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Kiu
Kiu
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Hal Schenck kindly provided a counter-example to the claim in question. Consider the twisted cubic: (the code is in Macaulay) 2 2 o3

o3 I = ideal (- yy^2 + xz, - yz + xw, - zz^2 + yw)

i4 : J = ideal leadTerm I

          2        2

o4 = ideal (zz^2 , y*z, yy^2 )

i5 : hilbertPolynomial coker gens ideal leadTerm I^2

o5 = - 16P + 9P 0 1

i6 : hilbertPolynomial coker gens ideal leadTerm J^2

o6 = - 20PP_0 + 10P 0 1 SoP_1

So, the quotient is actually positive dimensional.

Hal Schenck kindly provided a counter-example to the claim in question. Consider the twisted cubic: (the code is in Macaulay) 2 2 o3 I = ideal (- y + xz, - yz + xw, - z + yw)

i4 : J = ideal leadTerm I

          2        2

o4 = ideal (z , y*z, y )

i5 : hilbertPolynomial coker gens ideal leadTerm I^2

o5 = - 16P + 9P 0 1

i6 : hilbertPolynomial coker gens ideal leadTerm J^2

o6 = - 20P + 10P 0 1 So, the quotient is actually positive dimensional.

Hal Schenck kindly provided a counter-example to the claim in question. Consider the twisted cubic: (the code is in Macaulay)

o3 I = ideal (- y^2 + xz, - yz + xw, - z^2 + yw)

i4 : J = ideal leadTerm I

o4 = ideal (z^2 , y*z, y^2 )

i5 : hilbertPolynomial coker gens ideal leadTerm I^2

o5 = - 16P + 9P 0 1

i6 : hilbertPolynomial coker gens ideal leadTerm J^2

o6 = - 20P_0 + 10P_1

So, the quotient is actually positive dimensional.

Source Link
Kiu
Kiu
  • 893
  • 4
  • 15

Hal Schenck kindly provided a counter-example to the claim in question. Consider the twisted cubic: (the code is in Macaulay) 2 2 o3 I = ideal (- y + xz, - yz + xw, - z + yw)

i4 : J = ideal leadTerm I

          2        2

o4 = ideal (z , y*z, y )

i5 : hilbertPolynomial coker gens ideal leadTerm I^2

o5 = - 16P + 9P 0 1

i6 : hilbertPolynomial coker gens ideal leadTerm J^2

o6 = - 20P + 10P 0 1 So, the quotient is actually positive dimensional.