Proposition

Consider the system of equations

1. If $I$ is the ideal generated by these equations, find bases of $I \cap k[x]$ and $I \cap k[y]$.
2. find all solutions of the equations.
3. Which of the solutions are rational?
4. What is the smallest field $k$ containing $\mathbb{Q}$ such that all solutions lie in $k^2$?

Solution

1

from sympy import *
from sympy.polys.orderings import monomial_key

x, y = symbols('x y')

print(groebner([x**2 + 2*y**2 - 3, x**2 + x*y + y**2 - 3], x, y, order='lex'))
print(groebner([x**2 + 2*y**2 - 3, x**2 + x*y + y**2 - 3], y, x, order='lex'))

gives

GroebnerBasis([x**2 + 2*y**2 - 3, x*y - y**2, y**3 - y], x, y, domain='ZZ', order='lex')
GroebnerBasis([x**3 - 3*x + 2*y, x**4 - 4*x**2 + 3], y, x, domain='ZZ', order='lex')

By the Elimination Theorem (P.122, Ideals, Varieties, and Algorithms), $I \cap k[x]$ is generated by $x^4 - 4x^2 + 3$ and $I \cap k[y]$ is generated by $y^3 - y$.

2

It seems easier to solve $y^3 - y = 0$. This has 3 solutions.

• If $y = 0$, then $x^2 = 3$, so $x = \pm \sqrt{3}$. Then $(x, y) = (\pm \sqrt{3}, 0)$ are solutions.
• If $y = 1$, then $x^2 + 2 = 3$ and $x^2 + x = 2$. Then $(x, y) = (1, 1)$ is a solution.
• If $y = -1$, then $x^2 + 2 = 3$ and $x^2 - x = 2$. Then $(x, y) = (-1, 1)$ is a solution.

Therefore, $(\pm \sqrt{3}, 0), (\pm 1, 1)$ are all the solutions.

3

$(\pm 1, 1)$.

4

$k = \mathbb{Q}[\sqrt{3}]$.