Flatness of Residue Field












2












$begingroup$


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:



enter image description here



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:



enter image description here










share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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:



enter image description here



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:



enter image description here










share|cite|improve this question











$endgroup$












  • $begingroup$
    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














2












2








2


1



$begingroup$


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:



enter image description here



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:



enter image description here










share|cite|improve this question











$endgroup$




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:



enter image description here



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:



enter image description here







algebraic-geometry commutative-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 1:46







KarlPeter

















asked Jan 10 at 0:14









KarlPeterKarlPeter

3561315




3561315












  • $begingroup$
    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










  • $begingroup$
    @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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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.






share|cite|improve this answer









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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.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 10 at 3:11









        BenBen

        4,178617




        4,178617






























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