Manuel Kauers Hagenberg, December 17, 2002 RISC-Linz

Algorithmic Algebraic Geometry

Solution 3

a) Let R = ℚ[x, y, z] and F = x^{2} + y^{2} + z^{2} + 3 − 10xyz ∈ R. In order to find all lines on the surface F = 0, we write

F (at + b, ct + d, t) = A(a, b, c, d)t^{3} + B(a, b, c, d)t^{2} + C(a, b, c, d)t + D(a, b, c, d)

and hunt for (a, b, c, d) with A = B = C = D = 0. Using Singular, we can compute the primary decomposition of the ideal ⟨A, B, C, D⟩⊴ℚ[a, b, c, d] which gives the following set of prime ideals as result:

\begin{matrix}⟨d, c − 3, b − 1, a⟩, & ⟨5d − 3, c, 5b − 4, a − 1⟩, \\ ⟨d + 1, c, b, a + 3⟩, & ⟨5d − 4, c − 1, 5b − 3, a⟩, \\ ⟨5d + 3, c, 5b + 4, a + 1⟩, & ⟨d − 1, c, b, a − 3⟩, \\ ⟨5d + 4, c − 1, 5b − 3, a⟩, & ⟨d, c + 3, b + 1, a⟩, \\ ⟨d, 3c − 1, b − 1, a⟩, & ⟨d, 3c + 1, b + 1, a⟩, \\ ⟨5d − 3, c, 5b + 4, a − 1⟩, & ⟨d + 1, c, b, 3a + 1⟩, \\ ⟨5d − 4, c + 1, 5b + 3, a⟩, & ⟨5d + 3, c, 5b − 4, a + 1⟩, \\ ⟨d − 1, c, b, 3a − 1⟩, & ⟨5d + 4, c + 1, 5b + 3, a⟩. \end{matrix}

A first bunch of solutions is given by the above set of primary ideals, namely the lines

l: (\begin{matrix}b \\ d \\ 0 \end{matrix}) + t (\begin{matrix}a \\ c \\ 1 \end{matrix})

for the a, b, c, d given by one of the ideals. These are 16 lines.

Now, since F is symmetrix in x, y, z, i.e.

F (x, y, z) = F (x, z, y) = F (y, x, z) = F (y, z, x) = F (z, x, y) = F (z, y, x),

we obtain new solutions by coordinate permutation of the ones we already have. This gives 6 · 16 = 96 solutions.

Some of these 96 “solutions” discribe the same line. Given two lines g: u_{g} + tv_{g}, h: u_{h} + tv_{h} they are equal if and only if

Rank(v_{g}, v_{h}, u_{g} − u_{h}) = 1

as known from linear algebra. After deleting duplicates, 24 lines are remaining:

(\begin{matrix}1 \\ 0 \\ 0 \end{matrix}) + t (\begin{matrix}0 \\ 3 \\ 1 \end{matrix}) ,  (\begin{matrix}4/5 \\ 3/5 \\ 0 \end{matrix}) + t (\begin{matrix}1 \\ 0 \\ 1 \end{matrix}) ,  (\begin{matrix}0 \\ −1 \\ 0 \end{matrix}) + t (\begin{matrix}−3 \\ 0 \\ 1 \end{matrix}) ,

(\begin{matrix}3/5 \\ 4/5 \\ 0 \end{matrix}) + t (\begin{matrix}0 \\ 1 \\ 1 \end{matrix}) ,  (\begin{matrix}−4/5 \\ −3/5 \\ 0 \end{matrix}) + t (\begin{matrix}−1 \\ 0 \\ 1 \end{matrix}) ,  (\begin{matrix}0 \\ 1 \\ 0 \end{matrix}) + t (\begin{matrix}3 \\ 0 \\ 1 \end{matrix}) ,

(\begin{matrix}3/5 \\ −4/5 \\ 0 \end{matrix}) + t (\begin{matrix}0 \\ 1 \\ 1 \end{matrix}) ,  (\begin{matrix}−1 \\ 0 \\ 1 \end{matrix}) + t (\begin{matrix}0 \\ −3 \\ 0 \end{matrix}) ,  (\begin{matrix}1 \\ 0 \\ 0 \end{matrix}) + t (\begin{matrix}0 \\ 1/3 \\ 1 \end{matrix}) ,

(\begin{matrix}−1 \\ 0 \\ 0 \end{matrix}) + t (\begin{matrix}0 \\ −1/31 \\ , \end{matrix})  (\begin{matrix}−4/5 \\ 3/5 \\ 0 \end{matrix}) + t (\begin{matrix}1 \\ 0 \\ 1 \end{matrix}) ,  (\begin{matrix}0 \\ −1 \\ 0 \end{matrix}) + t (\begin{matrix}−1/3 \\ 0 \\ 1 \end{matrix}) ,

(\begin{matrix}−3/5 \\ 4/5 \\ 0 \end{matrix}) + t (\begin{matrix}0 \\ −1 \\ 1 \end{matrix}) ,  (\begin{matrix}4/5 \\ −3/5 \\ 0 \end{matrix}) + t (\begin{matrix}−1 \\ 0 \\ 1 \end{matrix}) ,  (\begin{matrix}0 \\ 1 \\ 0 \end{matrix}) + t (\begin{matrix}1/3 \\ 0 \\ 1 \end{matrix}) ,

(\begin{matrix}−3/5 \\ −4/5 \\ 0 \end{matrix}) + t (\begin{matrix}0 \\ −1 \\ 1 \end{matrix}) ,  (\begin{matrix}0 \\ 0 \\ 1 \end{matrix}) + t (\begin{matrix}3 \\ 1 \\ 0 \end{matrix}) ,  (\begin{matrix}0 \\ 0 \\ 1 \end{matrix}) + t (\begin{matrix}1 \\ 3 \\ 0 \end{matrix}) ,

(\begin{matrix}4/5 \\ 0 \\ 3/5 \end{matrix}) + t (\begin{matrix}1 \\ 1 \\ 0 \end{matrix}) ,  (\begin{matrix}0 \\ 4/5 \\ 3/5 \end{matrix}) + t (\begin{matrix}1 \\ 1 \\ 0 \end{matrix}) ,  (\begin{matrix}0 \\ 0 \\ −1 \end{matrix}) + t (\begin{matrix}−3 \\ 1 \\ 0 \end{matrix}) ,

(\begin{matrix}0 \\ 0 \\ −1 \end{matrix}) + t (\begin{matrix}1 \\ −3 \\ 0 \end{matrix}) ,  (\begin{matrix}−4/5 \\ 0 \\ −3/5 \end{matrix}) + t (\begin{matrix}−1 \\ 1 \\ 0 \end{matrix}) ,  (\begin{matrix}4/5 \\ 0 \\ −3/5 \end{matrix}) + t (\begin{matrix}−1 \\ 1 \\ 0 \end{matrix}) .

The number of solutions should be 27, so where are the three missing ones? Consider the homogenisation

F_{h}(x, y, z, w) := w^{3}F (x/w, y/w, z/w) = x^{2}w + y^{2}w + z^{2}w + 3w^{3} − 10xyz.

The points (x, y, z) ∈ ℚ^{3} with F_{h}(x, y, z, 0) = 0 build “solutions at infinity.” But F_{h}(x, y, z, 0) = −10xyz has obviously exactly the three solutions

t (\begin{matrix}1 \\ 0 \\ 0 \end{matrix}) + s (\begin{matrix}0 \\ 1 \\ 0 \end{matrix}) , t (\begin{matrix}1 \\ 0 \\ 0 \end{matrix}) + s (\begin{matrix}0 \\ 0 \\ 1 \end{matrix}) , t (\begin{matrix}0 \\ 1 \\ 0 \end{matrix}) + s (\begin{matrix}0 \\ 0 \\ 1 \end{matrix})

in ℚ^{3}, which correspond to lines at infinity in the projective space.

It follows that we have 24 real solutions and three more solutions at infinity, giving 27 solutions altogether as desired.