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Is there a method to compute the radical of an ideal? for example take $J=(xw-y^{2},xw^{2}-z^{3}) \subset k[x,y,z,w]$. I want to show $J$ is not radical, I guess the idea is to add and substract terms until we get something which is in the radical but not in $J$. However I don't see the trick for this particular case. Can you please help?

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Maybe you meant $x^2 - y$ and $x^3 - z$ ? – Lierre May 3 '12 at 8:59
What edition of the book do you have? In the second edition the affine twisted cubic curve is described in exercise 3.3.2 (set theoretically) by the non reduced ideal $(y-x^2, z^2-2xyz+y^3$. Anyway, whoever made a mistake, the polynomials $x-y^2$ and $x-z^3$ definitely don't define in any sense the affine twisted cubic curve: they define a singular affine curve. – Georges Elencwajg May 3 '12 at 13:23
up vote 1 down vote accepted

I think your ideal is radical. To prove that, you can compute the singular locus of the associated variety, using jacobian criterion. If its codimension at each point is at least one, then your variety is reduced, and hence your ideal is radical.

All these computation can be done with a commutative algebra software. For example, using Macaulay2 :

i18 : R = QQ[w .. z]

o18 = R

o18 : PolynomialRing

i19 : I = ideal( x*w - y^2, x*w^2-z^3 )

                    2   2     3
o19 = ideal (w*x - y , w x - z )

o19 : Ideal of R

i20 : primaryDecomposition I

                     2     2    3
o20 = {ideal (w*x - y , w*y  - z )}

o20 : List

i21 : radical I == I

o21 = true

i22 : dim I

o22 = 2

i23 : dim singularLocus I

o23 = 1
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How did you prove these equations for the twisted cubic ? I'm not convinced... – Lierre May 3 '12 at 8:57

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