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regular sequence $iff$ complete intersection

What is the connection between the definition of complete intersection variety and complete intersection ring?Problem about Complete Intersection in $textbf P^n$ (from Hartshorne).Complete intersectionIrreducibility of smooth intersectionAre complete intersection prime ideals of regular rings regular ideals?Locally complete intersection in a fiberMultiplicity and regular sequencesRegular Ring is a Complete Intersection RingLocal complete intersection

2

$begingroup$

Let $k$ be a field and $X subseteq mathbbP^n$ a closed subscheme of dimension $n-r$.
We say $X$ is complete intersection if $X = V(I)$, where $I$ is a homogeneous ideal which is generated by $r$ elements.

Next, for a closed immersion of locally noetherian schemes $f : X to Y$, we say that $f$ is a regular immersion of codimension $r$ if at every point $x in X$, the kernel of $f^#_x : mathscrO_Y, f(x) to mathscrO_X,x$ is generated by a regular sequence of $r$ elements.

Now for $Y = mathbbP^n$ and $i: X to Y$, are these two conditions equivalent?
In example 3.5. of Liu's "Algebraic geometry and arithmetic curves", the author says that a complete intersection is a regular immersion in this situation.

To show this, I think it's sufficient to show that:

Let $A$ be a noetherian local integral domain of dimension $n$, $I$ a proper ideal and suppose that $dim A/I = n-r$. Then if $I$ is generated by $r$ elements, then $I$ is generated by a regular sequence of $r$ elements.

For our purpose we can assume that $A$ is regular and is finitely generated over a field.

I showed that for a generator $f_1, dots, f_r$, $dim A/(f_1, dots, f_i) = n-i$.
So $f_i+1$ in $A/(f_1, dots, f_i)$ is not contained in any minimal prime ideal.
Therefore if these $A/(f_1, dots, f_i)$ are reduced, this sequence is a regular sequence.
And by this, it seems that we can assume these rings reduced.

Thank you very much.

algebraic-geometry commutative-algebra

share|cite|improve this question

edited Mar 20 at 20:53

user26857

39.5k124283

asked Mar 20 at 19:43

agababibuagababibu

413110

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The highlighted part is wrong unless $A$ is Cohen-Macaulay. A counterexample is given by $A=K[[X^4,X^3Y,XY^3,Y^4]]$, and $I=(X^4,Y^4)$.
  $endgroup$
  – user26857
  Mar 20 at 21:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user26857 Thanks for your comment. So, if $A$ is Cohen-Macaulay, is it true? If so could you show me a proof?
  $endgroup$
  – agababibu
  Mar 20 at 21:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  When $A$ is CM we have $mathrmgrade(I)=dim A-dim A/I=r$, and if $I=(a_1,dots,a_r)$ we get that $a_1,dots,a_r$ is an $A$-sequence. (No need to assume $A$ domain.)
  $endgroup$
  – user26857
  Mar 20 at 21:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user26857 Thank you very much!
  $endgroup$
  – agababibu
  Mar 20 at 22:30
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $k$ be a field and $X subseteq mathbbP^n$ a closed subscheme of dimension $n-r$.
We say $X$ is complete intersection if $X = V(I)$, where $I$ is a homogeneous ideal which is generated by $r$ elements.

Next, for a closed immersion of locally noetherian schemes $f : X to Y$, we say that $f$ is a regular immersion of codimension $r$ if at every point $x in X$, the kernel of $f^#_x : mathscrO_Y, f(x) to mathscrO_X,x$ is generated by a regular sequence of $r$ elements.

Now for $Y = mathbbP^n$ and $i: X to Y$, are these two conditions equivalent?
In example 3.5. of Liu's "Algebraic geometry and arithmetic curves", the author says that a complete intersection is a regular immersion in this situation.

To show this, I think it's sufficient to show that:

Let $A$ be a noetherian local integral domain of dimension $n$, $I$ a proper ideal and suppose that $dim A/I = n-r$. Then if $I$ is generated by $r$ elements, then $I$ is generated by a regular sequence of $r$ elements.

For our purpose we can assume that $A$ is regular and is finitely generated over a field.

I showed that for a generator $f_1, dots, f_r$, $dim A/(f_1, dots, f_i) = n-i$.
So $f_i+1$ in $A/(f_1, dots, f_i)$ is not contained in any minimal prime ideal.
Therefore if these $A/(f_1, dots, f_i)$ are reduced, this sequence is a regular sequence.
And by this, it seems that we can assume these rings reduced.

Thank you very much.

algebraic-geometry commutative-algebra

share|cite|improve this question

edited Mar 20 at 20:53

user26857

39.5k124283

asked Mar 20 at 19:43

agababibuagababibu

413110

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The highlighted part is wrong unless $A$ is Cohen-Macaulay. A counterexample is given by $A=K[[X^4,X^3Y,XY^3,Y^4]]$, and $I=(X^4,Y^4)$.
  $endgroup$
  – user26857
  Mar 20 at 21:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user26857 Thanks for your comment. So, if $A$ is Cohen-Macaulay, is it true? If so could you show me a proof?
  $endgroup$
  – agababibu
  Mar 20 at 21:36
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  When $A$ is CM we have $mathrmgrade(I)=dim A-dim A/I=r$, and if $I=(a_1,dots,a_r)$ we get that $a_1,dots,a_r$ is an $A$-sequence. (No need to assume $A$ domain.)
  $endgroup$
  – user26857
  Mar 20 at 21:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user26857 Thank you very much!
  $endgroup$
  – agababibu
  Mar 20 at 22:30
  

  
  
  
  

add a comment | 

$begingroup$

Let $k$ be a field and $X subseteq mathbbP^n$ a closed subscheme of dimension $n-r$.
We say $X$ is complete intersection if $X = V(I)$, where $I$ is a homogeneous ideal which is generated by $r$ elements.

Next, for a closed immersion of locally noetherian schemes $f : X to Y$, we say that $f$ is a regular immersion of codimension $r$ if at every point $x in X$, the kernel of $f^#_x : mathscrO_Y, f(x) to mathscrO_X,x$ is generated by a regular sequence of $r$ elements.

Now for $Y = mathbbP^n$ and $i: X to Y$, are these two conditions equivalent?
In example 3.5. of Liu's "Algebraic geometry and arithmetic curves", the author says that a complete intersection is a regular immersion in this situation.

To show this, I think it's sufficient to show that:

Let $A$ be a noetherian local integral domain of dimension $n$, $I$ a proper ideal and suppose that $dim A/I = n-r$. Then if $I$ is generated by $r$ elements, then $I$ is generated by a regular sequence of $r$ elements.

For our purpose we can assume that $A$ is regular and is finitely generated over a field.

I showed that for a generator $f_1, dots, f_r$, $dim A/(f_1, dots, f_i) = n-i$.
So $f_i+1$ in $A/(f_1, dots, f_i)$ is not contained in any minimal prime ideal.
Therefore if these $A/(f_1, dots, f_i)$ are reduced, this sequence is a regular sequence.
And by this, it seems that we can assume these rings reduced.

Thank you very much.

algebraic-geometry commutative-algebra

share|cite|improve this question

edited Mar 20 at 20:53

user26857

39.5k124283

asked Mar 20 at 19:43

agababibuagababibu

413110

$endgroup$

share|cite|improve this question

edited Mar 20 at 20:53

user26857

39.5k124283

asked Mar 20 at 19:43

agababibuagababibu

413110

share|cite|improve this question

edited Mar 20 at 20:53

user26857

39.5k124283

asked Mar 20 at 19:43

agababibuagababibu

413110

edited Mar 20 at 20:53

user26857

39.5k124283

edited Mar 20 at 20:53

user26857

39.5k124283

asked Mar 20 at 19:43

agababibuagababibu

413110

asked Mar 20 at 19:43

agababibuagababibu

413110