Proposition

Let $J = \ev{ x^2 + y^2 - 1, y - 1 }$. Find $f \in I(V(J))$ such that $f \notin J$.

Solution

GroebnerBasis([x**2, y - 1], x, y, domain='ZZ', order='lex')

Sympy shows that $\{ x^2, y - 1 \}$ is a Groebner basis of $J$. Thus $(x, y) = (0, 1)$ is the only point in $V(J)$. $(x, y) = (0, 1)$ vanishes on the function $h(x, y) = x$, so $h \in I(V(J))$. However, $x \notin J$ because it is not divisible by the Groebner basis.