Consider the union of a line and a plane in affine 3-space.
i1 : S = QQ[x,y,z];
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i2 : I = ideal(y*(x-1), z*(x-1));
o2 : Ideal of S
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The dimension is 2, the maximum of the dimensions of the two components. In order to find the dimension, Macaulay 2 requires the Groebner basis of I. It computes this behind the scenes, and caches the value with I.
i3 : dim I
o3 = 2
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i4 : gens gb I
o4 = | xz-z xy-y |
1 2
o4 : Matrix S <--- S
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Notice that
y is not in
I.
Now let's use a local order.
i6 : R = QQ[x,y,z,MonomialOrder=>{Weights=>{-1,-1,-1},RevLex},Global=>false];
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i7 : J = substitute(I,R)
o7 = ideal (- y + x*y, - z + x*z)
o7 : Ideal of R
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i8 : gens gb J
o8 = | y-xy z-xz |
1 2
o8 : Matrix R <--- R
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The dimension in this case is 1.
The following is WRONG. In this local ring,
y is in the ideal
J.
i10 : y % J
o10 = 0
o10 : R
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Translate the origin to
(1,0,0). The plane
x-1 = 0 goes through this new origin.
i11 : J = substitute(J, {x=>x+1})
o11 = ideal (x*y, x*z)
o11 : Ideal of R
|
i12 : dim J
o12 = 2
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Compute the global dimension after translation.
i13 : use ring I
o13 = S
o13 : PolynomialRing
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i14 : I1 = substitute(I, {x=>x+1})
o14 = ideal (x*y, x*z)
o14 : Ideal of S
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i15 : dim I1
o15 = 2
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See also
dim.