Self-Studying Algebraic Geometry: Finding $f$ in $I(V(J))$

Question

Problem 4.1.2 in Ideals, Varieties, and Algorithms asks:

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

I started by trying to get an idea of what $V(J)$ is- the set of common zeroes of $x^2+y^2-1$ and $y-1$. I figure both polynomials are zero at the point $(0,1)$.

Not quite sure where to go from here- any help is welcome.

Answer 1

As $x^2+y^2-1=x^2+(y-1)(y+1)$, we have $J=(x^2,y-1)$. Hence $I(V(J))=(x,y-1)$.