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I am interested in finding the Krull dimension of the quotient ring $A$ defined as follows: $$ A = \mathbb{C}[x,y,z] / (f_1, f_2, f_3), $$ where $$ f_1 = \frac12 y^3 z - (z-1) - xy $$ $$ f_2 = y^2 z^2 - xz $$ $$ f_3 = ( (z-1)+xy )yz - \frac12 xz y^2. $$ I at least know that $ \emptyset \not = V(I)$, where $I = (f_1, f_2, f_3)$, because it contains $(0,0,1)$.

Would anyone be able to figure out what the Krull dimension of this ring is by any chance? The ideal is somewhat complicated and I wasn't really sure where to start. I would greatly appreciate any input!

Thank you very much! PS This is not a homework question.

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  • $\begingroup$ oops yes . let me fix that $\endgroup$
    – Johnny T.
    Jul 4, 2015 at 2:18

1 Answer 1

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I guess this can be done by hand, but one has to be careful.

I typed the ideal into Macaulay2:

i1 : R = QQ[x,y,z]

o1 = R

o1 : PolynomialRing

i2 : I = ideal(1/2 * y^3*z-(z-1)-x*y, y^2*z^2-x*z, ((z-1)+x*y)*y*z-1/2 * x*z*y^2)

            1 3                  2 2        1   2       2
o2 = ideal (-y z - x*y - z + 1, y z  - x*z, -x*y z + y*z  - y*z)
            2                               2

o2 : Ideal of R

i3 : dim I

o3 = 1

. We can even decompose the ideal:

i15 : decompose I

                               2           2    2          3     3     2
o15 = {ideal (x*y + 2z - 2, - y z + x, 2y*z  + x  - 2y*z, x  - 4z  + 8z  - 4z),
      --------------------------------------------------------------------------
      ideal (z, x*y - 1)}

o15 : List

. Thus it decomposes into 2 components, both of which can be checked to have dimension one.

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  • $\begingroup$ So I guess the Krull dimension is $1$? $\endgroup$
    – Johnny T.
    Jul 4, 2015 at 22:19
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    $\begingroup$ @JohnnyT. Yes! . $\endgroup$ Jul 4, 2015 at 22:29
  • $\begingroup$ Thank you for the clarification! $\endgroup$
    – Johnny T.
    Jul 4, 2015 at 22:48

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