Integral closure error

[jatinm...@gmail.com]

Hi All,

I am trying to find the integral closure of a ring :

F = QQ[x,y,z,v,h,f]/(y^2*v-4*x^9-z^2,x*h-z*v^2+y^5,x*f-v^3+y^3*z).

It's showing that the ring is not normal but when I am trying to find the integral closure of F it's giving me F only and not the integral closure.

Please see the screenshot attached. Any help is highly appreciated.

Thank you.

Regards,

Jatin

[Douglas Leonard]

So I don't know what you were expecting as an answer to the integral closure of this particular quotient ring.

You can already solve for v, then h, then f explicitly as elements of frac(F[x,y,z]), so you shouldn't expect an answer other than what icFractions gives, namely x,y,z,v,h,f.

Doug

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From: maca...@googlegroups.com <maca...@googlegroups.com> on behalf of jatinm...@gmail.com <jatinm...@gmail.com>
Sent: Sunday, May 1, 2022 6:31 AM
To: Macaulay2 <maca...@googlegroups.com>
Subject: [EXT] [Macaulay2] Integral closure error

 

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[Irena Swanson]

I tried to run this integral closure in characteristics 3 and 7 with icFracP, but the calculation did not finish overnight.

You are going mod an ideal that is an intersection of two primes, P1 and P2, both of which have height 3.  The calculation mod one of them is fast, and it returns z/y, z/v, for example.  This means that if you represent the total ring of fractions as K1 \oplus K2, then (z/y, anything in the ring) and (z/v, anything in the ring) is integral over your ring and is not in your ring.

Some integral closure computations take too long, yours may be one of them.

Irena

[Douglas Leonard]

I give up. How about showing what P1 and P2 are explicitly?

Then how about what K1 and K2 are explicitly?

Then, if it is not already obvious, how about integral equations for z/y and z/v?

Doug

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From: maca...@googlegroups.com <maca...@googlegroups.com> on behalf of Irena Swanson <irenas...@gmail.com>
Sent: Tuesday, May 3, 2022 6:30 AM
To: Macaulay2 <maca...@googlegroups.com>
Subject: Re: [EXT] [Macaulay2] Integral closure error

 

To view this discussion on the web visit https://groups.google.com/d/msgid/macaulay2/661880c5-8197-4f6c-825e-b671128a7225n%40googlegroups.com.

[Irena Swanson]

You get P1 and P2 with running primaryDecomposition (of the ideal).  Then Ki = R/Pi. 

The integral closure of R equals the integral closure of R/P1 \oplus R/P2.

Once you see the presentation of R/P1, the equation of integral dependence of z/y and z/v is easy by observation of the one relevant binomials in the generating set.  Namely, one of the generators of P1 is y^2 v - z^2,

so an equation of integral dependence of z/y is (z/y)^2 - v.  Since a Groebner basis of P1 also contains yv^5-z^5, then an equation of integral dependence of z/v is (z/v)^5 - y.

i1 : input "mac/intclMay2022"

ii2 : R = ZZ/7[x,y,z,v,h,f]

oo2 = R

oo2 : PolynomialRing

ii3 : I = ideal (y^2*v-4*x^9-z^2,x*h-z*v^2+y^5,x*f-v^3+y^3*z)

               9    2     2   5      2         3     3
oo3 = ideal (3x  + y v - z , y  - z*v  + x*h, y z - v  + x*f)

oo3 : Ideal of R

ii4 : I == radical I -- true, intersection of two prime ideals, both of height 3

oo4 = true

ii5 :  primaryDecomposition I

                             2     2     3    4   3     3   5      2           3     3  
oo5 = {ideal (x, y v - z , y*z  - v , y z - v , y  - z*v ), ideal (y z - v  +
      --------------------------------------------------------------------------
              2 3    2 2      2            5      2           3 2    6        2  
      x*f, y v  - z v  - x*y f + x*z*h, y  - z*v  + x*h, y*z v  - v  - x*y*z h +
      --------------------------------------------------------------------------
             3     2 2   9     2      2   8 2     2           8 3
      2x*v f - x f , x  - 2y v + 2z , x v  - 2y f + 2z*h, x y  + 2v*h - 2z*f)}

oo5 : List

ii6 :

i7 : icFracP(R/oo5_0)

                     2   2
              z  z  y*z  y   v   v
o7 = {1, -, -, ---, --, --, -}
              y  v   v    v   z  y

o7 : List