$$0 to I/J to R[t_1, ..., t_n]/J to R[t_1, ..., t_n]/I to 0$$
and we tensor it with $k(s) = mathcal{O}_{S,s}/m_s$.
Why does it stay exact? Indeed, by assumption $mathcal{O}_{S,s}$ is flat so it's ok to tensor it with $mathcal{O}_{S,s}$ but what about $k(s)$? Why does it conserve the exactness?
In addition: Lemma 5.2.9:
algebraic-geometry commutative-algebra
asked Jan 10 at 0:14
KarlPeterKarlPeter
5461315
$endgroup$
My question refers to a step of in the proof of Corollary 8.5.17 in Bosch's "Commutative Algebra and Algebraic Geometry"; see page 395
See the red tagged line below:
We consider the exact sequence
$$0 to I/J to R[t_1, ..., t_n]/J to R[t_1, ..., t_n]/I to 0$$
and we tensor it with $k(s) = mathcal{O}_{S,s}/m_s$.
Why does it stay exact? Indeed, by assumption $mathcal{O}_{S,s}$ is flat so it's ok to tensor it with $mathcal{O}_{S,s}$ but what about $k(s)$? Why does it conserve the exactness?
In addition: Lemma 5.2.9:
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
asked Jan 10 at 0:14
KarlPeterKarlPeter
5461315
asked Jan 10 at 0:14
KarlPeterKarlPeter
5461315
edited Jan 10 at 1:46
KarlPeter
asked Jan 10 at 0:14
KarlPeterKarlPeter
5461315
asked Jan 10 at 0:14
KarlPeterKarlPeter
5461315
asked Jan 10 at 0:14
KarlPeterKarlPeter
5461315
5461315
$begingroup$
Isn't the assumption $R[t_1,ldots, t_n]/I$ flat over $R$? Then, tensoring over $R$ with $k(s)=R/M$, $M$ a maximal ideal, stays exact. Easiest way is to say that the kernel on the left is $Tor^1_R(k(s), R[t_1,ldots.t_n]/I)=0$ by flatness.
$endgroup$
– Mohan
Jan 10 at 1:05
$begingroup$
@Mohan: I'm not sure why $R$ should be flat over $R[t_1,ldots, t_n]/I$. Well, firstly we can indeed assume $S= Spec(R)$ and $X= Spec(R[t_1,ldots, t_n]/I)$ since problem is local. The assumption (ii) says that $f$ is only flat in $x$ not everywhere, so we know that $mathcal{O}_{X,x}$ is flat over $mathcal{O}_{S,s}$, right? I don't see why flatness is inherited for $R[t_1,ldots, t_n]/I$ over $R$. Please, correct me if I have overseen some argument.
$endgroup$
– KarlPeter
Jan 10 at 1:21
$begingroup$
What does 5.2/9 say?
$endgroup$
– Ben
Jan 10 at 1:32
$begingroup$
@Ben: I added it above. So the the core problem stays to see why $R[t_1,ldots, t_n]/I$ is flat over $R$ althought we only know that $f$ is flat in $x$.
$endgroup$
– KarlPeter
Jan 10 at 1:48
$begingroup$
Localize the sequence at $m_x$, which is still exact (localization is exact). Then by assumption the third term is flat over $R_s$, so apply 5.2/9 to the module $k(s) = R_s/m_s$. Does that do it?
$endgroup$
– Ben
Jan 10 at 2:40
|
show 1 more comment
$begingroup$
Isn't the assumption $R[t_1,ldots, t_n]/I$ flat over $R$? Then, tensoring over $R$ with $k(s)=R/M$, $M$ a maximal ideal, stays exact. Easiest way is to say that the kernel on the left is $Tor^1_R(k(s), R[t_1,ldots.t_n]/I)=0$ by flatness.
$endgroup$
– Mohan
Jan 10 at 1:05
$begingroup$
@Mohan: I'm not sure why $R$ should be flat over $R[t_1,ldots, t_n]/I$. Well, firstly we can indeed assume $S= Spec(R)$ and $X= Spec(R[t_1,ldots, t_n]/I)$ since problem is local. The assumption (ii) says that $f$ is only flat in $x$ not everywhere, so we know that $mathcal{O}_{X,x}$ is flat over $mathcal{O}_{S,s}$, right? I don't see why flatness is inherited for $R[t_1,ldots, t_n]/I$ over $R$. Please, correct me if I have overseen some argument.
$endgroup$
– KarlPeter
Jan 10 at 1:21
$begingroup$
What does 5.2/9 say?
$endgroup$
– Ben
Jan 10 at 1:32
$begingroup$
@Ben: I added it above. So the the core problem stays to see why $R[t_1,ldots, t_n]/I$ is flat over $R$ althought we only know that $f$ is flat in $x$.
$endgroup$
– KarlPeter
Jan 10 at 1:48
$begingroup$
Localize the sequence at $m_x$, which is still exact (localization is exact). Then by assumption the third term is flat over $R_s$, so apply 5.2/9 to the module $k(s) = R_s/m_s$. Does that do it?
$endgroup$
– Ben
Jan 10 at 2:40
$begingroup$
Isn't the assumption $R[t_1,ldots, t_n]/I$ flat over $R$? Then, tensoring over $R$ with $k(s)=R/M$, $M$ a maximal ideal, stays exact. Easiest way is to say that the kernel on the left is $Tor^1_R(k(s), R[t_1,ldots.t_n]/I)=0$ by flatness.
$endgroup$
– Mohan
Jan 10 at 1:05
$begingroup$
Isn't the assumption $R[t_1,ldots, t_n]/I$ flat over $R$? Then, tensoring over $R$ with $k(s)=R/M$, $M$ a maximal ideal, stays exact. Easiest way is to say that the kernel on the left is $Tor^1_R(k(s), R[t_1,ldots.t_n]/I)=0$ by flatness.
$endgroup$
– Mohan
Jan 10 at 1:05
$begingroup$
@Mohan: I'm not sure why $R$ should be flat over $R[t_1,ldots, t_n]/I$. Well, firstly we can indeed assume $S= Spec(R)$ and $X= Spec(R[t_1,ldots, t_n]/I)$ since problem is local. The assumption (ii) says that $f$ is only flat in $x$ not everywhere, so we know that $mathcal{O}_{X,x}$ is flat over $mathcal{O}_{S,s}$, right? I don't see why flatness is inherited for $R[t_1,ldots, t_n]/I$ over $R$. Please, correct me if I have overseen some argument.
$endgroup$
– KarlPeter
Jan 10 at 1:21
$begingroup$
@Mohan: I'm not sure why $R$ should be flat over $R[t_1,ldots, t_n]/I$. Well, firstly we can indeed assume $S= Spec(R)$ and $X= Spec(R[t_1,ldots, t_n]/I)$ since problem is local. The assumption (ii) says that $f$ is only flat in $x$ not everywhere, so we know that $mathcal{O}_{X,x}$ is flat over $mathcal{O}_{S,s}$, right? I don't see why flatness is inherited for $R[t_1,ldots, t_n]/I$ over $R$. Please, correct me if I have overseen some argument.
$endgroup$
– KarlPeter
Jan 10 at 1:21
$begingroup$
What does 5.2/9 say?
$endgroup$
– Ben
Jan 10 at 1:32
$begingroup$
What does 5.2/9 say?
$endgroup$
– Ben
Jan 10 at 1:32
$begingroup$
@Ben: I added it above. So the the core problem stays to see why $R[t_1,ldots, t_n]/I$ is flat over $R$ althought we only know that $f$ is flat in $x$.
$endgroup$
– KarlPeter
Jan 10 at 1:48
$begingroup$
@Ben: I added it above. So the the core problem stays to see why $R[t_1,ldots, t_n]/I$ is flat over $R$ althought we only know that $f$ is flat in $x$.
$endgroup$
– KarlPeter
Jan 10 at 1:48
$begingroup$
Localize the sequence at $m_x$, which is still exact (localization is exact). Then by assumption the third term is flat over $R_s$, so apply 5.2/9 to the module $k(s) = R_s/m_s$. Does that do it?
$endgroup$
– Ben
Jan 10 at 2:40
$begingroup$
Localize the sequence at $m_x$, which is still exact (localization is exact). Then by assumption the third term is flat over $R_s$, so apply 5.2/9 to the module $k(s) = R_s/m_s$. Does that do it?
$endgroup$
– Ben
Jan 10 at 2:40
|
show 1 more comment
1 Answer
1
active
oldest
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$begingroup$
Set $T = R[t_1,ldots, t_n]$. Our assumption that $f$ is flat at $x$ means that $(T/I)_x$ is flat over $R_s$.
We have an exact sequence:
$$ 0 to I/J to T/J to T/I to 0$$
Localization is exact so we again have the localized exact sequence:
$$0 to (I/J)_x to (T/J)_x to (T/I)_x to 0$$
Now in 5.2/9, we take $M''$ to be $(T/I)_x$ since it is flat over $R_s$, and take $N$ to be $k(s) = R_s/m_s$. We get the exact sequence we wanted:
$$0 to (I/J)_xotimes k(s) to (T/J)_xotimes k(s) to (T/I)_xotimes k(s) to 0$$
This justifies the statement "[the sequence] remains exact at $x$ when tensoring over $R$ it with $k(s)$".
(Sweeping it under the rug, but we just used that Localization commutes with tensor products to identify $Motimes_{R_s} k(s) = Motimes_{R}k(s)$ for an $R_s$-module $M$ by the way.)
Watch out for the typo in the next line though, $J/I$ should be $I/J$.
Per the title of this question: The residue field $k(s)$ is not going to be a flat module, see e.g. is residue field ever flat over its local ring? on MathOverflow, so it's important that $N$ in 5.2/9 can be any module.
answered Jan 10 at 3:11
BenBen
3,716616
$endgroup$
add a comment |
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$begingroup$
Set $T = R[t_1,ldots, t_n]$. Our assumption that $f$ is flat at $x$ means that $(T/I)_x$ is flat over $R_s$.
We have an exact sequence:
$$ 0 to I/J to T/J to T/I to 0$$
Localization is exact so we again have the localized exact sequence:
$$0 to (I/J)_x to (T/J)_x to (T/I)_x to 0$$
Now in 5.2/9, we take $M''$ to be $(T/I)_x$ since it is flat over $R_s$, and take $N$ to be $k(s) = R_s/m_s$. We get the exact sequence we wanted:
$$0 to (I/J)_xotimes k(s) to (T/J)_xotimes k(s) to (T/I)_xotimes k(s) to 0$$
This justifies the statement "[the sequence] remains exact at $x$ when tensoring over $R$ it with $k(s)$".
(Sweeping it under the rug, but we just used that Localization commutes with tensor products to identify $Motimes_{R_s} k(s) = Motimes_{R}k(s)$ for an $R_s$-module $M$ by the way.)
Watch out for the typo in the next line though, $J/I$ should be $I/J$.
Per the title of this question: The residue field $k(s)$ is not going to be a flat module, see e.g. is residue field ever flat over its local ring? on MathOverflow, so it's important that $N$ in 5.2/9 can be any module.
answered Jan 10 at 3:11
BenBen
3,716616
$endgroup$
add a comment |
$begingroup$
Set $T = R[t_1,ldots, t_n]$. Our assumption that $f$ is flat at $x$ means that $(T/I)_x$ is flat over $R_s$.
We have an exact sequence:
$$ 0 to I/J to T/J to T/I to 0$$
Localization is exact so we again have the localized exact sequence:
$$0 to (I/J)_x to (T/J)_x to (T/I)_x to 0$$
Now in 5.2/9, we take $M''$ to be $(T/I)_x$ since it is flat over $R_s$, and take $N$ to be $k(s) = R_s/m_s$. We get the exact sequence we wanted:
$$0 to (I/J)_xotimes k(s) to (T/J)_xotimes k(s) to (T/I)_xotimes k(s) to 0$$
This justifies the statement "[the sequence] remains exact at $x$ when tensoring over $R$ it with $k(s)$".
(Sweeping it under the rug, but we just used that Localization commutes with tensor products to identify $Motimes_{R_s} k(s) = Motimes_{R}k(s)$ for an $R_s$-module $M$ by the way.)
Watch out for the typo in the next line though, $J/I$ should be $I/J$.
Per the title of this question: The residue field $k(s)$ is not going to be a flat module, see e.g. is residue field ever flat over its local ring? on MathOverflow, so it's important that $N$ in 5.2/9 can be any module.
answered Jan 10 at 3:11
BenBen
3,716616
$endgroup$
add a comment |
$begingroup$
Set $T = R[t_1,ldots, t_n]$. Our assumption that $f$ is flat at $x$ means that $(T/I)_x$ is flat over $R_s$.
We have an exact sequence:
$$ 0 to I/J to T/J to T/I to 0$$
Localization is exact so we again have the localized exact sequence:
$$0 to (I/J)_x to (T/J)_x to (T/I)_x to 0$$
Now in 5.2/9, we take $M''$ to be $(T/I)_x$ since it is flat over $R_s$, and take $N$ to be $k(s) = R_s/m_s$. We get the exact sequence we wanted:
$$0 to (I/J)_xotimes k(s) to (T/J)_xotimes k(s) to (T/I)_xotimes k(s) to 0$$
This justifies the statement "[the sequence] remains exact at $x$ when tensoring over $R$ it with $k(s)$".
(Sweeping it under the rug, but we just used that Localization commutes with tensor products to identify $Motimes_{R_s} k(s) = Motimes_{R}k(s)$ for an $R_s$-module $M$ by the way.)
Watch out for the typo in the next line though, $J/I$ should be $I/J$.
Per the title of this question: The residue field $k(s)$ is not going to be a flat module, see e.g. is residue field ever flat over its local ring? on MathOverflow, so it's important that $N$ in 5.2/9 can be any module.
answered Jan 10 at 3:11
BenBen
3,716616
$endgroup$
Set $T = R[t_1,ldots, t_n]$. Our assumption that $f$ is flat at $x$ means that $(T/I)_x$ is flat over $R_s$.
We have an exact sequence:
$$ 0 to I/J to T/J to T/I to 0$$
Localization is exact so we again have the localized exact sequence:
$$0 to (I/J)_x to (T/J)_x to (T/I)_x to 0$$
Now in 5.2/9, we take $M''$ to be $(T/I)_x$ since it is flat over $R_s$, and take $N$ to be $k(s) = R_s/m_s$. We get the exact sequence we wanted:
$$0 to (I/J)_xotimes k(s) to (T/J)_xotimes k(s) to (T/I)_xotimes k(s) to 0$$
This justifies the statement "[the sequence] remains exact at $x$ when tensoring over $R$ it with $k(s)$".
(Sweeping it under the rug, but we just used that Localization commutes with tensor products to identify $Motimes_{R_s} k(s) = Motimes_{R}k(s)$ for an $R_s$-module $M$ by the way.)
Watch out for the typo in the next line though, $J/I$ should be $I/J$.
Per the title of this question: The residue field $k(s)$ is not going to be a flat module, see e.g. is residue field ever flat over its local ring? on MathOverflow, so it's important that $N$ in 5.2/9 can be any module.
answered Jan 10 at 3:11
BenBen
3,716616
answered Jan 10 at 3:11
BenBen
3,716616
answered Jan 10 at 3:11
BenBen
3,716616
answered Jan 10 at 3:11
BenBen
3,716616
3,716616
add a comment |
add a comment |
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$begingroup$
Isn't the assumption $R[t_1,ldots, t_n]/I$ flat over $R$? Then, tensoring over $R$ with $k(s)=R/M$, $M$ a maximal ideal, stays exact. Easiest way is to say that the kernel on the left is $Tor^1_R(k(s), R[t_1,ldots.t_n]/I)=0$ by flatness.
$endgroup$
– Mohan
Jan 10 at 1:05
$begingroup$
@Mohan: I'm not sure why $R$ should be flat over $R[t_1,ldots, t_n]/I$. Well, firstly we can indeed assume $S= Spec(R)$ and $X= Spec(R[t_1,ldots, t_n]/I)$ since problem is local. The assumption (ii) says that $f$ is only flat in $x$ not everywhere, so we know that $mathcal{O}_{X,x}$ is flat over $mathcal{O}_{S,s}$, right? I don't see why flatness is inherited for $R[t_1,ldots, t_n]/I$ over $R$. Please, correct me if I have overseen some argument.
$endgroup$
– KarlPeter
Jan 10 at 1:21
$begingroup$
What does 5.2/9 say?
$endgroup$
– Ben
Jan 10 at 1:32
$begingroup$
@Ben: I added it above. So the the core problem stays to see why $R[t_1,ldots, t_n]/I$ is flat over $R$ althought we only know that $f$ is flat in $x$.
$endgroup$
– KarlPeter
Jan 10 at 1:48
$begingroup$
Localize the sequence at $m_x$, which is still exact (localization is exact). Then by assumption the third term is flat over $R_s$, so apply 5.2/9 to the module $k(s) = R_s/m_s$. Does that do it?
$endgroup$
– Ben
Jan 10 at 2:40