Let’s first define the coordinate ring of $ \PP^3 $, where the twisted cubic lies:
i1 : kk = ZZ/32003;
i2 : R = kk[w, x, y, z]; -- this is a ring
This methods yields the twisted cubic as the ideal of a projective curve given parametrically by the map: \[ \begin{aligned} k[x,y,z] &\to k[t] \cr x &\mapsto t^3 \cr y &\mapsto t^2 \cr z &\mapsto t. \end{aligned} \]
i3 : monomialCurveIdeal(R, {1, 2, 3})
2 2
o3 = ideal (y - x*z, x*y - w*z, x - w*y)
o3 : Ideal of R
This method defines the twisted cubic as a determinantal ideal of $2\times 2$ minors, that is: \[ I = I_2 \begin{pmatrix} x & y & z \cr y & z & w \end{pmatrix} \]
i4 : minors(2, matrix {{x, y, z}, {y, z, w}})
2 2
o4 = ideal (- y + x*z, w*x - y*z, w*y - z )
o4 : Ideal of R
This method defines the twisted cubic as the kernel of the Veronese embedding of degree three on the projective line. That is:
i5 : kernel map(kk[s,t], R, {s^3, s^2*t, s*t^2, t^3})
2 2
o5 = ideal (y - x*z, x*y - w*z, x - w*y)
o5 : Ideal of R
A minimal free resolution of the ideal defining the twisted cubic: \[ 0 \gets \mathcal O_C \gets \mathcal O_{\PP^3} \gets 3\mathcal O_{\PP^3}(-2) \gets 2\mathcal O_{\PP^3}(-3) \gets 0 \]
i6 : res oo
1 3 2
o6 = R <-- R <-- R <-- 0
0 1 2 3
o6 : ChainComplex
i7 : betti oo
0 1 2
o7 = total: 1 3 2
0: 1 . .
1: . 3 2
o7 : BettiTally