Give answer to final (supplemental) question.

1 files changed, 61 insertions, 1 deletions

diff --git a/controle-20230412.tex b/controle-20230412.tex
index c7c55e6..65a37c0 100644
--- a/controle-20230412.tex
+++ b/controle-20230412.tex
@@ -101,7 +101,7 @@ Use of electronic devices of any kind is prohibited.
 Duration: 2 hours
 
 \ifcorrige
-This answer key has 6 pages (cover page included).
+This answer key has 7 pages (cover page included).
 \else
 This exam has 4 pages (cover page included).
 \fi
@@ -591,6 +591,66 @@ Plücker coordinates $w$.)
 The important part of this question is: how can we compute equations
 for $Y$ given the equation $h=0$ of $X$?
 
+\begin{answer}
+We have seen in question (6) that, so long as $w_{0,3} \neq 0$, the
+line $L_w$ defined by $(w_{0,1}{:}w_{0,2}{:}w_{0,3} {:} \penalty0
+w_{1,2}{:}w_{1,3}{:}w_{2,3})$ in $Q$ is the line through
+$(w_{0,3}{:}w_{1,3}{:}w_{2,3}{:}0)$ and
+$(0{:}w_{0,1}{:}w_{0,2}{:}w_{0,3})$.  In other words, $\lambda,\mu$ it
+is the line consisting of points $(\lambda w_{0,3} : \lambda w_{1,3} +
+\mu w_{0,1} : \lambda w_{2,3} + \mu w_{0,2} : \mu w_{0,3})$.  This
+line is included in $X$ iff $h(\lambda w_{0,3}, \lambda w_{1,3} + \mu
+w_{0,1}, \lambda w_{2,3} + \mu w_{0,2}, \mu w_{0,3}) = 0$ for all
+$\lambda,\mu$ in $k^{\alg}$, which just means that, seen as a
+polynomial in $\lambda,\mu$ now seen as two \emph{indeterminates},
+this is identically zero.  But this is a homogeneous polynomial in
+$\lambda,\mu$ whose coefficients are homogeneous polynomials in the
+$w_{i,j}$, so equating all these coefficients to $0$ gives homogeneous
+equations in the $w_{i,j}$ (coordinates in $\mathbb{P}^5$) for $L_w$
+to be included in $X$.  This only holds so long as $w_{0,3} \neq 0$
+(when it is zero, some of the resulting equation will be trivial);
+however, at least one $w_{i,j}$ must be zero in any case, so writing
+the corresponding equations for all permutations of coordinates,
+together with the equation ($\dagger$) of the Plücker quadric itself
+(to ensure that the $w_{i,j}$ do correspond to a line in
+$\mathbb{P}^3$) gives us the equations of the desired $Y$.
+
+To illustrate that this is actually algorithmic, the following Sage
+code computes the equations for the set $Y$ of lines inside the
+“diagonal cubic surface” $X := \{x_0^3 + x_1^3 + x_2^3 + x_3^3 = 0\}$:
+{\fontsize{8}{10}\relax
+\begin{verbatim}
+sage: R.<x0,x1,x2,x3,w01,w02,w03,w12,w13,w23,u,v> = PolynomialRing(QQ,12)
+sage: xvars = [x0,x1,x2,x3]
+sage: wvars = [[0,w01,w02,w03],[-w01,0,w12,w13],[-w02,-w12,0,w23],[-w03,-w13,-w23,0]]
+sage: # Plücker equation:
+sage: plucker = w01*w23 - w02*w13 + w03*w12
+sage: # Equation of the surface X:
+sage: h = x0^3 + x1^3 + x2^3 + x3^3
+sage: deg = h.degree()
+sage: # All possible permutations of (0,1,2,3):
+sage: perm4 = [(j0,j1,j2,j3) for j0 in range(4) for j1 in range(4) for j2 in range(4)
+....:  for j3 in range(4) if len(set([j0,j1,j2,j3]))==4]
+sage: # Generate the ideal I of the set Y of lines in X, as above:
+sage: I = R.ideal([plucker] + [h.subs(dict([(xvars[j0],wvars[j0][j3]*u), (xvars[j1],w
+....: vars[j1][j3]*u+wvars[j0][j1]*v), (xvars[j2],wvars[j2][j3]*u+wvars[j0][j2]*v), (
+....: xvars[j3],wvars[j0][j3]*v)])).coefficient({u:deg-k,v:k}) for k in range(deg) fo
+....: r (j0,j1,j2,j3) in perm4])
+sage: # Compute its radical:
+sage: I0 = I.radical()
+sage: # This really means Y is 0-dimensional in projective space:
+sage: I0.dimension()
+7
+sage: # This computes its number of geometric points (i.e., geometric lines on X):
+sage: hp = I0.hilbert_polynomial() ; hp.leading_coefficient()*factorial(hp.degree()) 
+27
+\end{verbatim}
+\par}\noindent (notation is as above except that $\lambda,\mu$ have
+been called \texttt{u},\texttt{v}); the above code proves that thare
+are $27$ geometric lines in the surface $\{x_0^3 + x_1^3 + x_2^3 +
+x_3^3 = 0\} \subseteq \mathbb{P}^3$ (over $\mathbb{Q}$).
+\end{answer}
+