Radical of $\ev{xy, (x - y)x}$
by Hidenori
Proposition
Let $J = \ev{xy, (x - y)x}$. Describe $V(J)$ and show that $\sqrt{J} = \ev{x}$.
Solution
from sympy import *
from sympy.polys.orderings import monomial_key
x, y, a = symbols('x y a')
print(groebner([x*y, (x - y)*x], x, y, a, order='lex'))
gives GroebnerBasis([x**2, x*y], x, y, a, domain='ZZ', order='lex')
.
Let $S = \{ (0, y) \mid y \in k \}$. Then $S \subset V(J)$ since $0^2 = 0 \cdot y = 0$.
On the other hand, let $(x, y) \in V(J)$. Then $(x, y)$ satisfies $x^2 = 0$, so $x = 0$. Therefore, $(x, y) \in S$.
Hence, $V(J) = S$.
Since $x^2 \in J$, $x \subset \sqrt{J}$. This implies that $\ev{x} \subset \sqrt{J}$.
Let $f \in \sqrt{J}$. Then $f^m \in J$. In other words, $f^m = ax^2 + bxy$ for some $a, b \in k[x, y]$. Since $f^m = (ax + by)x \in \ev{x}$, $f \in \sqrt{\ev{x}}$. By Proposition 9 on P.186 of Ideals, Varieties, and Algorithms, $\sqrt{\ev{x}} = \ev{x}$. Therefore, $\sqrt{J} \subset \ev{x}$.
Hence, $\sqrt{J} = \ev{x}$.
Subscribe via RSS